NCERT Solutions for Class 12 Maths In Hindi Chapter 4 Determinants

प्रश्नावली 4.1 

प्रश्न 1 से 2 तक मे सारणिकों का मान ज्ञात कीजिए: 

1. \[\left| {\begin{array}{*{20}{c}} { {2}}&{ {4}} \\  {{ { - 5}}}&{{ { - 1}}}  \end{array}} \right|\]

उत्तर: \[\left| {\begin{array}{*{20}{c}}  { {2}}&{ {4}} \\   {{ { - 5}}}&{{ { - 1}}} \end{array}} \right|\] = \[{ {2( - 1) - 4( - 5)  =   - 2 + 20  =  18}}\]

2. (i) \[\left| {\begin{array}{*{20}{c}}  {{ {cos\theta }}}&{{ { - sin\theta }}} \\   {{ {sin\theta }}}&{{ {cos\theta }}}  \end{array}} \right|\]

उत्तर: \[\left| {\begin{array}{*{20}{c}}  {{ {cos\theta }}}&{{ { - sin\theta }}} \\   {{ {sin\theta }}}&{{ {cos\theta }}}  \end{array}} \right|\] = \[ (cos\theta  \times cos\theta ) - (sin\theta  \times ( - sin\theta ))\]

\[{ { =  co}}{{ {s}}^{ {2}}}{ {\theta  + si}}{{ {n}}^{ {2}}}{ {\theta   =  1}}\]

(ii) \[\left| {\begin{array}{*{20}{c}}  {{{ {x}}^{ {2}}}{ { - x + 1}}}&{{ {x - 1}}} \\   {{ {x + 1}}}&{{ {x + 1}}}  \end{array}} \right|\]

उत्तर: \[\left| {\begin{array}{*{20}{c}}  {{{ {x}}^{ {2}}}{ { - x + 1}}}&{{ {x - 1}}} \\   {{ {x + 1}}}&{{ {x + 1}}}  \end{array}} \right|{ {  =  }}\left( {{{ {x}}^{ {2}}}{ { - x + 1}}} \right){ { \times (x + 1) - (x + 1) \times (x - 1)}}\]

\[\begin{align}  { { =  }}{{ {x}}^{ {3}}}{ { + }}{{ {x}}^{ {2}}}{ { - }}{{ {x}}^{ {2}}}{ { - x + x + 1 - }}\left( {{{ {x}}^{ {2}}}{ { + x - x - 1}}} \right) \hfill \\  { { =  }}{{ {x}}^{ {3}}}{ { - }}{{ {x}}^{ {2}}}{ { + 2}} \hfill \\  \end{align} \]

3. यदि \[{ {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {2}} \\   { {4}}&{ {2}} \end{array}} \right]\] तो दिखाइए \[{ {|2A|  =  4|A}}\mid \] 

उत्तर: \[{ {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {2}} \\   { {4}}&{ {2}}  \end{array}} \right]\]

\[{ {|2A|  =  }}\left| {\begin{array}{*{20}{l}}  { {2}}&{ {4}} \\   { {8}}&{ {4}}  \end{array}} \right|{ {  =  2(4) - 8(4)  =  8 - 32  =   - 24}}\] .......(i)

\[{ {4|A|  =  4}}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {2}} \\   { {4}}&{ {2}}  \end{array}} \right|{ {  =  4[1(2) - 2(4)]  =  4(2 - 8)  =   - 24}}\] .......(ii)

अतः (i) और (ii) से  \[{ {|2A|  =  4|A}}\mid \]

4. यदि \[{ {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {0}}&{ {1}} \\   { {0}}&{ {1}}&{ {2}} \\   { {0}}&{ {0}}&{ {4}}  \end{array}} \right]\] हो, तो दिखाइए \[{ {|3A|  =  27|A}}\mid \]

उत्तर: \[{ {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {0}}&{ {1}} \\   { {0}}&{ {1}}&{ {2}} \\   { {0}}&{ {0}}&{ {4}}  \end{array}} \right]\] 

\[{ {|3A|  =  }}\left| {\begin{array}{*{20}{c}}  { {3}}&{ {0}}&{ {3}} \\   { {0}}&{ {3}}&{ {6}} \\   { {0}}&{ {0}}&{{ {12}}}  \end{array}} \right|{ {  =  3(36 - 0) - 0(0 - 0) + 3(0 - 0)  =  108}}\]  .......(i)

\[{ {27|A|  =  27}}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}}&{ {1}} \\   { {0}}&{ {1}}&{ {2}} \\   { {0}}&{ {0}}&{ {4}}  \end{array}} \right|{ {  =  27[1(4 - 0) - 0(0 - 0) - 1(0 - 0)]  =  27(4)  =  108}}\] .......(ii)

अतः (i) और (ii) से \[{ {|3A|  =  27|A}}\mid \]

5. निम्नलिखित सारणिकों का मान ज्ञात कीजिए: 

(i) \[\left| {\begin{array}{*{20}{c}}  { {3}}&{{ { - 1}}}&{{ { - 2}}} \\   { {0}}&{ {0}}&{{ { - 1}}} \\   { {3}}&{{ { - 5}}}&{ {0}}  \end{array}} \right|\]

उत्तर: \[\left| {\begin{array}{*{20}{c}}  { {3}}&{{ { - 1}}}&{{ { - 2}}} \\   { {0}}&{ {0}}&{{ { - 1}}} \\   { {3}}&{{ { - 5}}}&{ {0}}  \end{array}} \right|{ {  =  3(0 - 5) + 1(0 + 3) + 2(0 - 0)}}\]

\[\begin{align}  { { =   - 15 + 3 - 0}} \hfill \\  { { =   - 12}} \hfill \\  \end{align} \]

(ii) \[\left| {\begin{array}{*{20}{c}}  { {3}}&{{ { - 4}}}&{ {5}} \\   { {1}}&{ {1}}&{{ { - 2}}} \\   { {2}}&{ {3}}&{ {1}}  \end{array}} \right|\]

उत्तर: \[\left| {\begin{array}{*{20}{c}}  { {3}}&{{ { - 4}}}&{ {5}} \\   { {1}}&{ {1}}&{{ { - 2}}} \\   { {2}}&{ {3}}&{ {1}}  \end{array}} \right|{ {  =  3(1 + 6) + 4(1 + 4) + 5(3 - 2)}}\]

\[\begin{align}  { { =  21 + 20 + 5}} \hfill \\  { { =  46}} \hfill \\  \end{align} \]

(iii) \[\left| {\begin{array}{*{20}{c}}  { {0}}&{ {1}}&{ {2}} \\   {{ { - 1}}}&{ {0}}&{{ { - 3}}} \\   {{ { - 2}}}&{ {3}}&{ {0}}  \end{array}} \right|\]

उत्तर: \[\left| {\begin{array}{*{20}{c}}  { {0}}&{ {1}}&{ {2}} \\   {{ { - 1}}}&{ {0}}&{{ { - 3}}} \\   {{ { - 2}}}&{ {3}}&{ {0}}  \end{array}} \right|\] = \[{ {0(0 + 9) - 1(0 - 6) + 2( - 3 - 0)}}\]

\[{ { =  0 + 6 - 6  =  0}}\]

(iv) \[\left| {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}}&{{ { - 2}}} \\   { {0}}&{ {2}}&{{ { - 1}}} \\   { {3}}&{{ { - 5}}}&{ {0}}  \end{array}} \right|\]

उत्तर: \[\left| {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}}&{{ { - 2}}} \\   { {0}}&{ {2}}&{{ { - 1}}} \\   { {3}}&{{ { - 5}}}&{ {0}}  \end{array}} \right|{ {  =  2(0 - 5) + 1(0 + 3) - 2(0 - 6)}}\]

\[\begin{align}  { { =   - 10 + 3 + 12}} \hfill \\  { { =  5}} \hfill \\  \end{align} \]

6. यदि \[{ {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {1}}&{ {2}} \\   { {2}}&{ {1}}&{ {3}} \\   { {5}}&{ {4}}&{ {9}}  \end{array}} \right]\] हो, तो ज्ञात कीजिए \[{ {|A|}}\]

उत्तर: \[{ {|A|  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {1}}&{ {2}} \\   { {2}}&{ {1}}&{ {3}} \\   { {5}}&{ {4}}&{ {9}}   \end{array}} \right|\]

\[\begin{align}  { { =  1( - 9 + 12) - 1( - 18 + 15) - 2(8 - 5)}} \hfill \\  { { =  3 + 3 - 6}} \hfill \\  { { =  0}} \hfill \\  \end{align} \]

7. \[{ {x}}\] के मान ज्ञात कीजिए यदि 

(i) \[\left| {\begin{array}{*{20}{l}}  { {2}}&{ {4}} \\   { {5}}&{ {1}}  \end{array}} \right|{ {  =  }}\left| {\begin{array}{*{20}{c}}  {{ {2x}}}&{ {4}} \\   { {6}}&{ {x}}  \end{array}} \right|\]

उत्तर: \[{ {2(1) - 5(4)  =  2x(x) - 6(4)}}\]

\[\begin{align}  { {2 - 20  =  2}}{{ {x}}^{ {2}}}{ { - 24}} \hfill \\  { { - 18  =  2}}{{ {x}}^{ {2}}}{ { - 24}} \hfill \\  { {2}}{{ {x}}^{ {2}}}{ { - 24 + 18  =  0}} \hfill \\  \end{align} \]

\[{ {2}}{{ {x}}^{ {2}}}{ { - 6  =  0}}\]

\[\begin{align}  {{ {x}}^{ {2}}}{ {  =  }}\dfrac{{ {6}}}{{ {2}}}{ {  =  3}} \hfill \\  { {x  =   \pm }}\sqrt { {3}}  \hfill \\  \end{align} \]

(ii) \[\left| {\begin{array}{*{20}{l}}  { {2}}&{ {3}} \\   { {4}}&{ {5}}  \end{array}} \right|{ {  =  }}\left| {\begin{array}{*{20}{c}}  { {x}}&{ {3}} \\   {{ {2x}}}&{ {5}}  \end{array}} \right|\]

उत्तर: \[{ {2(5)  -  4(3)  =  x(5) - 3(2x)}}\]

\[\begin{align}  { {10 - 12  =  5x - 6x}} \hfill \\  { { - 2  =   - x}} \hfill \\  { {x  =  2}} \hfill \\ \end{align} \]

8. यदि \[\left| {\begin{array}{*{20}{c}}  { {x}}&{ {2}} \\   {{ {18}}}&{ {x}}  \end{array}} \right|{ {  =  }}\left| {\begin{array}{*{20}{c}}  { {6}}&{ {2}} \\   {{ {18}}}&{ {6}} \end{array}} \right|\] हो तो \[{ {x}}\] बराबर है 

(a) \[6\]

(b) \[ \pm 6\]

(c) \[ - 6\]

(d) \[0\]

उत्तर: \[\left| {\begin{array}{*{20}{c}}  { {x}}&{ {2}} \\   {{ {18}}}&{ {x}}  \end{array}} \right|{ {  =  }}\left| {\begin{array}{*{20}{c}}  { {6}}&{ {2}} \\   {{ {18}}}&{ {6}}  \end{array}} \right|\]

\[\begin{align}  {{ {x}}^{ {2}}}{ { -  36  =  36  -  36}} \hfill \\  {{ {x}}^{ {2}}}{ {  =  36}} \hfill \\  { {x  =   \pm 6}} \hfill \\  \end{align} \]

इसलिए सही विकल्प (b) है।

प्रश्नावली 4.2 

बिना प्रसारण किए और सारणिकों के गुणधर्मों का प्रयोग करके निम्नलिखित प्रश्न 1 से 5 को सिद्ध कीजिए। 

1. \[\left| {\begin{array}{*{20}{l}}  { {x}}&{ {a}}&{{ {x + a}}} \\   { {y}}&{ {b}}&{{ {y + b}}} \\   { {z}}&{ {c}}&{{ {z + c}}}  \end{array}} \right|{ {  =  0}}\]

उत्तर: \[\left| {\begin{array}{*{20}{l}}  { {x}}&{ {a}}&{{ {x + a}}} \\   { {y}}&{ {b}}&{{ {y + b}}} \\   { {z}}&{ {c}}&{{ {z + c}}}  \end{array}} \right|\] = \[\left| {\begin{array}{*{20}{l}}  {{ {x + a}}}&{ {a}}&{{ {x + a}}} \\   {{ {y + b}}}&{ {b}}&{{ {y + b}}} \\   {{ {z + c}}}&{ {c}}&{{ {z + c}}} \end{array}} \right|\;\;\;\,\,\,\left[ {\because \;{{ {C}}_{ {1}}} \to {{ {C}}_{ {1}}}{ { + }}{{ {C}}_{ {2}}}} \right]\]

\[{ { =  0}}\;{ {        }}\left[ {\because \;{{ {C}}_{ {1}}}{ {  =  }}{{ {C}}_{ {3}}}} \right]\]

2. \[\left| {\begin{array}{*{20}{l}}  {{ {a - b}}}&{{ {b - c}}}&{{ {c - a}}} \\   {{ {b - c}}}&{{ {c - a}}}&{{ {a - b}}} \\   {{ {c - a}}}&{{ {a - b}}}&{{ {b - c}}}  \end{array}} \right|{ {  =  0}}\]

उत्तर: \[{ { =  }}\left| {\begin{array}{*{20}{l}}  { {0}}&{{ {b - c}}}&{{ {c - a}}} \\   { {0}}&{{ {c - a}}}&{{ {a - b}}} \\   { {0}}&{{ {a - b}}}&{{ {b - c}}}  \end{array}} \right|\;\;\;\;\;\;\;\;\,\left[ {{{ {C}}_{ {1}}} \to {{ {C}}_{ {1}}}{ { + }}{{ {C}}_{ {2}}}{ { + }}{{ {C}}_{ {3}}}} \right]\]

\[{ { =  0}}\;\;\;\,{ { }}\,\,\;{ { }}\left[ {{{ {C}}_{ {1}}}{ {  =  0}}} \right]\]

3. \[\left| {\begin{array}{*{20}{l}}  { {2}}&{ {7}}&{{ {65}}} \\   { {3}}&{ {8}}&{{ {75}}} \\   { {5}}&{ {9}}&{{ {86}}}  \end{array}} \right|{ {  =  0}}\]

उत्तर: \[\left| {\begin{array}{*{20}{l}}  { {2}}&{ {7}}&{{ {65}}} \\   { {3}}&{ {8}}&{{ {75}}} \\   { {5}}&{ {9}}&{{ {86}}}  \end{array}} \right|{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {2}}&{ {7}}&{{ {63}}} \\   { {3}}&{ {8}}&{{ {72}}} \\   { {5}}&{ {9}}&{{ {81}}}  \end{array}} \right|\;\;\;\;\;\,\;\;\;\;\;\left[ {{{ {C}}_{ {3}}} \to {{ {C}}_{ {3}}}{ { - }}{{ {C}}_{ {1}}}} \right]\]

\[{ { =  9}}\left| {\begin{array}{*{20}{l}}  { {2}}&{ {7}}&{ {7}} \\   { {3}}&{ {8}}&{ {8}} \\   { {5}}&{ {9}}&{ {9}}  \end{array}} \right|\]  \[{{C}_{3}}\] से \[9\] उभयनिष्ट लेने पर] 

\[{ { =  0}}\;\,{ {     }}\;\left[ {{{ {C}}_{ {2}}}{ {  =  }}{{ {C}}_{ {3}}}} \right]\]

4. \[\left| {\begin{array}{*{20}{l}}  { {1}}&{{ {bc}}}&{{ {a(b + c)}}} \\   { {1}}&{{ {ca}}}&{{ {b(c + a)}}} \\   { {1}}&{{ {ab}}}&{{ {c(a + b)}}}  \end{array}} \right|{ {  =  0}}\]

उत्तर: \[\left| {\begin{array}{*{20}{l}}  { {1}}&{{ {bc}}}&{{ {a(b + c)}}} \\   { {1}}&{{ {ca}}}&{{ {b(c + a)}}} \\   { {1}}&{{ {ab}}}&{{ {c(a + b)}}}  \end{array}} \right|{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{{ {bc}}}&{{ {ab + bc + ca}}} \\   { {1}}&{{ {ca}}}&{{ {ab + bc + ca}}} \\   { {1}}&{{ {ab}}}&{{ {ab + bc + ca}}}  \end{array}} \right|\]  \[\left[ {{{ {C}}_{ {3}}} \to {{ {C}}_{ {3}}}{ { + }}{{ {C}}_{ {2}}}} \right]\]

\[{ { =  (ab + bc + ca)}}\left| {\begin{array}{*{20}{l}}  { {1}}&{{ {bc}}}&{ {1}} \\   { {1}}&{{ {ca}}}&{ {1}} \\   { {1}}&{{ {ab}}}&{ {1}}  \end{array}} \right|\]         

(\[{{ {C}}_{ {3}}}\] से \[\left( {{ {ab + bc + }}} \right.{ {ca)}}\] उभयनिष्ठ लेने पर)

\[{ { =  0}}\;\,{ {  }}\;{ {    }}\;{ {   }}\left[ {{{ {C}}_{ {1}}}{ {  =  }}{{ {C}}_{ {3}}}} \right]\]

5. \[\left| {\begin{array}{*{20}{l}}  {{ {b + c}}}&{{ {q + r}}}&{{ {y + z}}} \\   {{ {c + a}}}&{{ {r + p}}}&{{ {z + x}}} \\   {{ {a + b}}}&{{ {p + q}}}&{{ {x + y}}}  \end{array}} \right|{ {  =  2}}\left| {\begin{array}{*{20}{l}}  { {a}}&{ {p}}&{ {x}} \\   { {b}}&{ {q}}&{ {y}} \\   { {c}}&{ {r}}&{ {z}}  \end{array}} \right|\]

उत्तर: \[\left| {\begin{array}{*{20}{l}}  {{ {b + c}}}&{{ {q + r}}}&{{ {y + z}}} \\   {{ {c + a}}}&{{ {r + p}}}&{{ {z + x}}} \\   {{ {a + b}}}&{{ {p + q}}}&{{ {x + y}}}  \end{array}} \right|{ {  =  2}}\left| {\begin{array}{*{20}{l}}  { {a}}&{ {p}}&{ {x}} \\   { {b}}&{ {q}}&{ {y}} \\   { {c}}&{ {r}}&{ {z}}  \end{array}} \right|\]

LHS \[{ { =  }}\left| {\begin{array}{*{20}{l}}  {{ {b + c}}}&{{ {q + r}}}&{{ {y + z}}} \\   {{ {c + a}}}&{{ {r + p}}}&{{ {z + x}}} \\   {{ {a + b}}}&{{ {p + q}}}&{{ {x + y}}}  \end{array}} \right|\]

\[{ { =  }}\left| {\begin{array}{*{20}{c}}  {{ {2c}}}&{{ {2r}}}&{{ {2z}}} \\   {{ {c + a}}}&{{ {r + p}}}&{{ {z + x}}} \\   {{ {a + b}}}&{{ {p + q}}}&{{ {x + y}}}  \end{array}} \right|\;\;\;\,\,\,\;\;\;\left[ {{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { + }}{{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {3}}}} \right]\]

\[{ { =  2}}\left| {\begin{array}{*{20}{c}}  { {c}}&{ {r}}&{ {z}} \\   {{ {c + a}}}&{{ {r + p}}}&{{ {z + x}}} \\   {{ {a + b}}}&{{ {p + q}}}&{{ {x + y}}}   \end{array}} \right|\]                 

\[{{R}_{1}}\] से \[ 2 \] उभयनिष्ट लेने पर]

\[{ { =  2}}\left| {\begin{array}{*{20}{c}}  { {c}}&{ {r}}&{ {z}} \\   { {a}}&{ {p}}&{ {x}} \\   {{ {a + b}}}&{{ {p + q}}\quad }&{{ {x + y}}}   \end{array}} \right|\;\;\;\;\;\;\left[ {{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {1}}}} \right]\]

\[{ { =  2}}\left| {\begin{array}{*{20}{l}}  { {c}}&{ {r}}&{ {z}} \\   { {a}}&{ {p}}&{ {x}} \\   { {b}}&{ {q}}&{ {y}}   \end{array}} \right|\;\;\;\;\;\;\;\,\;\;\left[ {{{ {R}}_{ {3}}} \to {{ {R}}_{ {3}}}{ { - }}{{ {R}}_{ {2}}}} \right]\]

\[{ { =  2}}\left| {\begin{array}{*{20}{l}}  { {a}}&{ {p}}&{ {x}} \\   { {c}}&{ {r}}&{ {z}} \\   { {b}}&{ {q}}&{ {y}}  \end{array}} \right|\;\;\,\;\,\,\;\;\;\,\;\left[ {{{ {R}}_{ {1}}} \leftrightarrow {{ {R}}_{ {2}}}} \right]\]

\[{ { =  2}}\left| {\begin{array}{*{20}{l}}  { {a}}&{ {p}}&{ {x}} \\   { {b}}&{ {q}}&{ {z}} \\   { {c}}&{ {r}}&{ {y}}   \end{array}} \right|\;\;\,\;\,\,\;\;\;\,\;\left[ {{{ {R}}_2} \leftrightarrow {{ {R}}_3}} \right]\]

= RHS

सारणिकों के गुणधर्मों का प्रयोग करके प्रश्न 6 से 14 तक को सिद्ध कीजिए। 

6. \[\left| {\begin{array}{*{20}{c}}  { {0}}&{ {a}}&{{ { - b}}} \\   {{ { - a}}}&{ {0}}&{{ { - c}}} \\   { {b}}&{ {c}}&{ {0}}  \end{array}} \right|{ {  =  0}}\]

उत्तर: \[\left| {\begin{array}{*{20}{c}}  { {0}}&{ {a}}&{{ { - b}}} \\   {{ { - a}}}&{ {0}}&{{ { - c}}} \\   { {b}}&{ {c}}&{ {0}}   \end{array}} \right|\]

LHS = \[\left| {\begin{array}{*{20}{c}}  { {0}}&{ {a}}&{{ { - b}}} \\   {{ { - a}}}&{ {0}}&{{ { - c}}} \\   { {b}}&{ {c}}&{ {0}}   \end{array}} \right|\]

\[{ { =  }}\left| {\begin{array}{*{20}{c}}  { {0}}&{ {a}}&{{ { - b}}} \\   {{ { - ab}}}&{ {0}}&{{ { - bc}}} \\   {{ {ab}}}&{{ {ac}}}&{ {0}}  \end{array}} \right|\;\,\,\;\;\;\;\,\;\;\;\left[ {{{ {R}}_{ {2}}} \to { {b}}{{ {R}}_{ {2}}}{ { , }}{{ {R}}_{ {3}}} \to { {a}}{{ {R}}_{ {3}}}} \right] \]

\[{ { =  }}\left| {\begin{array}{*{20}{c}}  { {0}}&{ {a}}&{{ { - b}}} \\   { {0}}&{{ {ac}}}&{{ { - bc}}} \\   {{ {ab}}}&{{ {ac}}}&{ {0}}  \end{array}} \right|\;\;\;\;\;\;\;\;\,\;\;\left[ {{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { + 3}}} \right]\]

\[=  ab( - abc + abc) \]

\[{ { =  ab(0)}} \]

\[{ { =  0}}\]

= RHS

7. \[\left| {\begin{array}{*{20}{c}}  {{ { - }}{{ {a}}^{ {2}}}}&{{ {ab}}}&{{ {ac}}} \\   {{ {ba}}}&{{ { - }}{{ {b}}^{ {2}}}}&{{ {bc}}} \\   {{ {ca}}}&{{ {cb}}}&{{ { - }}{{ {c}}^{ {2}}}} \end{array}} \right|{ {  =  4}}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}}  {{ { - }}{{ {a}}^{ {2}}}}&{{ {ab}}}&{{ {ac}}} \\   {{ {ba}}}&{{ { - }}{{ {b}}^{ {2}}}}&{{ {bc}}} \\   {{ {ca}}}&{{ {cb}}}&{{ { - }}{{ {c}}^{ {2}}}}  \end{array}} \right|\] 

\[{ { =  abc}}\left| {\begin{array}{*{20}{c}}  {{ { - a}}}&{ {a}}&{ {a}} \\   { {b}}&{ {-b}}&{ {b}} \\   { {c}}&{ {c}}&{{ { - c}}}  \end{array}} \right|\]                        

\[{{ {C}}_{ {1}}}{ {,}}\;{{ {C}}_{ {2}}}{ {,}}\;{{ {C}}_{ {3}}}\] से \[{ {a,}}\;{ {b,}}\;{ {c}}\] उभयनिष्ट लेने पर]

\[{ { =  }}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}}\left| {\begin{array}{*{20}{c}}  {{ { - 1}}}&{ {1}}&{ {1}} \\   { {1}}&{ {-1}}&{ {1}} \\   { {1}}&{ {1}}&{{ { - 1}}}  \end{array}} \right|\]            \[{{ {R}}_{ {1}}}{ {,}}\;{{ {R}}_{ {2}}}{ {,}}\;{{ {R}}_{ {3}}}\] से \[{ {a,}}\;{ {b,}}\;{ {c}}\] उभयनिष्ट लेने पर]

\[{ { =  }}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}}\left| {\begin{array}{*{20}{c}}  { {0}}&{ {1}}&{ {1}} \\   { {0}}&{{ { - 1}}}&{ {1}} \\   { {2}}&{ {1}}&{{ { - 1}}}  \end{array}} \right|\;\;\;\;\;\;\,\;\;\;\;\;\left[ {{{ {C}}_{ {1}}} \to {{ {C}}_{ {1}}}{ { + }}{{ {C}}_{ {2}}}} \right]\]

\[{ { =  }}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}}{ {\{ 2(1 + 1)\} }}\]         \[{{ {C}}_1}\] के अनुदिश प्रसरण करने पर]

\[{ { =  4}}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}}\]

= RHS

8. (i) \[\left| {\begin{array}{*{20}{l}}  { {1}}&{ {a}}&{{{ {a}}^{ {2}}}} \\   { {1}}&{ {b}}&{{{ {b}}^{ {2}}}} \\   { {1}}&{ {c}}&{{{ {c}}^{ {2}}}}  \end{array}} \right|{ {  =  (a - b)(b - c)(c - a)}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{l}}  { {1}}&{ {a}}&{{{ {a}}^{ {2}}}} \\   { {1}}&{ {b}}&{{{ {b}}^{ {2}}}} \\   { {1}}&{ {c}}&{{{ {c}}^{ {2}}}}  \end{array}} \right|\]

\[{ { =  }}\left| {\begin{array}{*{20}{c}}  { {0}}&{{ {a - b}}}&{{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}} \\   { {0}}&{{ {b - c}}}&{{{ {b}}^{ {2}}}{ { - }}{{ {c}}^{ {2}}}} \\   { {1}}&{ {c}}&{{{ {c}}^{ {2}}}}  \end{array}} \right|\;\,\;\;\;\;\,\;\;\,\left[ {{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { - }}{{ {R}}_{ {2}}}{ {, }}{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {3}}}} \right]\]

\[{ { =  (a - b)(b - c)}}\left| {\begin{array}{*{20}{c}}  { {0}}&{ {1}}&{{ {a + b}}} \\   { {0}}&{ {1}}&{{ {b + c}}} \\   { {1}}&{ {c}}&{{{ {c}}^{ {2}}}}  \end{array}} \right|\]                        \[{{ {R}}_{ {1}}}\] से \[{ {a - b, }}{{ {R}}_{ {2}}}\] से \[{ {b - c}}\] उभयनिष्ट लेने पर]

\[{ { =  (a - b)(b - c)\{ 1(b + c - a - b)\} }}\]            \[{{ {C}}_{ {1}}}\] के अनुदिश प्रसरण करने पर]

\[{ { =  (a - b)(b - c)(c - a)}}\]

(ii) \[\left| {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   { {a}}&{ {b}}&{ {c}} \\   {{{ {a}}^{ {3}}}}&{{{ {b}}^{ {3}}}}&{{{ {c}}^{ {3}}}}   \end{array}} \right|{ {  =  (a - b)(b - c)(c - a)(a + b + c)}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   { {a}}&{ {b}}&{ {c}} \\   {{{ {a}}^{ {3}}}}&{{{ {b}}^{ {3}}}}&{{{ {c}}^{ {3}}}}   \end{array}} \right|\]

\[{ { =  }}\left| {\begin{array}{*{20}{c}}  { {0}}&{ {0}}&{ {1}} \\   {{ {a - b}}}&{{ {b - c}}}&{ {c}} \\   {{{ {a}}^{ {3}}}{ { - }}{{ {b}}^{ {3}}}}&{{{ {b}}^{ {3}}}{ { - }}{{ {c}}^{ {3}}}}&{{{ {c}}^{ {3}}}}  \end{array}} \right|\;\;\;\;\;\,\;\;\left[ {{{ {C}}_{ {1}}} \to } \right.\left. {{{ {C}}_{ {1}}}{ { - }}{{ {C}}_{ {2}}}{ {, }}{{ {C}}_{ {2}}} \to {{ {C}}_{ {2}}}{ { - }}{{ {C}}_{ {3}}}} \right]\]

\[{ { =  (a - b)(b - c)}}\left| {\begin{array}{*{20}{c}}  { {0}}&{ {0}}&{ {1}} \\   { {1}}&{ {1}}&{ {c}} \\   {{{ {a}}^{ {2}}}{ { + ab + }}{{ {b}}^{ {2}}}}&{{{ {b}}^{ {2}}}{ { + bc + }}{{ {c}}^{ {2}}}}&{{{ {c}}^{ {3}}}}  \end{array}} \right|\]          \[{{ {C}}_{ {1}}}\] से \[{ {a - b}}\] , \[{{ {C}}_{ {2}}}\] से \[{ {b - c}}\] उभयनिष्ट लेने पर]

\[{ { =  (a - b)(b - c)}}\left\{ {{ {1}}\left( {{{ {b}}^{ {2}}}{ { + bc + }}{{ {c}}^{ {2}}}} \right){ { - }}\left( {{{ {a}}^{ {2}}}{ { + ab + }}{{ {b}}^{ {2}}}} \right)} \right\}\]           \[{{ {R}}_{ {1}}}\] के अनुदिश प्रसरण करने पर]

\[{ { =  (a - b)(b - c)(c - a)(a + b + c)}}\]

= RHS

9. \[\left| {\begin{array}{*{20}{l}}  { {x}}&{{{ {x}}^{ {2}}}}&{{ {yz}}} \\   { {y}}&{{{ {y}}^{ {2}}}}&{{ {zx}}} \\   { {z}}&{{{ {z}}^{ {2}}}}&{{ {xy}}}  \end{array}} \right|{ {  =  (x - y)(y - z)(z - x)(xy + yz + zx)}}\]

उत्तर: LHS =  \[\left| {\begin{array}{*{20}{l}}  { {x}}&{{{ {x}}^{ {2}}}}&{{ {yz}}} \\   { {y}}&{{{ {y}}^{ {2}}}}&{{ {zx}}} \\   { {z}}&{{{ {z}}^{ {2}}}}&{{ {xy}}}  \end{array}} \right|\]

\[{ { =  }}\left| {\begin{array}{*{20}{l}}  {{{ {x}}^{ {2}}}}&{{{ {x}}^{ {3}}}}&{{ {xyz}}} \\   {{{ {y}}^{ {2}}}}&{{{ {y}}^{ {3}}}}&{{ {yzx}}} \\   {{{ {z}}^{ {2}}}}&{{{ {z}}^{ {3}}}}&{{ {zxy}}}  \end{array}} \right|\;\;\;\,\,\,\,\,\;\;\;\left[ {{{ {R}}_{ {1}}} \to { {x}}{{ {R}}_{ {1}}},\;} \right.\left. {{{ {R}}_{ {2}}} \to { {y}}{{ {R}}_{ {2}}}{ {, }}{{ {R}}_{ {3}}} \to { {y}}{{ {R}}_{ {3}}}} \right]\]

\[{ { =  xyz}}\left| {\begin{array}{*{20}{l}}  {{{ {x}}^{ {2}}}}&{{{ {x}}^{ {3}}}}&{ {1}} \\   {{{ {y}}^{ {2}}}}&{{{ {y}}^{ {3}}}}&{ {1}} \\   {{{ {z}}^{ {2}}}}&{{{ {z}}^{ {3}}}}&{ {1}}   \end{array}} \right|\]    \[{{ {C}}_{ {3}}}\] से \[{ {xyz}}\] के उभयनिष्ट लेने पर]

\[{ { =  xyz}}\left| {\begin{array}{*{20}{c}}  {{{ {x}}^{ {2}}}{ { - }}{{ {y}}^{ {2}}}}&{{{ {x}}^{ {3}}}{ { - }}{{ {y}}^{ {3}}}}&{ {0}} \\   {{{ {y}}^{ {2}}}{ { - }}{{ {z}}^{ {2}}}}&{{{ {y}}^{ {3}}}{ { - }}{{ {z}}^{ {3}}}}&{ {0}} \\   {{{ {z}}^{ {2}}}}&{{{ {z}}^{ {2}}}}&{ {1}}  \end{array}} \right|\;\;\;\,\;\;\;\left[ {{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { - }}{{ {R}}_{ {2}}}{ {, }}{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {3}}}} \right] \]

\[{ { =  xyz(x - y)(y - z)}}\left| {\begin{array}{*{20}{c}}  {{ {x + y}}}&{{{ {x}}^{ {2}}}{ { + xy + }}{{ {y}}^{ {2}}}}&{ {0}} \\   {{ {y + z}}}&{{{ {y}}^{ {2}}}{ { + yz + }}{{ {z}}^{ {2}}}}&{ {0}} \\   {{{ {z}}^{ {2}}}}&{{{ {z}}^{ {2}}}}&{ {1}}   \end{array}} \right|\;\;\;\;\;\left[ {{{ {R}}_{ {1}}} \to { {x - y, }}{{ {R}}_{ {2}}} \to { {y - z}}} \right] \]  

\[{ { =  xyz(x - y)(y - z)\{ (x + y)(}}{{ {y}}^{ {2}}}{ { + yz + }}{{ {z}}^{ {2}}}{ {) - (y + z)(}}{{ {x}}^{ {2}}}{ { + xy + }}{{ {y}}^{ {2}}}{ {)\} }}\]      \[{{ {C}}_{ {3}}}\] के अनुदिश प्रसरण करने पर]

\[{ { =  (x - y)(y - z)(z - x)(xy + yz + zx)}}\]

= RHS

10. (i) \[\left| {\begin{array}{*{20}{c}}  {{ {x + 4}}}&{{ {2x}}}&{{ {2x}}} \\   {{ {2x}}}&{{ {x + 4}}}&{{ {2x}}} \\   {{ {2x}}}&{{ {2x}}}&{{ {x + 4}}}  \end{array}} \right|{ {  =  (5x + 4)(4 - x}}{{ {)}}^{ {2}}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}}  {{ {x + 4}}}&{{ {2x}}}&{{ {2x}}} \\   {{ {2x}}}&{{ {x + 4}}}&{{ {2x}}} \\   {{ {2x}}}&{{ {2x}}}&{{ {x + 4}}}  \end{array}} \right|\]

\[{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { + }}{{ {R}}_{ {2}}}{ { + }}{{ {R}}_{ {3}}} \]

= \[\left| {\begin{array}{*{20}{c}}  {{ {5x + 4}}}&{{ {5x + 4}}}&{{ {5x + 4}}} \\   {{ {2x}}}&{{ {x + 4}}}&{{ {2x}}} \\   {{ {2x}}}&{{ {2x}}}&{{ {x + 4}}}  \end{array}} \right| \]  

\[{ {(5x + 4)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   {{ {2x}}}&{{ {x + 4}}}&{{ {2x}}} \\   {{ {2x}}}&{{ {2x}}}&{{ {x + 4}}}   \end{array}} \right|\]  \[{{ {C}}_{ {2}}} \to {{ {C}}_{ {2}}}{ { - }}{{ {C}}_{ {1}}}{ {, }}{{ {C}}_{ {3}}} \to {{ {C}}_{ {3}}}{ { - }}{{ {C}}_{ {1}}}\] 

\[{ { = }}\;{ {(5x + 4)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   {{ {2x}}}&{{ { - x + 4}}}&{ {0}} \\   {{ {2x}}}&{ {0}}&{{ { - x + 4}}}  \end{array}} \right|\]

\[{ { = }}\;{ {(5x + 4)(4 - x)(4 - x)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   {{ {2x}}}&{ {1}}&{ {0}} \\   {{ {2x}}}&{ {0}}&{ {1}}  \end{array}} \right|\]

\[{ { = }}\;{ {(5x + 4)(4 - x}}{{ {)}}^{ {2}}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {0}} \\   {{ {2x}}}&{ {1}}   \end{array}} \right| \]

\[  { { = }}\;{ {(5x + 4)(4 - x}}{{ {)}}^{ {2}}}\]

= RHS

(ii) \[\left| {\begin{array}{*{20}{c}}  {{ {y + k}}}&{ {y}}&{ {y}} \\   { {y}}&{{ {y + k}}}&{ {y}} \\   { {y}}&{ {y}}&{{ {y + k}}}   \end{array}} \right|{ {  =  }}{{ {k}}^{ {2}}}{ {(3y + k)}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}}  {{ {y + k}}}&{ {y}}&{ {y}} \\   { {y}}&{{ {y + k}}}&{ {y}} \\   { {y}}&{ {y}}&{{ {y + k}}}   \end{array}} \right|\]

\[\begin{align}  {{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { + }}{{ {R}}_{ {2}}}{ { + }}{{ {R}}_{ {3}}} \hfill \\   = \;\left| {\begin{array}{*{20}{c}}  {{ {3y + k}}}&{{ {3y + k}}}&{{ {3y + k}}} \\   { {y}}&{{ {y + k}}}&{ {y}} \\   { {y}}&{ {y}}&{{ {y + k}}}   \end{array}} \right| \hfill \\   \end{align} \]

\[{ { = }}\;{ {(3y + k)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   { {y}}&{{ {y + k}}}&{ {y}} \\   { {y}}&{ {y}}&{{ {y + k}}}   \end{array}} \right|\]

\[\begin{align}  {{ {C}}_{ {2}}} \to {{ {C}}_{ {2}}}{ { - }}{{ {C}}_{ {1}}}{ {, }}{{ {C}}_{ {3}}} \to {{ {C}}_{ {3}}}{ { - }}{{ {C}}_{ {1}}} \hfill \\  { { = }}\;{ {(3y + k)}}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}}&{ {0}} \\   { {y}}&{ {k}}&{ {0}} \\   { {y}}&{ {0}}&{ {k}}   \end{array}} \right| \hfill \\   \end{align} \]

\[\begin{align}  { { = }}\;{ {(3y + k)}}{{ {k}}^{ {2}}}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}}&{ {0}} \\   { {y}}&{ {1}}&{ {0}} \\   { {y}}&{ {0}}&{ {1}}   \end{array}} \right| \hfill \\  { { = }}\;{ {(3y + k)}}{{ {k}}^{ {2}}}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}} \\   { {y}}&{ {1}}   \end{array}} \right| \hfill \\  { { = }}\;{ {(3y + k)}}{{ {k}}^{ {2}}} \hfill \\   \end{align} \]

= RHS

11. (i) \[\left| {\begin{array}{*{20}{c}}  {{ {a - b - c}}}&{{ {2a}}}&{{ {2a}}} \\   {{ {2b}}}&{{ {b - c - a}}}&{{ {2b}}} \\   {{ {2c}}}&{{ {2c}}}&{{ {c - a - b}}}  \end{array}} \right|{ {  =  (a + b + c}}{{ {)}}^3}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}}  {{ {a - b - c}}}&{{ {2a}}}&{{ {2a}}} \\   {{ {2b}}}&{{ {b - c - a}}}&{{ {2b}}} \\   {{ {2c}}}&{{ {2c}}}&{{ {c - a - b}}}  \end{array}} \right|\]

\[\begin{align}  {{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { + }}{{ {R}}_{ {2}}}{ { + }}{{ {R}}_{ {3}}} \hfill \\   = \;\left| {\begin{array}{*{20}{c}}  {{ {a + b + c}}}&{{ {a + b + c}}}&{{ {a + b + c}}} \\   {{ {2b}}}&{{ {b - c - a}}}&{{ {2b}}} \\   {{ {2c}}}&{{ {2c}}}&{{ {c - a - b}}}   \end{array}} \right| \hfill \\  { { = }}\;{ {(a + b + c)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   {{ {2b}}}&{{ {b - c - a}}}&{{ {2b}}} \\   {{ {2c}}}&{{ {2c}}}&{{ {c - a - b}}}   \end{array}} \right| \hfill \\  {{ {C}}_{ {2}}} \to {{ {C}}_{ {2}}}{ { - }}{{ {C}}_{ {1}}}{ {, }}{{ {C}}_{ {3}}} \to {{ {C}}_{ {3}}}{ { - }}{{ {C}}_{ {1}}} \hfill \\   \end{align} \]

\[\begin{align}  { { = }}\;{ {(a + b + c)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   {{ {2b}}}&{{ { - (a + b + c)}}}&{ {0}} \\   {{ {2c}}}&{ {0}}&{{ { - (a + b + c)}}}   \end{array}} \right| \hfill \\  { { = }}\;{{ {(a + b + c)}}^{ {3}}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   {{ {2b}}}&{{ { - 1}}}&{ {0}} \\   {{ {2c}}}&{ {0}}&{{ { - 1}}}   \end{array}} \right| \hfill \\  { { = }}\;{{ {(a + b + c)}}^{ {3}}}{ {( - 1) \times ( - 1)}}\;{ { =  (a + b + c}}{{ {)}}^{ {3}}} \hfill \\   \end{align} \]

= RHS

(ii) \[\left| {\begin{array}{*{20}{c}}  {{ {x + y + 2z}}}&{ {x}}&{ {y}} \\   { {z}}&{{ {y + z + 2x}}}&{ {y}} \\   { {z}}&{ {x}}&{{ {z + x + 2y}}}  \end{array}} \right|{ {  =  2(x + y + z}}{{ {)}}^{ {3}}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}}  {{ {x + y + 2z}}}&{ {x}}&{ {y}} \\   { {z}}&{{ {y + z + 2x}}}&{ {y}} \\   { {z}}&{ {x}}&{{ {z + x + 2y}}}  \end{array}} \right|\]

\[\begin{align}  {{ {C}}_{ {1}}} \to {{ {C}}_{ {1}}}{ { + }}{{ {C}}_{ {2}}}{ { + }}{{ {C}}_{ {3}}} \hfill \\   = \;\left| {\begin{array}{*{20}{c}}  {{ {2(x + y + z)}}}&{ {x}}&{ {y}} \\   {{ {2(x + y + z)}}}&{{ {y + z + 2x}}}&{ {y}} \\   {{ {2(x + y + z)}}}&{ {x}}&{{ {z + x + 2y}}}   \end{array}} \right| \hfill \\  { { = }}\;{ {2(x + y + z)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {x}}&{ {y}} \\   { {1}}&{{ {y + z + 2x}}}&{ {y}} \\   { {1}}&{ {x}}&{{ {z + x + 2y}}}   \end{array}} \right| \hfill \\  {{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {1}}}{ {, }}{{ {R}}_{ {3}}} \to {{ {R}}_{ {3}}}{ { - }}{{ {R}}_{ {1}}} \hfill \\  \end{align} \]

\[\begin{align}  { { = }}\;{ {2(x + y + z)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {x}}&{ {y}} \\   { {0}}&{{ {y + z + x}}}&{ {0}} \\   { {0}}&{ {0}}&{{ {z + x + y}}}  \end{array}} \right| \hfill \\  { { =  2(x + y + z}}{{ {)}}^{ {3}}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {x}}&{ {y}} \\   { {0}}&{ {1}}&{ {0}} \\   { {0}}&{ {0}}&{ {1}}   \end{array}} \right| \hfill \\  { { =  2(x + y + z}}{{ {)}}^{ {3}}} \hfill \\  \end{align} \]

= RHS

12. \[\left| {\begin{array}{*{20}{c}}  { {1}}&{ {x}}&{{{ {x}}^{ {2}}}} \\   {{{ {x}}^{ {2}}}}&{ {1}}&{ {x}} \\   { {x}}&{{{ {x}}^{ {2}}}}&{ {1}}   \end{array}} \right|{ {  =  }}{\left( {{ {1 + }}{{ {x}}^{ {3}}}} \right)^{ {2}}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}}  { {1}}&{ {x}}&{{{ {x}}^{ {2}}}} \\   {{{ {x}}^{ {2}}}}&{ {1}}&{ {x}} \\   { {x}}&{{{ {x}}^{ {2}}}}&{ {1}}   \end{array}} \right|\]

\[\begin{align}  {{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { + }}{{ {R}}_{ {2}}}{ { + }}{{ {R}}_{ {3}}} \hfill \\   = \;\left| {\begin{array}{*{20}{c}}  {{ {1 + x + }}{{ {x}}^{ {2}}}}&{{ {1 + x + }}{{ {x}}^{ {2}}}}&{{ {1 + x + }}{{ {x}}^{ {2}}}} \\   {{{ {x}}^{ {2}}}}&{ {1}}&{ {x}} \\   { {x}}&{{{ {x}}^{ {2}}}}&{ {1}}   \end{array}} \right| \hfill \\  \end{align} \]

\[\begin{align}   = \;\left( {{ {1 + x + }}{{ {x}}^{ {2}}}} \right)\left| {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   {{{ {x}}^{ {2}}}}&{ {1}}&{ {x}} \\   { {x}}&{{{ {x}}^{ {2}}}}&{ {1}}   \end{array}} \right| \hfill \\  {{ {C}}_{ {2}}} \to {{ {C}}_{ {2}}}{ { - }}{{ {C}}_{ {1}}}{ {, }}{{ {C}}_{ {3}}} \to {{ {C}}_{ {3}}}{ { - }}{{ {C}}_{ {1}}} \hfill \\   \end{align} \]

\[\begin{align}   = \;\left( {{ {1 + x + }}{{ {x}}^{ {2}}}} \right)\left| {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   {{{ {x}}^{ {2}}}}&{{ {1 - }}{{ {x}}^{ {2}}}}&{{ {x - }}{{ {x}}^{ {2}}}} \\   { {x}}&{{{ {x}}^{ {2}}}{ { - x}}}&{{ {1 - }}{{ {x}}^{ {2}}}}  \end{array}} \right| \hfill \\   = \,\left( {{ {1 + x + }}{{ {x}}^{ {2}}}} \right){ {(1 - x)(1 - x)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   {{{ {x}}^{ {2}}}}&{{ {1 + x}}}&{ {x}} \\   { {x}}&{{ { - x}}}&{ {1}}  \end{array}} \right| \hfill \\   = \,\left( {{ {1 - }}{{ {x}}^{ {3}}}} \right){ {(1 - x)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   {{{ {x}}^{ {2}}}}&{{ {1 + x}}}&{ {x}} \\   { {x}}&{{ { - x}}}&{ {1}}  \end{array}} \right| \hfill \\   = \;\left( {{ {1 - }}{{ {x}}^{ {3}}}} \right){ {(1 - x)(1)}}\left| {\begin{array}{*{20}{c}}  {{ {1 + x}}}&{ {x}} \\   {{ { - x}}}&{ {1}}  \end{array}} \right| \hfill \\   = \;\left( {{ {1 - }}{{ {x}}^{ {3}}}} \right){ {(1 - x)}}\left( {{ {1 + x + }}{{ {x}}^{ {2}}}} \right) \hfill \\   = \;{\left( {{ {1 - }}{{ {x}}^{ {3}}}} \right)^{ {2}}} \hfill \\   \end{align} \]

= RHS

13. \[\left| {\begin{array}{*{20}{c}}  {{ {1 + }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}}&{{ {2ab}}}&{{ { - 2b}}} \\   {{ {2ab}}}&{{ {1 - }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}}&{{ {2a}}} \\   {{ {2b}}}&{{ { - 2a}}}&{{ {1 - }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}}  \end{array}} \right|{ {  =  }}{\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)^{ {3}}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}}  {{ {1 + }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}}&{{ {2ab}}}&{{ { - 2b}}} \\   {{ {2ab}}}&{{ {1 - }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}}&{{ {2a}}} \\   {{ {2b}}}&{{ { - 2a}}}&{{ {1 - }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}}  \end{array}} \right|\]

\[{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - a}}{{ {R}}_{ {3}}}{ {, }}{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { + b}}{{ {R}}_{ {3}}}\]

\[ = \;\left| {\begin{array}{*{20}{c}}  {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}}&{ {0}}&{{ { - b}}\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)} \\   { {0}}&{{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}}&{{ {a}}\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)} \\   {{ {2b}}}&{{ { - 2a}}}&{{ {1 - }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}}   \end{array}} \right|\]

\[ = \;{\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)^{ {2}}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{{ { - b}}} \\   { {0}}&{ {1}}&{ {a}} \\   {{ {2b}}}&{{ { - 2a}}}&{{ {1 - }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}}  \end{array}} \right|\]

\[ = \;{\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)^{ {2}}}\left( {{ {1 - }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}{ { + 2}}{{ {a}}^{ {2}}}{ { - b( - 2b)}}} \right)\]

\[\begin{align}   = \,{\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)^{ {2}}}\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right) \hfill \\   = \,{\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)^{ {3}}} \hfill \\   \end{align} \]

= RHS

14. \[\left| {\begin{array}{*{20}{c}}  {{{ {a}}^{ {2}}}{ { + 1}}}&{{ {ab}}}&{{ {ac}}} \\   {{ {ab}}}&{{{ {b}}^{ {2}}}{ { + 1}}}&{{ {bc}}} \\   {{ {ca}}}&{{ {cb}}}&{{{ {c}}^{ {2}}}{ { + 1}}}  \end{array}} \right|{ {  =  1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}{ { + }}{{ {c}}^{ {2}}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}}  {{ {1 + }}{{ {a}}^{ {2}}}}&{{ {ab}}}&{{ {ac}}} \\   {{ {ab}}}&{{ {1 + }}{{ {b}}^{ {2}}}}&{{ {bc}}} \\   {{ {ca}}}&{{ {cb}}}&{{{ {c}}^{ {2}}}{ { + 1}}}   \end{array}} \right|\]

\[\begin{align}  { { = }}\;{ {abc}}\left| {\begin{array}{*{20}{c}}  {{ {a + }}\dfrac{{ {1}}}{{ {a}}}}&{ {b}}&{ {c}} \\   { {a}}&{{ {b + }}\dfrac{{ {1}}}{{ {b}}}}&{ {c}} \\   { {a}}&{ {b}}&{{ {c + }}\dfrac{{ {1}}}{{ {c}}}}   \end{array}} \right| \hfill \\  {{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {1}}}{ {, }}{{ {R}}_{ {3}}} \to {{ {R}}_{ {3}}}{ { - }}{{ {R}}_{ {1}}} \hfill \\   \end{align} \]

\[\begin{align}  { { = }}\;{ {abc}}\left| {\begin{array}{*{20}{c}}  {{ {a + }}\dfrac{{ {1}}}{{ {a}}}}&{ {b}}&{ {c}} \\   {{ { - }}\dfrac{{ {1}}}{{ {a}}}}&{\dfrac{{ {1}}}{{ {b}}}}&{ {0}} \\   {{ { - }}\dfrac{{ {1}}}{{ {a}}}}&{ {0}}&{\dfrac{{ {1}}}{{ {c}}}}   \end{array}} \right| \hfill \\  {{ {C}}_{ {1}}} \to { {a}}{{ {C}}_{ {1}}}{ {, }}{{ {C}}_{ {2}}} \to { {b}}{{ {C}}_{ {2}}}{ {, }}{{ {C}}_{ {3}}} \to { {c}}{{ {C}}_{ {3}}} \hfill \\   \end{align} \]

\[\begin{align}  { { = }}\;{ {abc \times }}\dfrac{{ {1}}}{{{ {abc}}}}\left| {\begin{array}{*{20}{c}}  {{ {1 + }}{{ {a}}^{ {2}}}}&{{{ {b}}^{ {2}}}}&{{{ {c}}^{ {2}}}} \\   {{ { - 1}}}&{ {1}}&{ {0}} \\   {{ { - 1}}}&{ {0}}&{ {1}}   \end{array}} \right| \hfill \\  { { = }}\;{ { - 1}}\left| {\begin{array}{*{20}{c}}  {{{ {b}}^{ {2}}}}&{{{ {c}}^{ {2}}}} \\   { {1}}&{ {0}}   \end{array}} \right|{ { + 1}}\left| {\begin{array}{*{20}{c}}  {{ {1 + }}{{ {a}}^{ {2}}}}&{{{ {b}}^{ {2}}}} \\   {{ { - 1}}}&{ {1}}   \end{array}} \right| \hfill \\  { { = }}\;{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}{ { + }}{{ {c}}^{ {2}}} \hfill \\  \end{align} \]

= RHS

प्रश्न संख्या 15 तथा 16 मे सही उत्तर चुनिये। 

15. यदि \[{ {A}}\] एक \[{ {3 \times 3}}\] कोटी का वर्ग आव्यूह है तो \[{ {|kA|}}\] का मान होगा: 

(a) \[{ {k|A|}}\]

(b) \[{{ {k}}^{ {2}}}{ {|\;A|}}\]

(c) \[{{ {k}}^{ {3}}}{ {|\;A|}}\]

(d) \[{ {3k|A|}}\]

उत्तर: \[{ {|A|  =  }}\left| {\begin{array}{*{20}{l}}  { {a}}&{ {b}}&{ {c}} \\   { {d}}&{ {e}}&{ {f}} \\   { {g}}&{ {h}}&{ {i}}   \end{array}} \right|\]

\[\begin{align}  { {|kA|  =  }}\left| {\begin{array}{*{20}{l}}  {{ {ka}}}&{{ {kb}}}&{{ {kc}}} \\   {{ {kd}}}&{{ {ke}}}&{{ {kf}}} \\   {{ {kg}}}&{{ {kh}}}&{{ {ki}}}  \end{array}} \right| \hfill \\  { {|kA|  =  }}{{ {k}}^{ {3}}}\left| {\begin{array}{*{20}{l}}  { {a}}&{ {b}}&{ {c}} \\   { {d}}&{ {e}}&{ {f}} \\   { {g}}&{ {h}}&{ {i}}  \end{array}} \right| \hfill \\  { {|kA|  =  }}{{ {k}}^{ {3}}}{ {|A|}} \hfill \\   \end{align} \]

अतः सही विकल्प (c) है।

16. निम्नलिखित मे से कौन सा कथन सही है। 

(a) सारणिक एक वर्ग आव्यूह है। 

(b) सारणिक एक आव्यूह से संबंध एक संख्या है। 

(c) सारणिक एक वर्ग आव्यूह से संबंध एक संख्या है। 

(d) इनमे से कोई नहीं है। 

उत्तर: (c) सारणिक एक वर्ग आव्यूह से संबंध संख्या है। क्योंकि सारणिक एक वर्ग आव्यूह का ही निकाला जा सकता है। 

प्रश्नावली 4.3 

1. निम्नलिखित प्रत्येक मे दिए गए शीर्ष बिन्दुओ वाले त्रिभुज का क्षेत्रफल ज्ञात कीजिए:

(i) \[{ {(1,0), (6,0), (4,3)}}\]

उत्तर: शीर्ष बिन्दुओ \[\left( {{{ {x}}_{ {1}}}{ {,}}{{ {y}}_{ {1}}}} \right){ {, }}\left( {{{ {x}}_{ {2}}}{ {,}}{{ {y}}_{ {2}}}} \right){ {, }}\left( {{{ {x}}_{ {3}}}{ {,}}{{ {y}}_{ {3}}}} \right)\] से होकर जाने वाले त्रिभुज का क्षेत्रफल, 

\[{ {\Delta   =  }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}}  {{{ {x}}_{ {1}}}}&{{{ {y}}_{ {1}}}}&{ {1}} \\   {{{ {x}}_{ {2}}}}&{{{ {y}}_{ {2}}}}&{ {1}} \\   {{{ {x}}_{ {3}}}}&{{{ {y}}_{ {3}}}}&{ {1}}   \end{array}} \right|\]

अभीष्ट त्रिभुज का क्षेत्रफल 

\[\begin{align}  { {\Delta   =  }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}}&{ {1}} \\   { {6}}&{ {0}}&{ {1}} \\   { {4}}&{ {3}}&{ {1}}   \end{array}} \right| \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ {[1(0 - 3) - 0(6 - 4) + 1(18 - 0)]}} \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 3 + 18]  =  }}\dfrac{{{ {15}}}}{{ {2}}} \hfill \\  \end{align} \]

(ii) \[{ {(2,7), (1,2), (10,8)}}\]

उत्तर: अभीष्ट त्रिभुज का क्षेत्रफल

\[{ {\Delta   =  }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{c}}  { {2}}&{ {7}}&{ {1}} \\   { {1}}&{ {1}}&{ {1}} \\   {{ {10}}}&{ {8}}&{ {1}}   \end{array}} \right|\]

\[\begin{align}  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ {[2(1 - 8) - 7(1 - 10) + 8(8 - 10)]}} \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 16 + 63]}} \hfill \\   \end{align} \]

\[{ { =  }}\dfrac{{{ {47}}}}{{ {2}}}\]

(iii) \[{ {( - 2, - 3), (3,2), ( - 1, - 8)}}\]

उत्तर: अभीष्ट त्रिभुज का क्षेत्रफल

\[{ {\Delta   =  }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{c}}  {{ { - 2}}}&{{ { - 3}}}&{ {1}} \\   { {3}}&{ {2}}&{ {1}} \\   {{ { - 1}}}&{{ { - 8}}}&{ {1}}   \end{array}} \right|\]

\[\begin{align}  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 2(2 + 8) + 3(3 + 1) + ( - 24 + 2)]}} \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 2(10) + 3(4) + ( - 22)]}} \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 20 + 12 - 22]}} \hfill \\  { { =   - }}\dfrac{{{ {30}}}}{{ {2}}}{ {  =   - 15}} \hfill \\  \end{align} \]

2. दर्शाइए की बिन्दु \[{ {A(a,b + c), B(b}}{ {.c + a), C(c,a + b)}}\] और संरेख है।

उत्तर: ज्ञात है त्रिभुज के शीर्ष \[{ {A(a,b + c), B(b}}{ {.c + a), C(c,a + b)}}\]

\[{ {\Delta }}\] का क्षेत्रफल = \[{ {\Delta   =  }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}}  {{{ {x}}_{ {1}}}}&{{{ {y}}_{ {1}}}}&{ {1}} \\   {{{ {x}}_{ {2}}}}&{{{ {y}}_{ {2}}}}&{ {1}} \\   {{{ {x}}_{ {3}}}}&{{{ {y}}_{ {3}}}}&{ {1}}  \end{array}} \right|\]

जहा \[{{ {x}}_{ {1}}}{ {  =  a, }}{{ {y}}_{ {1}}}{ {  =  b + c, }}{{ {x}}_{ {2}}}{ {  =  b, }}{{ {y}}_{ {2}}}{ {  =  c + a, }}{{ {x}}_{ {3}}}{ {  =  c, }}{{ {y}}_{ {3}}}{ {  =  a + b}}\]

\[{ { =  }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}}  { {a}}&{{ {b + c}}}&{ {1}} \\   { {b}}&{{ {c + a}}}&{ {1}} \\   { {c}}&{{ {a + b}}}&{ {1}}   \end{array}} \right|\,\;\,\;\;\;\;\;\;\,\;\left( {{{ {C}}_{ {1}}} \to {{ {C}}_{ {1}}}{ { + }}{{ {C}}_{ {2}}}} \right)\]

\[\begin{align}  { { =  }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}}  {{ {a + b + c}}}&{{ {b + c}}}&{ {1}} \\   {{ {a + b + c}}}&{{ {c + a}}}&{ {1}} \\   {{ {a + b + c}}}&{{ {a + b}}}&{ {1}}   \end{array}} \right| \hfill \\   { { =  }}\dfrac{{ {1}}}{{ {2}}}{ {(a + b + c)}}\left| {\begin{array}{*{20}{l}}  { {1}}&{{ {b + c}}}&{ {1}} \\   { {1}}&{{ {c + a}}}&{ {1}} \\   { {1}}&{{ {a + b}}}&{ {1}}  \end{array}} \right| \hfill \\   = \;\dfrac{{ {1}}}{{ {2}}}{ {(a + b + c) \times 0}} \hfill \\  \end{align} \]

\[{ {\Delta }}\] का क्षेत्रफल = \[{ {0}}\] 

अतः बिन्दु A, B, C संरेख है।

3. प्रत्येक मे का मान ज्ञात कीजिए यदि त्रिभुज का क्षेत्रफल चार वर्ग इकाई है जहा शरबिदुय निम्नलिखित है:

(i) \[{ {(k,0), (4,0), (0,2)}}\]

उत्तर: त्रिभुज का क्षेत्रफल = \[{ { =  }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}}  {{{ {x}}_{ {1}}}}&{{{ {y}}_{ {1}}}}&{ {1}} \\   {{{ {x}}_{ {2}}}}&{{{ {y}}_{ {2}}}}&{ {1}} \\   {{{ {x}}_{ {3}}}}&{{{ {y}}_{ {3}}}}&{ {1}}  \end{array}} \right|\]

\[\begin{align}  \dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}}  { {k}}&{ {0}}&{ {1}} \\   { {4}}&{ {0}}&{ {1}} \\   { {0}}&{ {2}}&{ {1}}   \end{array}} \right|\; = \; \pm 4 \hfill \\  \left| {\begin{array}{*{20}{l}}  { {k}}&{ {0}}&{ {1}} \\   { {4}}&{ {0}}&{ {1}} \\   { {0}}&{ {2}}&{ {1}}   \end{array}} \right|{ {  =   \pm 8}} \hfill \\  { {( - 2)}}\left| {\begin{array}{*{20}{l}}  { {k}}&{ {1}} \\   { {4}}&{ {1}}   \end{array}} \right|{ {  =   \pm 8}} \hfill \\  { {k - 4  =   \pm 4}} \hfill \\  { {k  =  0, 8}} \hfill \\   \end{align} \]

(ii) \[{ {( - 2,0), (0,4), (0,k)}}\]

उत्तर: त्रिभुज का क्षेत्रफल = \[{ { =  }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}}  {{{ {x}}_{ {1}}}}&{{{ {y}}_{ {1}}}}&{ {1}} \\   {{{ {x}}_{ {2}}}}&{{{ {y}}_{ {2}}}}&{ {1}} \\   {{{ {x}}_{ {3}}}}&{{{ {y}}_{ {3}}}}&{ {1}}   \end{array}} \right|\]

\[\begin{align}   = \;\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{c}}  {{ { - 2}}}&{ {0}}&{ {1}} \\   { {0}}&{ {4}}&{ {1}} \\   { {0}}&{ {k}}&{ {1}}  \end{array}} \right| \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 2(4 - k) - 0(0 - 0) + 1(0 - 0)}} \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ { \times ( - 2)(4 - k)  =  k - 4  =   \pm 4}} \hfill \\  { { + ve, k - 4  =  4 ; k  =  8}} \hfill \\  { { - ve, k - 4  =  4 ; k  =  0}} \hfill \\   \end{align} \]

4. (i) सारणिकों का प्रयोग करके \[{ {(1,2), (3,6)}}\] को मिलाने वाली रेखा का समीकरण ज्ञात कीजिए। 

उत्तर: माना कोई बिन्दु \[{ {(x,y)}}\] है 

इसलिए त्रिभुज के शीर्ष \[{ {(x,y), (1,2), (3,6)}}\] होंगे। 

\[{ {\Delta   =  }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}}  {{{ {x}}_{ {1}}}}&{{{ {y}}_{ {1}}}}&{ {1}} \\   {{{ {x}}_{ {2}}}}&{{{ {y}}_{ {2}}}}&{ {1}} \\   {{{ {x}}_{ {3}}}}&{{{ {y}}_{ {3}}}}&{ {1}}  \end{array}} \right|\]

जहा \[{{ {x}}_{ {1}}}{ {  =  x, }}{{ {y}}_{ {1}}}{ {  =  y, }}{{ {x}}_{ {2}}}{ {  =  1, }}{{ {y}}_{ {2}}}{ {  =  2, }}{{ {x}}_{ {3}}}{ {  =  3, }}{{ {y}}_{ {3}}}{ {  =  6}}\]

\[\begin{align}  { { =  }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{c}}  { {x}}&{ {y}}&{ {1}} \\   { {1}}&{ {2}}&{ {1}} \\   { {3}}&{ {6}}&{ {1}}  \end{array}} \right| \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ {[x(2 - 6) - y(1 - 3) + 1(6 - 6)]}} \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ {[x \times ( - 4) - y( - 2) + 1 \times 0]}} \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 4x + 2y]}} \hfill \\  \end{align} \]

\[\begin{align}  { { =  }}\dfrac{{ {1}}}{{ {2}}}{ { \times 2( - 2x + y)}} \hfill \\  { { =   - 2x + y}} \hfill \\   \end{align} \]

बिन्दु संरेख है। 

इसलिए \[{ {\Delta }}\]  त्रिभुज का क्षेत्रफल शून्य होगा। 

\[\begin{align}  { {0  =  - 2x + y}} \hfill \\  { {2x - y  =  0}} \hfill \\   \end{align} \]

यही अभीष्ट समीकरण है। 

(ii) सारणिकों का प्रयोग करके \[{ {(3,1), (9,3)}}\] को मिलाने वाली रेखा का समीकरण ज्ञात कीजिए।

उत्तर: माना बिन्दुओ \[{ {A(3,1), B(9,3)}}\] को मिलाने वाली रेखा पर बिन्दु \[{ {P(x,y)}}\] है। तब बिन्दु A, B, P संरेख है। 

\[\begin{align}  { {APB  =  0}} \hfill \\  \dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}}  { {3}}&{ {1}}&{\mid { {1}}} \\   { {9}}&{ {3}}&{ {1}} \\   { {x}}&{ {y}}&{ {1}}   \end{array}} \right|{ {  =  0}} \hfill \\  \left| {\begin{array}{*{20}{l}}  { {3}}&{ {1}}&{ {1}} \\   { {9}}&{ {3}}&{ {1}} \\   { {x}}&{ {y}}&{ {1}}   \end{array}} \right|{ {  =  0}} \hfill \\  \end{align} \]

5. यदि शीर्ष \[{ {(2, - 6), (5,4), (k,4)}}\] वाले त्रिभुज का क्षेत्रफल \[{ {35}}\] वर्ग इकाई हो तो का मान है। 

(a) \[{ {12}}\]

(b) \[{ { - 2}}\]

(c) \[{ { - 12,  - 2}}\]

(d) \[{ {12,  - 2}}\]

उत्तर: दिया है, त्रिभुज के शीर्ष \[{ {(2, - 6), (5,4), (k,4)}}\] 

\[{ {\Delta   =  }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}}  {{{ {x}}_{ {1}}}}&{{{ {y}}_{ {1}}}}&{ {1}} \\   {{{ {x}}_{ {2}}}}&{{{ {y}}_{ {2}}}}&{ {1}} \\   {{{ {x}}_{ {3}}}}&{{{ {y}}_{ {3}}}}&{ {1}}  \end{array}} \right|\]

जहा \[{{ {x}}_{ {1}}}{ {  =  2, }}{{ {y}}_{ {1}}}{ {  =  6, }}{{ {x}}_{ {2}}}{ {  =  5, }}{{ {y}}_{ {2}}}{ {  =  4, }}{{ {x}}_{ {3}}}{ {  =  k, }}{{ {y}}_{ {3}}}{ {  =  4}}\]

ज्ञात \[{ {\Delta }}\] का क्षेत्रफल = \[{ { \pm 35}}\]

\[\begin{align}  { { \pm 35  =  }}\dfrac{{ {1}}}{{ {2}}}{ {[2(4 - 4) + 6(5 - k) + k(120 - 4k)]}} \hfill \\  { { \pm 35  =  }}\dfrac{{ {1}}}{{ {2}}}{ {[2 \times 0 + 6(5 - k) + 1(20 - 4k)]}} \hfill \\  { { \pm 70  =  6(5 - k) + 20 - 4k}} \hfill \\  { { \pm 70  =  30 - 6k + 20 - 4k}} \hfill \\  { { \pm 70  =  50 - 10k}} \hfill \\  { { \pm 70  =  5 - k}} \hfill \\  { { + ve, 7  =  5 - k ; k  =  5 - 7  =   - 2}} \hfill \\  { { - ve,  - 7  =  5 - k ;  - 12  =   - k ; k  =  12}} \hfill \\  \end{align} \] 

अतः \[{ {k  =  12, - 2}}\]

अतः विकल्प (d) सही है।

प्रश्नावली 4.4

निम्नलिखित सारणिकों के अवयवों के उपसारणिक एवं सहखंड लिखिए: 

1. (i) \[\left| {\begin{array}{*{20}{c}}  { {2}}&{{ { - 4}}} \\   { {0}}&{ {3}}  \end{array}} \right|\]

उत्तर: उपसारणिक \[{{ {M}}_{11}}{ {  =  3,  }}{{ {M}}_{12}}{ {  =  0,  }}{{ {M}}_{21}}{ {  =   - 4,  }}{{ {M}}_{22}}{ {  =  2}}\]

तथा सहखंड \[{{ {A}}_{11}}{ {  =  3,  }}{{ {A}}_{12}}{ {  =  0,  }}{{ {A}}_{21}}{ {  =   - ( - 4),  }}{{ {A}}_{22}}{ {  =  2}}\]

(ii) \[\left| {\begin{array}{*{20}{l}}  { {a}}&{ {c}} \\   { {b}}&{ {d}}  \end{array}} \right|\]

उत्तर: उपसारणिक \[{{ {M}}_{11}}{ {  =  d,  }}{{ {M}}_{12}}{ { =  b,  }}{{ {M}}_{21}}{ {  =  c, }}{{ {M}}_{22}}{ {  =  a}}\]

तथा सहखंड \[{{ {A}}_{11}}{ {  =  d, }}{{ {A}}_{12}}{ {  =   - b, }}{{ {A}}_{21}}{ {  =   - c, }}{{ {A}}_{22}}{ {  =  a}}\]

2. (i) \[\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}}&{ {0}} \\   { {0}}&{ {1}}&{ {0}} \\   { {0}}&{ {0}}&{ {1}}  \end{array}} \right|\]

उत्तर: उपसारणिक

\[\begin{align}  {{ {M}}_{{ {11}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}} \\   { {0}}&{ {1}}  \end{array}} \right|{ {  =  1 - 0  =  1 , }}{{ {M}}_{{ {12}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {0}}&{ {0}} \\   { {0}}&{ {1}} \end{array}} \right|{ {  =  0}} \hfill \\  {{ {M}}_{{ {13}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {0}}&{ {1}} \\   { {0}}&{ {0}}  \end{array}} \right|{ {  =  0 , }}{{ {M}}_{{ {21}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {0}}&{ {0}} \\   { {0}}&{ {1}}  \end{array}} \right|{ {  =  0}} \hfill \\  {{ {M}}_{{ {22}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}} \\   { {0}}&{ {1}}  \end{array}} \right|{ {  =  1 , }}{{ {M}}_{{ {23}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}} \\   { {0}}&{ {0}}  \end{array}} \right|{ {  =  0}} \hfill \\  {{ {M}}_{{ {31}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {0}}&{ {0}} \\   { {0}}&{ {1}}  \end{array}} \right|{ {  =  0 , }}{{ {M}}_{{ {32}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}} \\   { {0}}&{ {0}}  \end{array}} \right|{ {  =  0}} \hfill \\  {{ {M}}_{{ {31}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}} \\   { {0}}&{ {1}}  \end{array}} \right|{ {  =  0}} \hfill \\  \end{align} \]

तथा सहखंड

\[\begin{align}  {{ {A}}_{{ {11}}}}{ {  =  ( - 1}}{{ {)}}^{{ {1 + 1}}}}{{ {M}}_{{ {11}}}}{ {  =  1 , }}{{ {A}}_{{ {12}}}}{ {  =  ( - 1}}{{ {)}}^{{ {1 + 2}}}}{{ {M}}_{{ {12}}}}{ {  =  ( - 1) \times 0  =  0}} \hfill \\  {{ {A}}_{{ {13}}}}{ {  =  ( - 1}}{{ {)}}^{{ {1 + 3}}}}{{ {M}}_{{ {12}}}}{ {  =  1 \times 0  =  0 , }}{{ {A}}_{{ {21}}}}{ {  =  ( - 1}}{{ {)}}^{{ {2 + 1}}}}{{ {M}}_{{ {21}}}}{ {  =  0}} \hfill \\  \end{align} \]

\[\begin{align}  {{ {A}}_{{ {22}}}}{ {  =   ( - 1}}{{ {)}}^{{ {2 + 2}}}}{{ {M}}_{{ {22}}}}{ {  =  1 , }}{{ {A}}_{{ {23}}}}{ {  =  ( - 1}}{{ {)}}^{{ {2 + 3}}}}{{ {M}}_{{ {23}}}}{ {  =  0}} \hfill \\  {{ {A}}_{{ {31}}}}{ {  =  ( - 1}}{{ {)}}^{{ {3 + 1}}}}{{ {M}}_{{ {31}}}}{ {  =  0 , }}{{ {A}}_{{ {32}}}}{ {  =  ( - 1}}{{ {)}}^{{ {3 + 2}}}}{{ {M}}_{{ {32}}}}{ {  =  0}} \hfill \\  {{ {A}}_{{ {33}}}}{ {  =  ( - 1}}{{ {)}}^{{ {3 + 3}}}}{{ {M}}_{{ {33}}}}{ {  =  1}}{ {.1  =  1}} \hfill \\  \end{align} \]

(ii) \[\left| {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {4}} \\   { {3}}&{ {5}}&{{ { - 1}}} \\   { {0}}&{ {1}}&{ {2}}  \end{array}} \right|\]

उत्तर: उपसारणिक

\[\begin{align}  {{ {M}}_{{ {11}}}}{ {  =  }}\left| {\begin{array}{*{20}{c}}  { {5}}&{{ { - 1}}} \\   { {1}}&{ {2}}  \end{array}} \right|{ {  =  5 \times 2 - 1(1)  =  11 , }}{{ {M}}_{{ {12}}}}{ {  =  }}\left| {\begin{array}{*{20}{c}}  { {3}}&{{ { - 1}}} \\   { {0}}&{ {2}}  \end{array}} \right|{ {  =  3 \times 2 - 0  =  6}} \hfill \\  {{ {M}}_{{ {13}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {3}}&{ {5}} \\   { {0}}&{ {1}}  \end{array}} \right|{ {  =  3 \times 1 - 0  =  3 , }}{{ {M}}_{{ {21}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {0}}&{ {4}} \\   { {1}}&{ {2}}  \end{array}} \right|{ {  =  0 \times 2 - 1 \times 4  =   - 4}} \hfill \\  {{ {M}}_{{ {22}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {4}} \\   { {0}}&{ {2}}  \end{array}} \right|{ {  =  1 \times 2 - 0  =  2 , }}{{ {M}}_{{ {23}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}} \\   { {0}}&{ {2}}  \end{array}} \right|{ {  =  1 \times 1  =  1}} \hfill \\  {{ {M}}_{{ {32}}}}{ {  =  }}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {4}} \\   { {3}}&{{ { - 1}}}  \end{array}} \right|{ {  =  1 \times  - (1) - 3 \times 4  =  13 , }}{{ {M}}_{{ {33}}}}{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}} \\   { {3}}&{ {5}}  \end{array}} \right|{ {  =  5 - 0  =  5}} \hfill \\  \end{align} \]

तथा सहखंड

\[\begin{align}  {{ {A}}_{{ {11}}}}{ {  =  ( - 1}}{{ {)}}^{{ {1 + 1}}}}{{ {M}}_{{ {11}}}}{ {  =  11 , }}{{ {A}}_{{ {12}}}}{ {  =  ( - 1}}{{ {)}}^{{ {1 + 2}}}}{{ {M}}_{{ {12}}}}{ {  =  ( - 1) \times 6  =   - 6}} \hfill \\  {{ {A}}_{{ {13}}}}{ {  =  ( - 1}}{{ {)}}^{{ {1 + 3}}}}{{ {M}}_{{ {12}}}}{ {  =  1}}{ {.3  =  3 , }}{{ {A}}_{{ {21}}}}{ {  =  ( - 1}}{{ {)}}^{{ {2 + 1}}}}{{ {M}}_{{ {21}}}}{ {  =  ( - 1) \times ( - 4)  =  4}} \hfill \\ \end{align} \] \[\begin{align}  {{ {A}}_{{ {22}}}}{ {  =  ( - 1}}{{ {)}}^{{ {2 + 2}}}}{{ {M}}_{{ {22}}}}{ {  =  1 \times 2  =  2 , }}{{ {A}}_{{ {23}}}}{ {  =  ( - 1}}{{ {)}}^{{ {2 + 3}}}}{{ {M}}_{{ {23}}}}{ {  =  ( - 1)1  =   - 1}} \hfill \\  {{ {A}}_{{ {31}}}}{ {  =  ( - 1}}{{ {)}}^{{ {3 + 1}}}}{{ {M}}_{{ {31}}}}{ {  =  1}}{ {.( - 20)  =   - 20 , }}{{ {A}}_{{ {32}}}}{ {  =  ( - 1}}{{ {)}}^{{ {3 + 2}}}}{{ {M}}_{{ {32}}}}{ {  =  ( - 1)( - 13)  =  13}} \hfill \\  {{ {A}}_{{ {33}}}}{ {  =  ( - 1}}{{ {)}}^{{ {3 + 3}}}}{{ {M}}_{{ {33}}}}{ {  =  1}}{ {.5  =  1}} \hfill \\  \end{align} \]

3. दूसरी पंक्ति के अवयवों के सहखंडों का प्रयोग करके  का मान ज्ञात कीजिए। 

उत्तर: \[{{ {A}}_{{ {21}}}}{ {  =  ( - 1}}{{ {)}}^{{ {2 + 1}}}}\left| {\begin{array}{*{20}{l}}  { {3}}&{ {8}} \\   { {2}}&{ {3}}  \end{array}} \right|{ {  =  ( - 1) \times [3 \times 3 - 2 \times 8]  =  7}}\]

\[\begin{align}  {{ {A}}_{{ {22}}}}{ {  =  ( - 1}}{{ {)}}^{{ {2 + 2}}}}\left| {\begin{array}{*{20}{l}}  { {5}}&{ {8}} \\   { {1}}&{ {3}}  \end{array}} \right|{ {  =  1[5 \times 3 - 2 \times 8]  =  7}} \hfill \\  {{ {A}}_{{ {23}}}}{ {  =  ( - 1}}{{ {)}}^{{ {2 + 3}}}}\left| {\begin{array}{*{20}{l}}  { {5}}&{ {3}} \\   { {1}}&{ {2}}  \end{array}} \right|{ {  =  ( - 1)[5 \times 2 - 3 \times 1]  =   - 7}} \hfill \\  { {\Delta   =  }}{{ {a}}_{{ {21}}}}{{ {A}}_{{ {21}}}}{ { + }}{{ {a}}_{{ {22}}}}{{ {A}}_{{ {22}}}}{ { + }}{{ {a}}_{{ {23}}}}{{ {A}}_{{ {23}}}} \hfill \\  { { =  2 \times 7 + 0 \times 7 + 1 \times ( - 7)}} \hfill \\  { { =  14 - 7  =  7}} \hfill \\  \end{align} \]

4. तीसरे स्तम्भ के अवयवों के सहखंडों का प्रयोग करके \[{ {\Delta   =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {x}}&{{ {yz}}} \\   { {1}}&{ {y}}&{{ {zx}}} \\   { {1}}&{ {2}}&{{ {xy}}}  \end{array}} \right|\] का मान ज्ञात कीजिए। 

उत्तर: \[{{ {A}}_{{ {13}}}}{ {  =   - }}{{ {1}}^{{ {1 + 3}}}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {y}} \\   { {1}}&{ {z}}  \end{array}} \right|{ {  =  (1)(z - y)  =  (z - y)}}\]

\[{{ {A}}_{{ {23}}}}{ {  =   - }}{{ {1}}^{{ {2 + 3}}}}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {x}} \\   { {1}}&{ {z}}  \end{array}} \right|{ {  =  (1)(z - x)  =   - (x - z)}}\]

\[\begin{align}  {{ {A}}_{{ {33}}}}{ {  =   - }}{{ {1}}^{{ {1 + 3}}}}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {x}} \\   { {1}}&{ {y}}  \end{array}} \right|{ {  =  (1)(y - x)  =  (y - x)}} \hfill \\  { {\Delta   =  }}{{ {a}}_{{ {13}}}}{{ {A}}_{{ {13}}}}{ { + }}{{ {a}}_{{ {23}}}}{{ {A}}_{{ {23}}}}{ { + }}{{ {a}}_{{ {33}}}}{{ {A}}_{{ {33}}}} \hfill \\  { { =  yz(z - y) + zx(x - z) + xy(y - x)}} \hfill \\  { { =  y}}{{ {z}}^{ {2}}}{ { - }}{{ {y}}^{ {2}}}{ {z + z}}{{ {x}}^{ {2}}}{ { - }}{{ {z}}^{ {2}}}{ {x + x}}{{ {y}}^{ {2}}}{ { - }}{{ {x}}^{ {2}}}{ {y}} \hfill \\  { { =  z}}{{ {x}}^{ {2}}}{ { - }}{{ {x}}^{ {2}}}{ {y + x}}{{ {y}}^{ {2}}}{ { - }}{{ {z}}^{ {2}}}{ {y + y}}{{ {z}}^{ {2}}}{ { - }}{{ {y}}^{ {2}}}{ {z}} \hfill \\  { { =  }}{{ {x}}^{ {2}}}{ {((z - y) + z(y - z)(y + z) + yz(z - y)}} \hfill \\  { { =  (z - y)}}\left[ {{{ {x}}^{ {2}}}{ { - x(y + z) + yz}}} \right] \hfill \\  { { =  (z - y)}}\left[ {{{ {x}}^{ {2}}}{ { - xy - xz + yz}}} \right] \hfill \\  { { =  (z - y)[x(x - y) - z(x - y)]}} \hfill \\  { { =  (z - y)(x - y)(x - z)}} \hfill \\  { { =  (x - y)(y - z)(z - x)}} \hfill \\  \end{align} \]

5. यदि \[{ {\Delta   =  }}\left| {\begin{array}{*{20}{l}}  {{{ {a}}_{{ {11}}}}}&{{{ {a}}_{{ {12}}}}}&{{{ {a}}_{{ {13}}}}} \\   {{{ {a}}_{{ {21}}}}}&{{{ {a}}_{{ {22}}}}}&{{{ {a}}_{{ {23}}}}} \\   {{{ {a}}_{{ {31}}}}}&{{{ {a}}_{{ {32}}}}}&{{{ {a}}_{{ {33}}}}}  \end{array}} \right|\] और \[{{ {a}}_{ {y}}}\] का सहखंड \[{{ {A}}_{ {y}}}\] हो तो \[{ {\Delta }}\] शन का मान निम्नलिखित रूप मे व्यक्त किया जाता है: 

(a) \[{{ {a}}_{{ {11}}}}{ { \times }}{{ {A}}_{{ {31}}}}{ { + }}{{ {a}}_{{ {12}}}}{{ {A}}_{{ {32}}}}{ { + }}{{ {a}}_{{ {13}}}}{{ {A}}_{{ {13}}}}\]

(b) \[{{ {a}}_{{ {11}}}}{ { \times }}{{ {A}}_{{ {11}}}}{ { + }}{{ {a}}_{{ {12}}}}{{ {A}}_{{ {21}}}}{ { + }}{{ {a}}_{{ {13}}}}{{ {A}}_{{ {31}}}}\]

(c) \[{{ {a}}_{{ {21}}}}{ { \times }}{{ {A}}_{{ {11}}}}{ { + }}{{ {a}}_{{ {22}}}}{{ {A}}_{{ {12}}}}{ { + }}{{ {a}}_{{ {23}}}}{{ {A}}_{{ {13}}}}\]

(d) \[{{ {a}}_{{ {11}}}}{ { \times }}{{ {A}}_{{ {11}}}}{ { + }}{{ {a}}_{{ {21}}}}{{ {A}}_{{ {21}}}}{ { + }}{{ {a}}_{{ {31}}}}{{ {A}}_{{ {31}}}}\]

उत्तर: \[{ {\Delta }}\] =  किसी पंक्ति अथवा स्तम्भ के अवयवों तथा उनके संगत महखंडों के गुणा का योग 

\[{{ {C}}_{ {1}}}\] = स्तम्भ के अवयवों 

= \[{{ {A}}_{{ {11}}}}{ {,}}\;{{ {A}}_{{ {21}}}}{ {,}}\;{{ {A}}_{{ {31}}}}\]

\[ \Rightarrow \;{{ {a}}_{{ {11}}}}{{ {A}}_{{ {11}}}}{ { + }}{{ {a}}_{{ {21}}}}{{ {A}}_{{ {21}}}}{ { + }}{{ {a}}_{{ {31}}}}{{ {A}}_{{ {31}}}}\]

अतः विकल्प (d) सही है।

प्रश्नावली 4.5 

प्रश्न 1 और 2 मे प्रत्येक आव्यूह का सहखंडज ज्ञात कीजिए। 

1. \[\left[ \begin{align}  1\,\,\,\,\,2 \hfill \\  3\;\,\,4 \hfill \\  \end{align}  \right]\]

उत्तर: यहा \[{ {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {2}} \\   { {3}}&{ {4}}  \end{array}} \right]\]

इसलिए \[{ {adjA  =  }}\left[ {\begin{array}{*{20}{c}}  { {4}}&{{ { - 2}}} \\   {{ { - 3}}}&{ {1}}  \end{array}} \right]\]

2. \[\left[ \begin{align}  1\,\;\,\;\,\;\; - 1\;\;\;\;\;\;\;2 \hfill \\  2\;\;\;\;\;\;\;3\;\;\;\;\;\;\;\;5 \hfill \\   - 2\;\;\;\,\;\;\;0\;\;\;\;\;\;1 \hfill \\ \end{align}  \right]\]

उत्तर: यहा \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{{ { - 1}}}&{ {2}} \\   { {2}}&{ {3}}&{ {5}} \\   {{ { - 2}}}&{ {0}}&{ {1}}  \end{array}} \right]\]

इसलिए \[{ {adjA  =  }}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\   {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\   {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{ {1}}&{{ { - 11}}} \\   {{ { - 12}}}&{ {5}}&{{ { - 1}}} \\   { {6}}&{ {2}}&{ {5}} \end{array}} \right]\]

प्रश्न 3 और 4 मे सत्यापित कीजिए की \[{ {A(adjA)  =  (adjA) \times A  =  |A|}}{ {.I}}\] है।

3. \[\left[ \begin{align}  2\;\;\;\;\;\;\;3 \hfill \\   - 4\;\;\; - 6 \hfill \\  \end{align}  \right]\]

उत्तर: यहा \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{ {3}} \\   {{ { - 4}}}&{{ { - 6}}}  \end{array}} \right]\]

इसलिए \[{ {adjA  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 6}}}&{{ { - 3}}} \\   { {4}}&{ {2}}  \end{array}} \right]\]

\[\begin{align}  { {A(adjA)  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{ {3}} \\   {{ { - 4}}}&{{ { - 6}}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  {{ { - 6}}}&{{ { - 3}}} \\   { {4}}&{ {2}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 12 + 12}}}&{{ { - 6 + 6}}} \\   {{ {24 - 24}}}&{{ {12 - 12}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {0}}&{ {0}} \\   { {0}}&{ {0}}  \end{array}} \right] \hfill \\  \end{align} \]

\[\begin{align}  { {(adjA)A  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 6}}}&{{ { - 3}}} \\   { {4}}&{ {2}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {2}}&{ {3}} \\   {{ { - 4}}}&{{ { - 6}}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 12 + 12}}}&{{ { - 18 + 18}}} \\   {{ {8 - 8}}}&{{ {12 - 12}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  { {0}}&{ {0}} \\   { {0}}&{ {0}}  \end{array}} \right] \hfill \\  \end{align} \]

\[\begin{align}  { {|A| I  =  0 \times }}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {0}} \\   { {0}}&{ {1}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  { {0}}&{ {0}} \\   { {0}}&{ {0}}  \end{array}} \right] \hfill \\  { {A(adjA)  =  (adjA)A  =  |A| I  =  }}\left[ {\begin{array}{*{20}{c}}  { {0}}&{ {0}} \\   { {0}}&{ {0}}  \end{array}} \right] \hfill \\  \end{align} \]

4. \[\left[ \begin{align}  1\,\;\,\;\,\;\; - 1\;\;\;\;\;\;\;2 \hfill \\  3\;\;\;\;\;\;\;\;0\;\;\;\;\; - 2 \hfill \\  1\;\;\;\,\;\;\;\;0\;\;\;\;\;\;\;\;3 \hfill \\ \end{align}  \right]\]

उत्तर: \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{{ { - 1}}}&{ {2}} \\   { {3}}&{ {0}}&{{ { - 2}}} \\   { {1}}&{ {0}}&{ {3}}  \end{array}} \right]\]

\[\begin{align}  { {adjA  =  }}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\   {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\   {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {0}}&{ {3}}&{ {2}} \\   {{ { - 11}}}&{ {1}}&{ {8}} \\   { {0}}&{{ { - 1}}}&{ {3}}  \end{array}} \right] \hfill \\  { {A(adjA)  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{{ { - 1}}}&{ {2}} \\   { {3}}&{ {0}}&{{ { - 2}}} \\   { {1}}&{ {0}}&{ {3}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {0}}&{ {3}}&{ {2}} \\   {{ { - 11}}}&{ {1}}&{ {8}} \\   { {0}}&{{ { - 1}}}&{ {3}} \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  {{ {11}}}&{ {0}}&{ {0}} \\   { {0}}&{{ {11}}}&{ {0}} \\   { {0}}&{ {0}}&{{ {11}}}  \end{array}} \right] \hfill \\  { {(adjA)A  =  }}\left[ {\begin{array}{*{20}{c}}  { {0}}&{ {3}}&{ {2}} \\   {{ { - 11}}}&{ {1}}&{ {8}} \\   { {0}}&{{ { - 1}}}&{ {3}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {1}}&{{ { - 1}}}&{ {2}} \\   { {3}}&{ {0}}&{{ { - 2}}} \\   { {1}}&{ {0}}&{ {3}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ {11}}}&{ {0}}&{ {0}} \\   { {0}}&{{ {11}}}&{ {0}} \\   { {0}}&{ {0}}&{{ {11}}}  \end{array}} \right] \hfill \\  \end{align} \]

\[\begin{align}  { {|A| I  =  11 }}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {0}}&{ {0}} \\   { {0}}&{ {1}}&{ {0}} \\   { {0}}&{ {0}}&{ {1}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ {11}}}&{ {0}}&{ {0}} \\   { {0}}&{{ {11}}}&{ {0}} \\   { {0}}&{ {0}}&{{ {11}}}  \end{array}} \right] \hfill \\  { {A(adjA)  =  (adjA)A  =  |A| I  =  }}\left[ {\begin{array}{*{20}{c}}  {{ {11}}}&{ {0}}&{ {0}} \\   { {0}}&{{ {11}}}&{ {0}} \\   { {0}}&{ {0}}&{{ {11}}}  \end{array}} \right] \hfill \\ \end{align} \]

प्रश्न 5 से 11 मे दिए गए प्रत्येक आवयहो के व्युत्क्रम ज्ञात कीजिए। 

5. \[\left[ \begin{align}  2\;\;\;\,\;\;\, - 2 \hfill \\  4\;\,\;\;\;\;\;\;\,3 \hfill \\ \end{align}  \right]\]

उत्तर: यहा \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 2}}} \\   { {4}}&{ {3}}  \end{array}} \right]\]

इसलिए \[{ {adjA  =  }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{ {2}} \\   {{ { - 4}}}&{ {2}}  \end{array}} \right]\]

\[{ {|A|  =  6 + 8  =  14}}\; \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है। 

\[{{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA)  =  }}\dfrac{{ {1}}}{{{ {14}}}}\left[ {\begin{array}{*{20}{c}}  { {3}}&{ {2}} \\   {{ { - 4}}}&{ {2}} \end{array}} \right]\]

6. \[\left[ \begin{align}   - 1\;\;\;\,\;\;\,\;\;5 \hfill \\   - 3\;\,\;\;\;\;\;\;\,2 \hfill \\  \end{align}  \right]\]

उत्तर: यहा \[{ {A  =  }}\left[ {\begin{array}{*{20}{l}}  {{ { - 1}}}&{ {5}} \\   {{ { - 3}}}&{ {2}}  \end{array}} \right]\]

इसलिए \[{ {adjA  =  }}\left[ {\begin{array}{*{20}{l}}  { {2}}&{{ { - 5}}} \\   { {3}}&{{ { - 1}}}  \end{array}} \right]\]

\[{ {|A|  =   - 2 + 15  =  13 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है। 

\[{{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA)  =  }}\dfrac{{ {1}}}{{{ {13}}}}\left[ {\begin{array}{*{20}{l}}  { {2}}&{{ { - 5}}} \\   { {3}}&{{ { - 1}}} \end{array}} \right]\]

7. \[\left[ \begin{align}  1\;\;\;\,\;\;\,\;\;2\;\;\,\;\;\,\;\;3 \hfill \\  0\;\;\,\;\;\;\;\;2\;\;\;\;\;\;\,\;4 \hfill \\  0\;\,\;\;\;\;\;\;\,0\;\;\;\;\;\;\;5 \hfill \\  \end{align}  \right]\]

उत्तर: यहा \[{ {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {2}}&{ {3}} \\   { {0}}&{ {2}}&{ {4}} \\   { {0}}&{ {0}}&{ {5}}  \end{array}} \right]\]

इसलिए \[{ {adjA  =  }}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\   {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\   {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ {10}}}&{{ { - 10}}}&{ {2}} \\   { {0}}&{ {5}}&{{ { - 4}}} \\   { {0}}&{ {0}}&{ {2}} =  \end{array}} \right]\]

\[{ {|A|  =  1(10 - 0) - 2(0 - 0) + 3(0 - 0)  =  10 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है। 

\[\begin{align}  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA)  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\   {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\   {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right] \hfill \\  { { =  }}\dfrac{{ {1}}}{{{ {10}}}}\left[ {\begin{array}{*{20}{c}}  {{ {10}}}&{{ { - 10}}}&{ {2}} \\   { {0}}&{ {5}}&{{ { - 4}}} \\   { {0}}&{ {0}}&{ {2}}  \end{array}} \right] \hfill \\ \end{align} \]

8. \[\left[ \begin{align}  1\;\;\;\,\;\;\,\;\;0\;\;\,\;\;\,\;\;0 \hfill \\  3\;\;\,\;\;\;\;\;3\;\;\;\;\;\;\,\;0 \hfill \\  5\;\,\;\;\;\;\;\;\,2\;\;\;\;\;\; - 1 \hfill \\  \end{align}  \right]\]

उत्तर: यहा \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   { {3}}&{ {3}}&{ {0}} \\   { {5}}&{ {2}}&{{ { - 1}}}  \end{array}} \right]\]

इसलिए \[{ {adjA  =  }}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\   {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\   {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 3}}}&{ {0}}&{ {0}} \\   { {3}}&{{ { - 1}}}&{ {0}} \\   {{ { - 9}}}&{{ { - 2}}}&{ {3}} \end{array}} \right]\]

\[{ {|A|  =  1( - 3 - 0) - 0( - 3 - 0) + 0(6 - 15)  =   - 3 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है। 

\[\begin{align}  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA)  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\   {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\   {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right] \hfill \\  { { =  }}\dfrac{{ {1}}}{{{ { - 3}}}}\left[ {\begin{array}{*{20}{c}}  {{ { - 3}}}&{ {0}}&{ {0}} \\   { {3}}&{{ { - 1}}}&{ {0}} \\   {{ { - 9}}}&{{ { - 2}}}&{ {3}}  \end{array}} \right] \hfill \\ \end{align} \]

9. \[\left[ \begin{align}  2\;\;\;\,\;\;\,\;\;\;\,\;\;1\;\,\;\;\,\;\;\;\;\;3 \hfill \\  4\;\;\,\;\;\;\;\;\;\; - 1\;\;\;\;\,\;\;\;0 \hfill \\   - 7\;\,\;\;\;\;\;\;\;\;\,2\;\;\;\;\;\,\;\;1 \hfill \\  \end{align}  \right]\]

उत्तर: यह \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{ {1}}&{ {3}} \\   { {4}}&{{ { - 1}}}&{ {0}} \\   {{ { - 7}}}&{ {2}}&{ {1}}  \end{array}} \right]\]

इसलिए \[{ {adjA  =  }}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\   {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\   {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 1}}}&{ {5}}&{ {3}} \\   {{ { - 4}}}&{{ {23}}}&{{ {12}}} \\   { {1}}&{{ { - 11}}}&{{ { - 6}}}  \end{array}} \right]\]

\[{ {|A|  =  2( - 1 - 0) - 1(4 - 0) + 3(8 - 7)  =   - 3 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है। 

\[\begin{align}  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA)  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\   {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\   {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right] \hfill \\  { { =  }}\dfrac{{ {1}}}{{{ { - 3}}}}\left[ {\begin{array}{*{20}{c}}  {{ { - 1}}}&{ {5}}&{ {3}} \\   {{ { - 4}}}&{{ {23}}}&{{ {12}}} \\   { {1}}&{{ { - 11}}}&{{ { - 6}}}  \end{array}} \right] \hfill \\  \end{align} \]

10. \[\left[ \begin{align}  1\;\;\;\,\;\;\,\;\; - 1\;\,\;\;\,\;\;\;\;\;2 \hfill \\  0\;\;\,\;\;\;\;\;\;\;2\;\;\;\;\,\;\;\; - 3 \hfill \\  3\;\,\;\;\;\;\;\, - 2\;\;\;\;\;\,\;\;4 \hfill \\  \end{align}  \right]\]

उत्तर: यहा \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{{ { - 1}}}&{ {2}} \\   { {0}}&{ {2}}&{{ { - 3}}} \\   { {3}}&{{ { - 2}}}&{ {4}}  \end{array}} \right]\]

इसलिए \[{ {adjA  =  }}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\   {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\   {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{ {0}}&{{ { - 1}}} \\   {{ { - 9}}}&{{ { - 2}}}&{ {3}} \\   {{ { - 6}}}&{{ { - 1}}}&{ {2}}  \end{array}} \right]\]

\[{ {|A|  =  1(8 - 6) + 1(0 + 9) + 2(0 - 6)  =   - 1 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है। 

\[\begin{align}  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA)  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\   {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\   {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right] \hfill \\  { { =  }}\dfrac{{ {1}}}{{{ { - 1}}}}\left[ {\begin{array}{*{20}{c}}  { {2}}&{ {0}}&{{ { - 1}}} \\   {{ { - 9}}}&{{ { - 2}}}&{ {3}} \\   {{ { - 6}}}&{{ { - 1}}}&{ {2}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 2}}}&{ {0}}&{ {1}} \\   { {9}}&{ {2}}&{{ { - 3}}} \\   { {6}}&{ {1}}&{{ { - 2}}}  \end{array}} \right] \hfill \\  \end{align} \]

11. \[\left[ \begin{align}  { {1}}\;\;\;\,\;\;\,\;\;{ {0}}\;\,\;\;\,\;\;\;\;\;\;\;\;\;\,{ {0}} \hfill \\  { {0}}\;\;\,\;\;\;\;{ {cos\alpha }}\;\;\;\;\,\;\;\;{ {sin\alpha }} \hfill \\  { {0}}\;\,\;\;\;\;\,{ {sin\alpha }}\;\;\;\;\;\,\;\;{ {cos\alpha }} \hfill \\   \end{align}  \right]\]

उत्तर: यहा \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   { {0}}&{{ {cos\alpha }}}&{{ {sin\alpha }}} \\   { {0}}&{{ {sin\alpha }}}&{{ { - cos\alpha }}}  \end{array}} \right]\]

इसलिए \[{ {adjA  =  }}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\   {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\   {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   { {0}}&{{ { - cos\alpha }}}&{{ { - sin\alpha }}} \\   { {0}}&{{ { - sin\alpha }}}&{{ {cos\alpha }}}  \end{array}} \right]\]

\[{ {|A|  =  1}}\left( {{ { - co}}{{ {s}}^{ {2}}}{ {\alpha  - si}}{{ {n}}^{ {2}}}{ {\alpha }}} \right){ { + 0(0 - 0) + 0(0 - 0)  =   - 1 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है। 

\[\begin{align}  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA)  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\   {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\   {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right] \hfill \\  { { =  }}\dfrac{{ {1}}}{{{ { - 1}}}}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   { {0}}&{{ { - cos\alpha }}}&{{ { - sin\alpha }}} \\   { {0}}&{{ { - sin\alpha }}}&{{ {cos\alpha }}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 1}}}&{ {0}}&{ {0}} \\   { {0}}&{{ {cos\alpha }}}&{{ {sin\alpha }}} \\   { {0}}&{{ {sin\alpha }}}&{{ { - cos\alpha }}}  \end{array}} \right] \hfill \\  \end{align} \]

12. यदि \[{ {A}}\;{ { = }}\;\left[ \begin{align}  { {3}}\;\;\;\,\;\;\,\;\;{ {7}} \hfill \\  { {2}}\;\,\;\;\;\;\;\;\,{ {5}} \hfill \\  \end{align}  \right]\] और \[{ {B}}\;{ { = }}\;\left[ \begin{align}  { {6}}\;\;\;\,\;\;\,\;\;{ {8}} \hfill \\  { {7}}\;\,\;\;\;\;\;\;\,{ {9}} \hfill \\  \end{align}  \right]\] है तो सत्यापित कीजिए की \[{{ {(AB)}}^{{ { - 1}}}}\,{ { = }}\,{{ {B}}^{{ { - 1}}}}{{ {A}}^{{ { - 1}}}}\]

उत्तर: यहा \[{ {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {3}}&{ {7}} \\   { {2}}&{ {5}}  \end{array}} \right]\]

\[{ {|A|  =  15 - 14  =  1 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है। 

\[{{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA)  =  }}\dfrac{{ {1}}}{{ {1}}}\left[ {\begin{array}{*{20}{c}}  { {5}}&{{ { - 7}}} \\   {{ { - 2}}}&{ {3}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {5}}&{{ { - 7}}} \\   {{ { - 2}}}&{ {3}}  \end{array}} \right]\]

यहा \[{ {B  =  }}\left[ {\begin{array}{*{20}{l}}  { {6}}&{ {8}} \\   { {7}}&{ {9}} \end{array}} \right]\]

\[{ {|B|  =  54 - 56  =   - 2 }} \ne { { 0 }} \to { { }}{{ {B}}^{{ { - 1}}}}\] का अस्तित्व है। 

\[{{ {B}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|B|}}}}{ {(adjB)  =  }}\dfrac{{ {1}}}{{{ { - 2}}}}\left[ {\begin{array}{*{20}{c}}  { {9}}&{{ { - 8}}} \\   {{ { - 7}}}&{ {6}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - }}\dfrac{{ {9}}}{{ {2}}}}&{ {4}} \\   {\dfrac{{ {7}}}{{ {2}}}}&{{ { - 3}}}  \end{array}} \right]\]

\[\begin{align}  {{ {B}}^{{ { - 1}}}}{{ {A}}^{{ { - 1}}}}{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - }}\dfrac{{ {9}}}{{ {2}}}}&{ {4}} \\   {\dfrac{{ {7}}}{{ {2}}}}&{{ { - 3}}}   \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {5}}&{{ { - 7}}} \\   {{ { - 2}}}&{ {3}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - }}\dfrac{{{ {45}}}}{{ {2}}}{ { - 8}}}&{\dfrac{{{ {63}}}}{{ {2}}}{ { + 12}}} \\   {\dfrac{{{ {35}}}}{{ {2}}}{ { + 6}}}&{{ { - }}\dfrac{{{ {49}}}}{{ {2}}}{ { - 9}}}   \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - }}\dfrac{{{ {81}}}}{{ {2}}}}&{\dfrac{{{ {87}}}}{{ {2}}}} \\   {\dfrac{{{ {47}}}}{{ {2}}}}&{{ { - }}\dfrac{{{ {67}}}}{{ {2}}}}  \end{array}} \right] \hfill \\  { {AB  =  }}\left[ {\begin{array}{*{20}{l}}  { {3}}&{ {7}} \\   { {2}}&{ {5}}  \end{array}} \right]\left[ {\begin{array}{*{20}{l}}  { {6}}&{ {8}} \\   { {7}}&{ {9}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  {{ {18 + 49}}}&{{ {24 + 63}}} \\   {{ {12 + 35}}}&{{ {16 + 45}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  {{ {67}}}&{{ {87}}} \\   {{ {47}}}&{{ {61}}}  \end{array}} \right] \hfill \\  \end{align} \]

\[{ {|AB|  =  4087 - 4089  =   - 2 }} \ne { { 0 }} \to { { (AB}}{{ {)}}^{{ { - 1}}}}\] का अस्तित्व है। 

\[\begin{align}  {{ {(AB)}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|AB|}}}}{ {(adjAB)  =  }}\dfrac{{ {1}}}{{{ { - 2}}}}\left[ {\begin{array}{*{20}{c}}  {{ {61}}}&{{ { - 87}}} \\   {{ { - 47}}}&{{ {67}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - }}\dfrac{{{ {61}}}}{{ {2}}}}&{\dfrac{{{ {87}}}}{{ {2}}}} \\   {\dfrac{{{ {47}}}}{{ {2}}}}&{{ { - }}\dfrac{{{ {67}}}}{{ {2}}}}  \end{array}} \right] \hfill \\  {{ {(AB)}}^{{ { - 1}}}}{ {  =  }}{{ {B}}^{{ { - 1}}}}{ {\;}}{{ {A}}^{{ { - 1}}}} \hfill \\  \end{align} \]

13. यदि \[{ {A}}\;{ { = }}\;\left[ \begin{align}  3\;\;\;\,\;\;\,\;\;\;1 \hfill \\   - 1\;\,\;\;\;\;\;\;2 \hfill \\  \end{align}  \right]\] है तो दर्शाइए की \[{{ {A}}^{ {2}}}{ { - 5A + 7I}}\;{ { = }}\;{ {0}}\] है इसकी सहायता से \[{{ {A}}^{{ { - 1}}}}\] ज्ञात कीजिए। 

उत्तर: LHS = \[{{ {A}}^{ {2}}}{ { - 5A + 7I  =  AA - 5A + 7I}}\]

\[\begin{align}  { { =  }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{ {1}} \\   {{ { - 1}}}&{ {2}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {3}}&{ {1}} \\   {{ { - 1}}}&{ {2}}  \end{array}} \right]{ { - 5}}\left[ {\begin{array}{*{20}{c}}  { {3}}&{ {1}} \\   {{ { - 1}}}&{ {2}}  \end{array}} \right]{ { + 7}}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {0}} \\   { {0}}&{ {1}}   \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  {{ {9 - 1}}}&{{ {3 + 2}}} \\   {{ { - 3 - 2}}}&{{ { - 1 + 4}}}   \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}}  {{ {15}}}&{ {5}} \\   {{ { - 5}}}&{{ {10}}}   \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}}  { {7}}&{ {0}} \\   { {0}}&{ {7}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  {{ {8 - 15 + 7}}}&{{ {5 - 5 + 0}}} \\   {{ { - 5 + 5 + 0}}}&{{ {3 - 10 + 7}}}   \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  { {0}}&{ {0}} \\   { {0}}&{ {0}}  \end{array}} \right] \hfill \\  { { =  0}} \hfill \\  \end{align} \]

= RHS

\[\begin{align}  {{ {A}}^{ {2}}}{ { - 5A + 7I  =  0}} \hfill \\  {{ {A}}^{ {2}}}{ { - 5A  =   - 7I}} \hfill \\  { {AA}}{{ {A}}^{{ { - 1}}}}{ { - 5A}}{{ {A}}^{{ { - 1}}}}{ {  =   - 7I}}{{ {A}}^{{ { - 1}}}} \hfill \\  { {7}}{{ {A}}^{{ { - 1}}}}{ {  =  5I - AI  =  5}}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {0}} \\   { {0}}&{ {1}}  \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{ {1}} \\   {{ { - 1}}}&{ {2}}   \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{l}}  { {5}}&{ {0}} \\   { {0}}&{ {5}}   \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{ {1}} \\   {{ { - 1}}}&{ {2}}   \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}} \\   { {1}}&{ {3}}   \end{array}} \right] \hfill \\  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{ {7}}}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}} \\   { {1}}&{ {3}}   \end{array}} \right] \hfill \\   \end{align} \]

14. आव्यूह \[{ {A}}\;{ { = }}\;\left[ \begin{align}  3\;\;\;\,\;\;\,\;\;\;2 \hfill \\  1\;\,\;\;\;\;\;\;\;\;1 \hfill \\  \end{align}  \right]\] के लिए और ऐसी संख्या ज्ञात कीजिए ताकि \[{{ {A}}^{ {2}}}{ { + aA + bI}}\;{ { = }}\;{ {0}}\] हो। 

उत्तर: \[{{ {A}}^{ {2}}}{ { + aA + bI}}\;{ { = }}\;{ {0}}\]

\[\begin{align}  { { =  }}\left[ {\begin{array}{*{20}{l}}  { {3}}&{ {2}} \\   { {1}}&{ {1}}  \end{array}} \right]\left[ {\begin{array}{*{20}{l}}  { {3}}&{ {2}} \\   { {1}}&{ {1}}  \end{array}} \right]{ { + a}}\left[ {\begin{array}{*{20}{l}}  { {3}}&{ {2}} \\   { {1}}&{ {1}}  \end{array}} \right]{ { + b}}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {0}} \\   { {0}}&{ {1}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  { {0}}&{ {0}} \\   { {0}}&{ {0}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{l}}  {{ {9 + 2}}}&{{ {6 + 2}}} \\   {{ {3 + 1}}}&{{ {2 + 1}}}  \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}}  {{ {3a}}}&{{ {2a}}} \\   { {a}}&{ {a}}  \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}}  { {b}}&{ {0}} \\   { {0}}&{ {b}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  { {0}}&{ {0}} \\   { {0}}&{ {0}}   \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{l}}  {{ {11 + 3a + b}}}&{{ {8 + 2a + 0}}} \\   {{ {4 + a + 0}}}&{{ {3 + a + b}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  { {0}}&{ {0}} \\   { {0}}&{ {0}}  \end{array}} \right] \hfill \\  { {4 + a  =  0}} \hfill \\  { {a  =   - 4}} \hfill \\  { {3 + a + b  =  0}} \hfill \\  { {3 - 4 + b  =  0}} \hfill \\  { {b  =  1}} \hfill \\  { {a  =   - 4, b  =  1}} \hfill \\  \end{align} \]

15. आव्यूह \[{ {A}}\;{ { = }}\;\left[ \begin{align}  1\;\;\;\,\;\;\;1\;\,\;\;\;\;\;1 \hfill \\  1\;\;\;\;\;\;\,\;2\;\;\; - 3 \hfill \\  2\;\,\;\;\; - 1\;\;\;\;\;\;3 \hfill \\  \end{align}  \right]\] के लिए दर्शाइए की \[{{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 5A + 11 I}}\;{ { = }}\;{ {0}}\] है। इसकी सहायता से \[{{ {A}}^{{ { - 1}}}}\] ज्ञात कीजिए। 

उत्तर: \[{ {A}}\;{ { = }}\;\left[ \begin{align}  1\;\;\;\,\;\;\;1\;\,\;\;\;\;\;1 \hfill \\  1\;\;\;\;\;\;\,\;2\;\;\; - 3 \hfill \\  2\;\,\;\;\; - 1\;\;\;\;\;\;3 \hfill \\  \end{align}  \right]\]

\[\begin{align}  {{ {A}}^2}{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   { {1}}&{ {2}}&{{ { - 3}}} \\   { {2}}&{{ { - 1}}}&{ {3}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   { {1}}&{ {2}}&{{ { - 3}}} \\   { {2}}&{{ { - 1}}}&{ {3}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {4}}&{ {2}}&{ {1}} \\   {{ { - 3}}}&{ {8}}&{{ { - 14}}} \\   { {7}}&{{ { - 3}}}&{{ {14}}}   \end{array}} \right] \hfill \\  {{ {A}}^3}{ {  =  }}{{ {A}}^2}{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {4}}&{ {2}}&{ {1}} \\   {{ { - 3}}}&{ {8}}&{{ { - 14}}} \\   { {7}}&{{ { - 3}}}&{{ {14}}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   { {1}}&{ {2}}&{{ { - 3}}} \\   { {2}}&{{ { - 1}}}&{ {3}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  { {8}}&{ {7}}&{ {1}} \\   {{ { - 23}}}&{{ {27}}}&{{ { - 69}}} \\   {{ {32}}}&{{ { - 13}}}&{{ {58}}}  \end{array}} \right] \hfill \\  \end{align} \]

LHS = \[{{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 5A + 11 I}}\;{ { = }}\;{ {0}}\]

\[\begin{align}  { { =  }}\left[ {\begin{array}{*{20}{c}}  { {8}}&{ {7}}&{ {1}} \\   {{ { - 23}}}&{{ {27}}}&{{ { - 69}}} \\   {{ {32}}}&{{ { - 13}}}&{{ {58}}}  \end{array}} \right]{ { - 6}}\left[ {\begin{array}{*{20}{c}}  { {4}}&{ {2}}&{ {1}} \\   {{ { - 3}}}&{ {8}}&{{ { - 14}}} \\   { {7}}&{{ { - 3}}}&{{ {14}}}   \end{array}} \right]{ { + 5}}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   { {1}}&{ {2}}&{{ { - 3}}} \\   { {2}}&{{ { - 1}}}&{ {3}}   \end{array}} \right]{ { + 11}}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   { {0}}&{ {1}}&{ {0}} \\   { {0}}&{ {0}}&{ {1}}   \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  { {8}}&{ {7}}&{ {1}} \\   {{ { - 23}}}&{{ {27}}}&{{ { - 69}}} \\   {{ {32}}}&{{ { - 13}}}&{{ {58}}}   \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}}  {{ {24}}}&{{ {12}}}&{ {6}} \\   {{ { - 18}}}&{{ {48}}}&{{ { - 84}}} \\   {{ {42}}}&{{ { - 18}}}&{{ {84}}}  \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}}  { {5}}&{ {5}}&{ {5}} \\   { {5}}&{{ {10}}}&{{ { - 15}}} \\   {{ {10}}}&{{ { - 5}}}&{{ {15}}}  \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}}  {{ {11}}}&{ {0}}&{ {0}} \\   { {0}}&{{ {11}}}&{ {0}} \\   { {0}}&{ {0}}&{{ {11}}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  {{ {8 - 24 + 5 + 11}}}&{{ {7 - 12 + 5 + 0}}}&{{ {1 - 6 + 5 + 0}}} \\   {{ { - 23 + 18 + 5 + 0}}}&{{ {27 - 48 + 10 + 11}}}&{{ { - 69 + 84 - 15 + 0}}} \\   {{ {32 - 42 + 10 + 0}}}&{{ { - 13 + 18 - 5 + 0}}}&{{ {58 - 84 + 15 + 11}}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{l}}  { {0}}&{ {0}}&{ {0}} \\   { {0}}&{ {0}}&{ {0}} \\   { {0}}&{ {0}}&{ {0}}  \end{array}} \right]{ {  =  0}} \hfill \\  \end{align} \]

= RHS

\[\begin{align}  {{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 5A + 11 I  =  0}} \hfill \\  {{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 5A  =   - 11 I}} \hfill \\  {{ {A}}^{ {2}}}{ {A}}{{ {A}}^{{ { - 1}}}}{ { - 6AA}}{{ {A}}^{{ { - 1}}}}{ { + 5A}}{{ {A}}^{{ { - 1}}}}{ {  =   - 11 I }}{{ {A}}^{{ { - 1}}}} \hfill \\  \end{align} \]

\[{ {11 }}{{ {A}}^{{ { - 1}}}}{ {  =   - }}{{ {A}}^{ {2}}}{ { + 6A - 5I  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 4}}}&{{ { - 2}}}&{{ { - 1}}} \\   { {3}}&{{ { - 8}}}&{{ {14}}} \\   {{ { - 7}}}&{ {3}}&{{ { - 14}}}  \end{array}} \right]{ { + 6}}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   { {1}}&{ {2}}&{{ { - 3}}} \\   { {2}}&{{ { - 1}}}&{ {3}}  \end{array}} \right]{ { - 5}}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   { {0}}&{ {1}}&{ {0}} \\   { {0}}&{ {0}}&{ {1}}  \end{array}} \right]\] \[\begin{align}  { { =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 4}}}&{{ { - 2}}}&{{ { - 1}}} \\   { {3}}&{{ { - 8}}}&{{ {14}}} \\   {{ { - 7}}}&{ {3}}&{{ { - 14}}}  \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}}  { {6}}&{ {6}}&{ {6}} \\   { {6}}&{{ {12}}}&{{ { - 18}}} \\   {{ {12}}}&{{ { - 6}}}&{{ {18}}}  \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}}  { {5}}&{ {0}}&{ {0}} \\   { {0}}&{ {5}}&{ {0}} \\   { {0}}&{ {0}}&{ {5}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 3}}}&{ {4}}&{ {5}} \\   { {9}}&{{ { - 1}}}&{{ { - 4}}} \\   { {5}}&{{ { - 3}}}&{{ { - 1}}}  \end{array}} \right] \hfill \\  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {11}}}}\left[ {\begin{array}{*{20}{c}}  {{ { - 3}}}&{ {4}}&{ {5}} \\   { {9}}&{{ { - 1}}}&{{ { - 4}}} \\   { {5}}&{{ { - 3}}}&{{ { - 1}}}  \end{array}} \right] \hfill \\  \end{align} \]

16. यदि \[{ {A}}\;{ { = }}\;\left[ \begin{align}  \;2\;\;\;\,\;\; - 1\;\,\;\;\;\;\;1 \hfill \\   - 1\;\;\;\;\;\,\;2\;\;\; - 1 \hfill \\  1\;\,\;\;\;\,\;\; - 1\;\;\;\;\;\;2 \hfill \\   \end{align}  \right]\] तो सत्यापित कीजिए की \[{{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 9A - 4 I}}\;{ { = }}\;{ {0}}\] है तथा इसकी सहायता से \[{{ {A}}^{{ { - 1}}}}\]  ज्ञात कीजिए। 

उत्तर: \[{ {A}}\;{ { = }}\;\left[ \begin{align}  \;2\;\;\;\,\;\; - 1\;\,\;\;\;\;\;1 \hfill \\   - 1\;\;\;\;\;\,\;2\;\;\; - 1 \hfill \\  1\;\,\;\;\;\,\;\; - 1\;\;\;\;\;\;2 \hfill \\  \end{align}  \right]\]

\[\begin{align}  {{ {A}}^2}{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}}&{ {1}} \\   {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\   { {1}}&{{ { - 1}}}&{ {2}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}}&{ {1}} \\   {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\   { {1}}&{{ { - 1}}}&{ {2}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {6}}&{{ { - 5}}}&{ {5}} \\   {{ { - 5}}}&{ {6}}&{{ { - 5}}} \\   { {5}}&{{ { - 5}}}&{ {6}}  \end{array}} \right] \hfill \\  {{ {A}}^3}{ {  =  }}{{ {A}}^2}{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {6}}&{{ { - 5}}}&{ {5}} \\   {{ { - 5}}}&{ {6}}&{{ { - 5}}} \\   { {5}}&{{ { - 5}}}&{ {6}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}}&{ {1}} \\   {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\   { {1}}&{{ { - 1}}}&{ {2}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ {22}}}&{{ { - 21}}}&{{ {21}}} \\   {{ { - 21}}}&{{ {22}}}&{{ { - 21}}} \\   {{ {21}}}&{{ { - 21}}}&{{ {22}}}   \end{array}} \right] \hfill \\   \end{align} \]

LHS = \[{{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 9A - 4 I}}\;{ { = }}\;{ {0}}\]

\[{ { =  }}\left[ {\begin{array}{*{20}{c}}  {{ {22}}}&{{ { - 21}}}&{{ {21}}} \\   {{ { - 21}}}&{{ {22}}}&{{ { - 21}}} \\   {{ {21}}}&{{ { - 21}}}&{{ {22}}}  \end{array}} \right]{ { - 6}}\left[ {\begin{array}{*{20}{c}}  { {6}}&{{ { - 5}}}&{ {5}} \\   {{ { - 5}}}&{ {6}}&{{ { - 5}}} \\   { {5}}&{{ { - 5}}}&{ {6}}   \end{array}} \right]{ { + 9}}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}}&{ {1}} \\   {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\   { {1}}&{{ { - 1}}}&{ {2}}   \end{array}} \right]{ { - 4}}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   { {0}}&{ {1}}&{ {0}} \\   { {0}}&{ {0}}&{ {1}}  \end{array}} \right]\]

\[{ { =  }}\left[ {\begin{array}{*{20}{c}}  {{ {22}}}&{{ { - 21}}}&{{ {21}}} \\   {{ { - 21}}}&{{ {22}}}&{{ { - 21}}} \\   {{ {21}}}&{{ { - 21}}}&{{ {22}}}  \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}}  {{ {36}}}&{{ { - 30}}}&{{ {30}}} \\   {{ { - 30}}}&{{ {36}}}&{{ { - 30}}} \\   {{ {30}}}&{{ { - 30}}}&{{ {36}}} \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}}  {{ {18}}}&{{ { - 9}}}&{ {9}} \\   {{ { - 9}}}&{{ {18}}}&{{ { - 9}}} \\   { {9}}&{{ { - 9}}}&{{ {18}}}  \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{l}}  { {4}}&{ {0}}&{ {0}} \\   { {0}}&{ {4}}&{ {0}} \\   { {0}}&{ {0}}&{ {4}}   \end{array}} \right]\]

\[\begin{align}  { { =  }}\left[ {\begin{array}{*{20}{l}}  {{ {22 - 36 + 18 - 4}}}&{{ { - 21 + 30 - 9 - 0}}}&{{ { - 21 - 30 + 9 - 0}}} \\   {{ { - 21 + 30 - 9 - 0}}}&{{ {22 - 36 + 18 - 4}}}&{{ { - 21 + 30 - 9 - 0}}} \\   {{ {21 - 30 + 9 - 0}}}&{{ { - 21 + 30 - 9 - 0}}}&{{ {22 - 36 + 18 - 4}}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{l}}  { {0}}&{ {0}}&{ {0}} \\   { {0}}&{ {0}}&{ {0}} \\   { {0}}&{ {0}}&{ {0}}  \end{array}} \right]{ {  =  0}} \hfill \\  \end{align} \]

= RHS

\[\begin{align}  {{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 9A - 4I  =  0}} \hfill \\  {{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 9A  =  4I}} \hfill \\  {{ {A}}^{ {2}}}{ {A}}{{ {A}}^{{ { - 1}}}}{ { - 6AA}}{{ {A}}^{{ { - 1}}}}{ { + 9A}}{{ {A}}^{{ { - 1}}}}{ {  =  4 I }}{{ {A}}^{{ { - 1}}}} \hfill \\  { {4}}{{ {A}}^{{ { - 1}}}}{ {  =  }}{{ {A}}^{ {2}}}{ { - 6A + 9I  =  }}\left[ {\begin{array}{*{20}{c}}  { {6}}&{{ { - 5}}}&{ {5}} \\   {{ { - 5}}}&{ {6}}&{{ { - 5}}} \\   { {5}}&{{ { - 5}}}&{ {6}}  \end{array}} \right]{ { - 6}}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}}&{ {1}} \\   {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\   { {1}}&{{ { - 1}}}&{ {2}}   \end{array}} \right]{ { + 9}}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   { {0}}&{ {1}}&{ {0}} \\   { {0}}&{ {0}}&{ {1}}  \end{array}} \right] \hfill \\   \end{align} \]

\[\begin{align}  { { =  }}\left[ {\begin{array}{*{20}{c}}  { {6}}&{{ { - 5}}}&{ {5}} \\   {{ { - 5}}}&{ {6}}&{{ { - 5}}} \\   { {5}}&{{ { - 5}}}&{ {6}}  \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}}  {{ {12}}}&{{ { - 6}}}&{ {6}} \\   {{ { - 6}}}&{{ {12}}}&{{ { - 6}}} \\   { {6}}&{{ { - 6}}}&{{ {12}}}  \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}}  { {9}}&{ {0}}&{ {0}} \\   { {0}}&{ {9}}&{ {0}} \\   { {0}}&{ {0}}&{ {9}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{ {1}}&{{ { - 1}}} \\   { {1}}&{ {3}}&{ {1}} \\   {{ { - 1}}}&{ {1}}&{ {3}}  \end{array}} \right] \hfill \\  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}}  { {3}}&{ {1}}&{{ { - 1}}} \\   { {1}}&{ {3}}&{ {1}} \\   {{ { - 1}}}&{ {1}}&{ {3}}  \end{array}} \right] \hfill \\ \end{align} \]

17. यदि \[{ {A,}}\;{ {3 \times 3}}\] कोटी का वर्ग आव्यूह है तो \[\left| {{ {adj}}\;{ {A}}} \right|\] का मान है। 

(a) \[\left| { {A}} \right|\]

(b) \[{\left| { {A}} \right|^2}\]

(c) \[{\left| { {A}} \right|^3}\]

(d) \[3\left| { {A}} \right|\]

उत्तर: (b) \[{\left| { {A}} \right|^2}\]

18. यदि \[{ {A}}\] कोटी दो का व्युत्क्रमिय आव्यूह है तो \[{ {det(}}{{ {A}}^{{ { - 1}}}}{ {)}}\] बराबर: 

(a) \[{ {det(A)}}\]

(b) \[\dfrac{1}{{{ {det(A)}}}}\]

(c) \[1\]

(d) \[0\]

उत्तर: (b) \[\dfrac{1}{{{ {det(A)}}}}\]

प्रश्नावली 4.6 

निम्नलिखित प्रश्नों 1 से 6 तक दी गई समीकरण निकायों का संगत अथवा असंगत के वर्गीकरण कीजिए: 

1. \[{ {x + 2y}}\;{ { = }}\;{ {2,}}\,\;{ {2x + 3y}}\;{ { = }}\;{ {3}}\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {2}} \\   { {2}}&{ {3}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}}  \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{l}}  { {2}} \\   { {3}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {2}} \\   { {2}}&{ {3}}  \end{array}} \right|{ {  =  1 \times 3 - 2 \times 2  =   - 1 }} \ne { { 0}} \hfill \\  \end{align} \]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। तो यह समीकरण निकाय संगत है। 

2. \[{ {2x - y}}\;{ { = }}\;5{ {,}}\,\;{ {x + y}}\;{ { = }}\;4\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}} \\   { {1}}&{ {1}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}}  \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{l}}  { {5}} \\   { {4}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}} \\   { {1}}&{ {1}}  \end{array}} \right|{ {  =  2 \times 1 - ( - 1) \times 1  =  3 }} \ne { { 0}} \hfill \\  \end{align} \]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। तो यह समीकरण निकाय संगत है। 

3. \[{ {x + 3y}}\;{ { = }}\;5{ {,}}\,\;2{ {x + 6y}}\;{ { = }}\;8\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {3}} \\   { {2}}&{ {6}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}}  \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{l}}  { {5}} \\   { {8}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {3}} \\   { {2}}&{ {6}}  \end{array}} \right|{ {  =  1 \times 6 - 3 \times 2  =  0}} \hfill \\  \end{align} \]

\[{ {AdjA  =  }}\left[ {\begin{array}{*{20}{c}}  { {6}}&{{ { - 2}}} \\   {{ { - 3}}}&{ {1}}  \end{array}} \right]\]

\[{ {(AdjA)B  =  }}\left[ {\begin{array}{*{20}{c}}  { {6}}&{{ { - 2}}} \\   {{ { - 3}}}&{ {1}}  \end{array}} \right]\left[ {\begin{array}{*{20}{l}}  { {5}} \\   { {8}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  {{ {14}}} \\   {{ { - 7}}}   \end{array}} \right]\; \ne \;0\]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व नहीं है। तो यह समीकरण निकाय संगत है। 

4. \[{ {x + y + z}}\;{ { = }}\;{ {1,}}\,\;{ {2x + 3y + 2z}}\;{ { = }}\;{ {2,}}\;{ {ax + ay + 2az}}\;{ { = }}\;{ {4}}\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   { {2}}&{ {3}}&{ {2}} \\   { {a}}&{ {a}}&{{ {2a}}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}} \\   { {z}}   \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{l}}  { {1}} \\   { {2}} \\   { {4}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   { {2}}&{ {3}}&{ {2}} \\   { {a}}&{ {a}}&{{ {2a}}}  \end{array}} \right|{ {  =  1(3 \times 2a - 2 \times a) - 1(2 \times 2a - 2 \times a) + (2 \times a - 3 \times a)}} \hfill \\  { { =  a }} \ne { { 0}} \hfill \\   \end{align} \]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। तो यह समीकरण निकाय संगत है। 

5. \[{ {3x - y - 2z}}\;{ { = }}\;{ {2,}}\,\;{ {2y - z}}\;{ { = }}\;{ { - 1,}}\;{ {3x - 5y}}\;{ { = }}\;{ {3}}\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{{ { - 1}}}&{{ { - 2}}} \\   { {0}}&{ {2}}&{{ { - 1}}} \\   { {3}}&{{ { - 5}}}&{ {0}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}} \\   { {z}}   \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}} \\  {{ { - 1}}} \\   { {3}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{c}}  { {3}}&{{ { - 1}}}&{{ { - 2}}} \\   { {0}}&{ {2}}&{{ { - 1}}} \\   { {3}}&{{ { - 5}}}&{ {0}}  \end{array}} \right|{ {  =   - 15 + 3 + 12  =  0}} \hfill \\  { {Adj A  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 5}}}&{{ {10}}}&{ {5}} \\  {{ { - 3}}}&{ {6}}&{ {3}} \\   {{ { - 6}}}&{{ {12}}}&{ {6}}   \end{array}} \right] \hfill \\  { {(AdjA)B  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 5}}}&{{ {10}}}&{ {5}} \\   {{ { - 3}}}&{ {6}}&{ {3}} \\   {{ { - 6}}}&{{ {12}}}&{ {6}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {2}} \\   {{ { - 1}}} \\   { {3}}   \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 10 - 10 + 15}}} \\   {{ { - 6 - 6 + 9}}} \\   {{ { - 12 - 12 + 18}}}  \end{array}} \right] \hfill \\  \end{align} \]

\[{ { =  }}\left[ {\begin{array}{*{20}{l}}  {{ { - 5}}} \\   {{ { - 3}}} \\   {{ { - 6}}}  \end{array}} \right]{ { }} \ne { { 0}}\]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व नहीं है। तो यह समीकरण निकाय असंगत है। 

6. \[{ {5x - y + 4z}}\;{ { = }}\;{ {5,}}\,\;{ {2x + 3y - 5z}}\;{ { = }}\;{ {2,}}\;{ {5x - 2y + 6z}}\;{ { = }}\;{ { - 1}}\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {5}}&{{ { - 1}}}&{ {4}} \\   { {2}}&{ {3}}&{ {5}} \\   { {5}}&{{ { - 2}}}&{ {6}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}} \\   { {z}}   \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{c}}  { {5}} \\   { {2}} \\   {{ { - 1}}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{c}}  { {5}}&{{ { - 1}}}&{ {4}} \\   { {2}}&{ {3}}&{ {5}} \\   { {5}}&{{ { - 2}}}&{ {6}}  \end{array}} \right|{ {  =  5(18 + 3) - 1(12 - 25) + 4( - 4 - 15)}} \hfill \\  { { =  5 \times 28 - 13 - 4 \times 5  =  67 }} \ne { { 0}} \hfill \\  \end{align} \]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। तो यह समीकरण निकाय संगत है।

निम्नलिखित प्रश्न 7 से 17 तक प्रत्येक समीकरण निकाय को आव्यूह विधि से हल कीजिए: 

7. \[{ {5x + 2y}}\;{ { = }}\;4{ {,}}\,\;7{ {x + 3y}}\;{ { = }}\;5\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है: 

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {5}}&{ {2}} \\   { {7}}&{ {3}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}}   \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{l}}  { {4}} \\   { {5}}   \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{l}}  { {5}}&{ {2}} \\   { {7}}&{ {3}}   \end{array}} \right|{ {  =  5 \times 3 - 7 \times 2  =  15 - 14  =  1 }} \ne { { 0}} \hfill \\  \end{align} \]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। 

\[\begin{align}  { {Adj A  =  }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{{ { - 2}}} \\   {{ { - 7}}}&{ {5}}   \end{array}} \right] \hfill \\  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{{ { - 2}}} \\   {{ { - 7}}}&{ {5}}   \end{array}} \right] \hfill \\   \end{align} \]

\[{ {X  =  }}{{ {A}}^{{ { - 1}}}}{ {B  =  }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{{ { - 2}}} \\   {{ { - 7}}}&{ {5}}   \end{array}} \right]\left[ {\begin{array}{*{20}{l}}  { {4}} \\   { {5}}   \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ {(12 - 10)}}} \\   {{ {( - 28 + 25)}}}   \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}} \\   {{ { - 3}}}   \end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}}   \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}} \\   {{ { - 3}}}   \end{array}} \right]\]

इसलिए इस समीकरण निकाय का हल है: 

\[{ {x  =  2, y  =   - 3}}\]

8. \[{ {2x - y}}\;{ { = }}\; - 2{ {,}}\,\;3{ {x + 4y}}\;{ { = }}\;3\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है: 

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}} \\   { {3}}&{ {4}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}}   \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 2}}} \\   { {3}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{c}}  { {2}}&{{ { - 1}}} \\   { {3}}&{ {4}}  \end{array}} \right|{ {  =  2 \times 4 + 1 \times 3  =  8 + 3  =  11 }} \ne { { 0}} \hfill \\  \end{align} \]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। 

\[\begin{align}  { {Adj A  =  }}\left[ {\begin{array}{*{20}{c}}  { {4}}&{ {1}} \\   {{ { - 3}}}&{ {2}}  \end{array}} \right] \hfill \\  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {11}}}}\left[ {\begin{array}{*{20}{c}}  { {4}}&{ {1}} \\   {{ { - 3}}}&{ {2}}   \end{array}} \right] \hfill \\  \end{align} \]

\[\begin{align}  { {X  =  }}{{ {A}}^{{ { - 1}}}}{ {B  =  }}\dfrac{{ {1}}}{{{ {11}}}} \hfill \\  \left[ {\begin{array}{*{20}{c}}  { {4}}&{ {1}} \\   {{ { - 3}}}&{ {2}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  {{ { - 2}}} \\   { {3}}   \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ {(4 \times  - 2 + 1 \times 3)}}} \\   {{ {( - 3 \times  - 2 + 2 \times 3)}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {11}}}}\left[ {\begin{array}{*{20}{c}}  {{ { - 5}}} \\   {{ {12}}}  \end{array}} \right] \hfill \\  \left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {\dfrac{{{ { - 5}}}}{{{ {11}}}}} \\   {\dfrac{{{ {12}}}}{{{ {11}}}}}  \end{array}} \right] \hfill \\  \end{align} \]

इसलिए इस समीकरण निकाय का हल है: 

\[{ {x  =   - }}\dfrac{{ {5}}}{{{ {11}}}}{ {, y  =  }}\dfrac{{{ {12}}}}{{{ {11}}}}\]

9. \[{ {4x - 3y}}\;{ { = }}\;3{ {,}}\,\;3{ {x - 5y}}\;{ { = }}\;7\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है: 

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {4}}&{{ { - 3}}} \\   { {3}}&{{ { - 5}}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}}  \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{l}}  { {3}} \\   { {7}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{l}}  { {4}}&{{ { - 3}}} \\   { {3}}&{{ { - 5}}}  \end{array}} \right|{ {  =  4 \times  - 5 + 3 \times 3  =   - 20 + 9  =   - 11 }} \ne { { 0}} \hfill \\  \end{align} \]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। 

\[\begin{align}  { {Adj A  =  }}\left[ {\begin{array}{*{20}{l}}  {{ { - 5}}}&{ {3}} \\   {{ { - 3}}}&{ {4}}  \end{array}} \right] \hfill \\  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ {  =   - }}\dfrac{{ {1}}}{{{ {11}}}}\left[ {\begin{array}{*{20}{l}}  {{ { - 5}}}&{ {3}} \\   {{ { - 3}}}&{ {4}}  \end{array}} \right] \hfill \\  { {X  =  }}{{ {A}}^{{ { - 1}}}}{ {B  =  }}\dfrac{{ {1}}}{{{ {11}}}} \hfill \\  \left[ {\begin{array}{*{20}{l}}  {{ { - 5}}}&{ {3}} \\   {{ { - 3}}}&{ {4}}  \end{array}} \right]\left[ {\begin{array}{*{20}{l}}  { {3}} \\   { {7}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  {{ {( - 5 \times 3 + 3 \times 7)}}} \\   {{ {( - 3 \times 3 + 4 \times 7)}}}  \end{array}} \right]{ {  =   - }}\dfrac{{ {1}}}{{{ {11}}}}\left[ {\begin{array}{*{20}{c}}  { {6}} \\   {{ {19}}}  \end{array}} \right] \hfill \\  \left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  {{ { - }}\dfrac{{ {6}}}{{{ {11}}}}} \\   {{ { - }}\dfrac{{{ {19}}}}{{{ {11}}}}}  \end{array}} \right] \hfill \\  \end{align} \]

इसलिए इस समीकरण निकाय का हल है: 

\[{ {x  =  }}\dfrac{{ {6}}}{{{ {11}}}}{ {, y  =   - }}\dfrac{{{ {19}}}}{{{ {11}}}}\]

10. \[{ {5x + 2y}}\;{ { = }}\;3{ {,}}\,\;3{ {x + 2y}}\;{ { = }}\;5\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है: 

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {5}}&{ {2}} \\   { {3}}&{ {2}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}}  \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{l}}  { {3}} \\   { {7}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{l}}  { {5}}&{ {2}} \\   { {3}}&{ {2}}  \end{array}} \right|{ {  =  5 \times 2 - 3 \times 2}} \hfill \\  \end{align} \]

\[{ { =  10 - 6  =  4 }} \ne { { 0}}\]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। 

\[{ {Adj A  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 2}}} \\   {{ { - 3}}}&{ {5}}  \end{array}} \right]\]

\[{{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ {  =  }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 2}}} \\   {{ { - 3}}}&{ {5}}  \end{array}} \right]\]

\[\begin{align}  { {X  =  }}{{ {A}}^{{ { - 1}}}}{ {B  =  }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 2}}} \\   {{ { - 3}}}&{ {5}}  \end{array}} \right]\left[ {\begin{array}{*{20}{l}}  { {3}} \\   { {5}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ {(2 \times 3 - 2 \times 5)}}} \\   {{ {( - 3 \times 3 + 5 \times 5)}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}}  {{ { - 4}}} \\   {{ {16}}}  \end{array}} \right] \hfill \\  \left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - 1}}} \\   {{ {16}}}  \end{array}} \right] \hfill \\  \end{align} \]

इसलिए इस समीकरण निकाय का हल है: 

\[{ {x  =   - 1, y  =  4}}\]

11. \[{ {2x + y + z}}\;{ { = }}\;{ {1,}}\,\;{ {x - 2y - z}}\;{ { = }}\;\dfrac{{ {3}}}{{ {2}}}{ {,}}\;{ {3y - 5z}}\;{ { = }}\;{ {9}}\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है: 

\[\begin{align}  { {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{ {1}}&{ {1}} \\   { {1}}&{{ { - 2}}}&{{ { - 1}}} \\   { {0}}&{ {3}}&{{ { - 5}}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}} \\   { {z}}  \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{l}}  { {1}} \\   {\dfrac{{ {3}}}{{ {2}}}} \\   { {9}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{c}}  { {2}}&{ {1}}&{ {1}} \\   { {1}}&{{ { - 2}}}&{{ { - 1}}} \\   { {0}}&{ {3}}&{{ { - 5}}}  \end{array}} \right|{ {  =  2(10 + 3) - 1( - 5 - 3) + 0( - 1 + 3)}} \hfill \\  { { =  26 + 8 + 0  =  34 }} \ne { { 0}} \hfill \\  \end{align} \]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। 

\[\begin{align}  { {AdjA  =  }}\left[ {\begin{array}{*{20}{c}}  {{ {13}}}&{ {8}}&{ {1}} \\   { {5}}&{{ { - 10}}}&{ {3}} \\   { {3}}&{{ { - 6}}}&{{ { - 5}}}  \end{array}} \right] \hfill \\  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {34}}}}\left[ {\begin{array}{*{20}{c}}  {{ {13}}}&{ {8}}&{ {1}} \\   { {5}}&{{ { - 10}}}&{ {3}} \\   { {3}}&{{ { - 6}}}&{{ { - 5}}}  \end{array}} \right] \hfill \\  \end{align} \]

\[{ {X  =  }}{{ {A}}^{{ { - 1}}}}{ {B}}\]

\[{ { =  }}\dfrac{{ {1}}}{{{ {34}}}}\left[ {\begin{array}{*{20}{c}}  {{ {13}}}&{ {8}}&{ {1}} \\   { {5}}&{{ { - 10}}}&{ {3}} \\   { {3}}&{{ { - 6}}}&{{ { - 5}}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {1}} \\   {\dfrac{{ {3}}}{{ {2}}}} \\   { {9}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {34}}}}\left[ {\begin{array}{*{20}{c}}  {\left( {{ {13 \times 1 + 8 \times }}\dfrac{{ {3}}}{{ {2}}}{ { + 1 \times 9}}} \right)} \\   {\left( {{ {5 \times 1 - 10 \times }}\dfrac{{ {3}}}{{ {2}}}{ { + 3 \times 9}}} \right)} \\   {\left( {{ {3 \times 1 - 6 \times }}\dfrac{{ {3}}}{{ {2}}}{ { - 5 \times 9}}} \right)}  \end{array}} \right]\]

\[{ { =  }}\dfrac{{ {1}}}{{{ {34}}}}\left[ {\begin{array}{*{20}{c}}  {{ {34}}} \\   {{ {17}}} \\   {{ { - 51}}}  \end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}} \\   { {z}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}} \\   {\dfrac{{ {1}}}{{ {2}}}} \\   {{ { - }}\dfrac{{ {3}}}{{ {2}}}}  \end{array}} \right]\]

इसलिए इस समीकरण निकाय का हल है: 

\[{ {x  =  1, y  =  }}\dfrac{{ {1}}}{{ {2}}}{ {, Z  =   - }}\dfrac{{ {3}}}{{ {2}}}\]

12. \[{ {x - y + z}}\;{ { = }}\;{ {4,}}\,\;{ {2x + y - 3z}}\;{ { = }}\;{ {0,}}\;{ {x + y + z}}\;{ { = }}\;{ {2}}\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है: 

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{{ { - 1}}}&{ {1}} \\   { {2}}&{ {1}}&{{ { - 3}}} \\   { {1}}&{ {1}}&{ {1}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}} \\   { {z}}  \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{l}}  { {4}} \\   { {0}} \\   { {2}}   \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{c}}  { {1}}&{{ { - 1}}}&{ {1}} \\   { {2}}&{ {1}}&{{ { - 3}}} \\   {{ {11}}}&{ {1}}&{ {1}}  \end{array}} \right|{ {  =  1(1 + 3) + 1(2 + 3) + 1(2 - 1)}} \hfill \\  { { =  4 + 5 + 1  =  10 }} \ne { { 0}} \hfill \\  \end{align} \]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। 

\[\begin{align}  { {Adj A  =  }}\left[ {\begin{array}{*{20}{c}}  { {4}}&{ {2}}&{ {2}} \\   {{ { - 5}}}&{ {0}}&{ {5}} \\   { {1}}&{{ { - 2}}}&{ {3}}  \end{array}} \right] \hfill \\  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {10}}}}\left[ {\begin{array}{*{20}{c}}  { {4}}&{ {2}}&{ {2}} \\   {{ { - 5}}}&{ {0}}&{ {5}} \\   { {1}}&{{ { - 2}}}&{ {3}} \end{array}} \right] \hfill \\  { {X  =  }}{{ {A}}^{{ { - 1}}}}{ {B}} \hfill \\  { { =  }}\dfrac{{ {1}}}{{{ {10}}}}\left[ {\begin{array}{*{20}{c}}  { {4}}&{ {2}}&{ {2}} \\   {{ { - 5}}}&{ {0}}&{ {5}} \\   { {1}}&{{ { - 2}}}&{ {3}}  \end{array}} \right]\left[ {\begin{array}{*{20}{l}}  { {4}} \\   { {0}} \\   { {2}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {10}}}}\left[ {\begin{array}{*{20}{c}}  {{ {(4 \times 4 + 2 \times 0 + 2 \times 2)}}} \\   {{ {( - 5 \times 4 + 0 \times 0 + 5 \times 2)}}} \\   {{ {(1 \times 4 - 2 \times 0 - 3 \times 2)}}}  \end{array}} \right] \hfill \\  { { =  }}\dfrac{{ {1}}}{{{ {10}}}}\left[ {\begin{array}{*{20}{c}}  {{ {16 + 0 + 4}}} \\   {{ { - 20 + 0 + 10}}} \\   {{ {4 + 0 + 6}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {10}}}}\left[ {\begin{array}{*{20}{c}}  {{ {20}}} \\   {{ { - 10}}} \\   {{ {10}}}  \end{array}} \right] \hfill \\  \end{align} \]

\[\left[ {\begin{array}{*{20}{c}}  { {x}} \\   { {y}} \\   { {z}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}} \\   {{ { - 1}}} \\   { {1}}  \end{array}} \right]\]

इसलिए इस समीकरण निकाय का हल है: 

\[{ {x  =  2, y  =   - 1, Z  =  1}}\]

13. \[{ {2x + 3y + 3z}}\;{ { = }}\;5{ {,}}\,\;{ {x - 2y + z}}\;{ { = }}\; - 4{ {,}}\;3{ {x - y - 2z}}\;{ { = }}\;3\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है: 

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{ {3}}&{ {3}} \\   { {1}}&{{ { - 2}}}&{ {1}} \\   { {3}}&{{ { - 1}}}&{{ { - 2}}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}} \\   { {z}}  \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{c}}  { {5}} \\   {{ { - 4}}} \\   { {3}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{c}}  { {2}}&{ {3}}&{ {3}} \\   { {1}}&{{ { - 2}}}&{ {1}} \\   { {3}}&{{ { - 1}}}&{{ { - 2}}}  \end{array}} \right|{ {  =  2(4 + 1) - 3( - 2 - 3) + 3( - 1 + 6)  =  10 + 15 + 15  =  40 }} \ne { { 0}} \hfill \\  \end{align} \]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। 

\[\begin{align}  { {AdjA  =  }}\left[ {\begin{array}{*{20}{c}}  { {5}}&{ {3}}&{ {9}} \\   { {5}}&{{ { - 13}}}&{ {1}} \\   { {5}}&{{ {11}}}&{{ { - 7}}}  \end{array}} \right] \hfill \\  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {40}}}}\left[ {\begin{array}{*{20}{c}}  { {5}}&{ {3}}&{ {9}} \\   { {5}}&{{ { - 13}}}&{ {1}} \\   { {5}}&{{ {11}}}&{{ { - 7}}}  \end{array}} \right] \hfill \\  { {X  =  }}{{ {A}}^{{ { - 1}}}}{ {B  =  }}\dfrac{{ {1}}}{{{ {40}}}}\left[ {\begin{array}{*{20}{c}}  { {5}}&{ {3}}&{ {9}} \\   { {5}}&{{ { - 13}}}&{ {1}} \\   { {5}}&{{ {11}}}&{{ { - 7}}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {5}} \\   {{ { - 4}}} \\   { {3}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {40}}}}\left[ {\begin{array}{*{20}{c}}  {{ {(5 \times 5 + 3 \times  - 4 + 9 \times 3)}}} \\   {{ {(5 \times 5 - 13 \times  - 4 + 1 \times 3)}}} \\   {{ {(5 \times 5 + 11 \times  - 4 - 7 \times 3)}}}  \end{array}} \right] \hfill \\  \end{align} \]

\[\begin{align}  { { =  }}\dfrac{{ {1}}}{{{ {40}}}}\left[ {\begin{array}{*{20}{c}}  {{ {25 - 12 + 27}}} \\   {{ {25 + 52 + 3}}} \\   {{ {25 - 44 - 21}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {40}}}}\left[ {\begin{array}{*{20}{c}}  {{ {40}}} \\   {{ {80}}} \\   {{ { - 40}}}  \end{array}} \right] \hfill \\  \left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}} \\   { {z}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}} \\   { {2}} \\   {{ { - 1}}}  \end{array}} \right] \hfill \\  \end{align} \]

इसलिए इस समीकरण निकाय का हल है: 

\[{ {x  =  1, y  =  2, Z  =   - 1}}\]

14. \[{ {x - y + 2z}}\;{ { = }}\;{ {7,}}\,\;{ {3x + 4y - 5z}}\;{ { = }}\;{ { - 5,}}\;{ {2x - y + 3z}}\;{ { = }}\;{ {12}}\]

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है: 

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{{ { - 1}}}&{ {2}} \\   { {3}}&{ {4}}&{{ { - 5}}} \\   { {2}}&{{ { - 1}}}&{{ { - 3}}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}} \\   { {z}}  \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{c}}  { {7}} \\   {{ { - 5}}} \\   {{ {12}}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{c}}  { {1}}&{{ { - 1}}}&{ {2}} \\   { {3}}&{ {4}}&{{ { - 5}}} \\   { {2}}&{{ { - 1}}}&{{ { - 3}}}  \end{array}} \right| \hfill \\  { { =  1(12 - 5) + 1(9 + 10) + 2( - 3 - 8)}} \hfill \\  { { =  7 + 19 - 22  =  4 }} \ne { { 0}} \hfill \\  \end{align} \]

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। 

\[\begin{align}  { {Adj A  =  }}\left[ {\begin{array}{*{20}{c}}  { {7}}&{ {1}}&{{ { - 3}}} \\   {{ { - 19}}}&{{ { - 1}}}&{{ {11}}} \\   {{ { - 11}}}&{{ { - 1}}}&{ {7}}  \end{array}} \right] \hfill \\  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ {  =  }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}}  { {7}}&{ {1}}&{{ { - 3}}} \\   {{ { - 19}}}&{{ { - 1}}}&{{ {11}}} \\   {{ { - 11}}}&{{ { - 1}}}&{ {7}}  \end{array}} \right] \hfill \\  { {X  =  }}{{ {A}}^{{ { - 1}}}}{ {B}} \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}}  { {7}}&{ {1}}&{{ { - 3}}} \\   {{ { - 19}}}&{{ { - 1}}}&{{ {11}}} \\   {{ { - 11}}}&{{ { - 1}}}&{ {7}}  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}  { {7}} \\   {{ { - 5}}} \\   {{ {12}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}}  {{ {(7 \times 7 + 1 \times  - 5 - 3 \times 12)}}} \\   {{ {( - 19 \times 7 - 1 \times  - 5 + 11 \times 12)}}} \\   {{ {( - 11 \times 7 - 1 \times  - 5 + 7 \times 12)}}}  \end{array}} \right] \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}}  {{ {49 - 5 - 36}}} \\   {{ { - 133 + 5 + 132}}} \\   {{ { - 77 + 5 + 84}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}}  { {8}} \\   { {4}} \\   {{ {12}}}  \end{array}} \right] \hfill \\  \left[ {\begin{array}{*{20}{c}}  { {x}} \\   { {y}} \\   { {z}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  { {2}} \\   { {1}} \\   { {3}}  \end{array}} \right] \hfill \\  \end{align} \]

इसलिए इस समीकरण निकाय का हल है: 

\[{ {x  =  2, y  =  1, Z  =  3}}\]

15. यदि \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 3}}}&{ {5}} \\   { {3}}&{ {2}}&{{ { - 4}}} \\   { {1}}&{ {1}}&{{ { - 2}}}  \end{array}} \right]\] है तो \[{{ {A}}^{{ { - 1}}}}\] ज्ञात कीजिए, \[{{ {A}}^{{ { - 1}}}}\] का प्रयोग करके निम्नलिखित समीकरण निकाय को हल कीजिए:

\[{ {2x - 3y + 5z  =  11 , 3x + 2y - 4z  =   - 5 , x + y - 2z  =   - 3}}\]

उत्तर: \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{{ { - 3}}}&{ {5}} \\   { {3}}&{ {2}}&{{ { - 4}}} \\   { {1}}&{ {1}}&{{ { - 2}}}  \end{array}} \right]\]

\[\begin{align}  { {|A|  =  }}\left| {\begin{array}{*{20}{c}}  { {2}}&{{ { - 3}}}&{ {5}} \\   { {3}}&{ {2}}&{{ { - 4}}} \\   { {1}}&{ {1}}&{{ { - 2}}}  \end{array}} \right|{ {  =  2( - 4 + 4) + 3( - 6 + 4) + 5(3 - 2)}} \hfill \\  { { =  0 - 6 + 5  =   - 1}} \hfill \\  { {AdjA  =  }}\left[ {\begin{array}{*{20}{c}}  { {0}}&{{ { - 1}}}&{ {2}} \\   { {2}}&{{ { - 9}}}&{{ {23}}} \\   { {1}}&{{ { - 5}}}&{{ {13}}}  \end{array}} \right] \hfill \\  {{ {A}}^{{ { - 1}}}}{ {  =   - }}\left[ {\begin{array}{*{20}{l}}  { {0}}&{{ { - 1}}}&{ {2}} \\   { {2}}&{{ { - 9}}}&{{ {23}}} \\   { {1}}&{{ { - 5}}}&{{ {13}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{r}}  {{ { - 0}}}&{ {1}}&{{ { - 2}}} \\   {{ { - 2}}}&{ {9}}&{{ { - 23}}} \\   {{ { - 1}}}&{ {5}}&{{ { - 13}}}  \end{array}} \right] \hfill \\  \because \;{ {X  =  }}{{ {A}}^{{ { - 1}}}}{ {B}} \hfill \\  \therefore \;{ {B  =  }}\left[ {\begin{array}{*{20}{c}}  {{ {11}}} \\   {{ { - 5}}} \\   {{ { - 3}}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}} \\   { {z}}  \end{array}} \right] \hfill \\  { {2x - 3y + 5z  =  11, 3x + 2y - 4z  =   - 5, x + y - 2z  =   - 3}} \hfill \\  \end{align} \]

\[\begin{align}  { {X  =  }}\left[ {\begin{array}{*{20}{l}}  {{ { - 0}}}&{ {1}}&{{ { - 2}}} \\   {{ { - 2}}}&{ {9}}&{{ { - 23}}} \\   {{ { - 1}}}&{ {5}}&{{ { - 13}}}  \end{array}} \right]\left[ {\begin{array}{*{20}{l}}  {{ {11}}} \\   {{ { - 5}}} \\   {{ { - 3}}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{l}}  {{ {( - 0 \times 11 - 1 \times 5 + 2 \times 3)}}} \\   {{ {( - 2 \times 11 - 9 \times 5 + 23 \times 3)}}} \\   {{ {( - 1 \times 11 - 5 \times 5 + 13 \times 3)}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ {0 - 5 + 6}}} \\   {{ { - 22 - 45 + 69}}} \\   {{ { - 11 - 25 + 39}}}  \end{array}} \right] \hfill \\  \end{align} \]

\[{ { =  }}\left[ {\begin{array}{*{20}{l}}  { {1}} \\   { {2}} \\   { {3}}  \end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{l}}  { {y}} \\   { {z}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  { {1}} \\   { {2}} \\   { {3}}  \end{array}} \right]\]

इसलिए, इस समीकरण निकाय का हल है 

\[{ {x  =  1, y  =  2, Z  =  3}}\]

16. \[{ {4\;kg}}\] प्याज, \[{ {3\;kg}}\] गेहू और \[{ {2\;kg}}\] चावल का मूल्य \[{ {Rs}}{ {. 60}}\] है। \[{ {2\;kg}}\] प्याज, \[{ {4\;kg}}\] गेहू और \[{ {6\;kg}}\] चावल का मूल्य \[{ {Rs}}{ {. 90}}\] है। \[{ {6\;kg}}\] प्याज, \[{ {2\;kg}}\] गेहू और \[{ {3 kg}}\] चावल का मूल्य \[{ {Rs}}{ {. 70}}\] है। आव्यूह विधि द्वारा प्रत्येक का मूल्य प्रति \[{ {kg}}\] ज्ञात कीजिए। 

उत्तर: मान लेते है की प्याज का मूल्य प्रति \[{ {kg Rs}}{ {. x}}\] , गेहू का मूल्य प्रति \[{ {kg Rs}}{ {. y}}\] , तथा चवक का मूल्य प्रति \[{ {kg Rs}}{ {. z}}\] है। 

इसलिए हम फी गई जानकारी को समीकरणों के रूप मे कुछ इस प्रकार रख सकते है: 

\[{ {4x + 3y + 2z  =  60}}\]

\[\begin{align}  { {2x + 4y + 6z  =  90}} \hfill \\  { {6x + 2y + 3y  =  70}} \hfill \\  \end{align} \]

दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है। 

\[\begin{align}  { {AX  =  B}} \hfill \\  { {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {4}}&{ {3}}&{ {2}} \\   { {2}}&{ {4}}&{ {6}} \\   { {6}}&{ {2}}&{ {3}}  \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}} \\   { {z}}  \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{l}}  {{ {60}}} \\   {{ {90}}} \\   {{ {70}}}  \end{array}} \right] \hfill \\  { {|A|  =  }}\left| {\begin{array}{*{20}{l}}  { {4}}&{ {3}}&{ {2}} \\   { {2}}&{ {4}}&{ {6}} \\   { {6}}&{ {2}}&{ {3}}  \end{array}} \right|{ {  =  4(12 - 12) - 3(6 - 36) + 2(4 - 24)  =  0 + 90 - 40  =  50 }} \ne { { 0}} \hfill \\  { {AdjA  =  }}\left[ {\begin{array}{*{20}{c}}  { {0}}&{{ { - 5}}}&{{ {10}}} \\   {{ {30}}}&{ {0}}&{{ { - 20}}} \\   {{ { - 20}}}&{{ {10}}}&{{ {10}}}  \end{array}} \right] \hfill \\  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {50}}}}\left[ {\begin{array}{*{20}{c}}  { {0}}&{{ { - 5}}}&{{ {10}}} \\   {{ {30}}}&{ {0}}&{{ { - 20}}} \\   {{ { - 20}}}&{{ {10}}}&{{ {10}}}  \end{array}} \right] \hfill \\  { {X  =  }}{{ {A}}^{{ { - 1}}}}{ {B  =  }}\dfrac{{ {1}}}{{{ {50}}}}\left[ {\begin{array}{*{20}{c}}  { {0}}&{{ { - 5}}}&{{ {10}}} \\   {{ {30}}}&{ {0}}&{{ { - 20}}} \\   {{ { - 20}}}&{{ {10}}}&{{ {10}}}  \end{array}} \right]\left[ {\begin{array}{*{20}{l}}  {{ {60}}} \\   {{ {90}}} \\   {{ {70}}}  \end{array}} \right] \hfill \\  \end{align} \]

\[{ { =  }}\dfrac{{ {1}}}{{{ {50}}}}\left[ {\begin{array}{*{20}{c}}  {{ {(0 \times 60 - 5 \times 90 + 10 \times 70)}}} \\   {{ {(30 \times 60 + 0 \times 90 - 20 \times 70)}}} \\   {{ {( - 20 \times 60 + 10 \times 90 + 10 \times 70)}}}   \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {50}}}}\left[ {\begin{array}{*{20}{c}}  {{ {0 - 450 + 700}}} \\   {{ {1800 + 0 - 1400}}} \\   {{ { - 1200 + 900 + 700}}}  \end{array}} \right]\]

\[\begin{align}  { { =  }}\dfrac{{ {1}}}{{{ {50}}}}\left[ {\begin{array}{*{20}{l}}  {{ {250}}} \\   {{ {400}}} \\   {{ {400}}}  \end{array}} \right] \hfill \\  \left[ {\begin{array}{*{20}{l}}  { {x}} \\   { {y}} \\   { {z}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  { {5}} \\   { {8}} \\   { {8}}  \end{array}} \right] \hfill \\  \end{align} \]

इसलिए इस समीकरण निकाय का हल है: 

\[{ {x  =  5, y  =  8, Z  =  8}}\]

प्याज का मूल्य प्रति \[{ {kg Rs}}{ {. 5}}\] , गेहू का मूल्य प्रति \[{ {kg Rs}}{ {. 8}}\] , तथा चावल का मूल्य प्रति \[{ {kg Rs}}{ {. 8}}\] है। 

प्रश्नावली A4

1. सिद्ध कीजिए की सारणिक \[\left| {\begin{array}{*{20}{c}}  { {x}}&{{ {sin\theta }}}&{{ {cos\theta }}} \\   {{ { - sin\theta }}}&{{ { - x}}}&{ {1}} \\   {{ {cos\theta }}}&{ {1}}&{ {x}}  \end{array}} \right|{ { \theta }}\] से स्वतंत्र है। 

उत्तर: \[{ {\Delta   =  }}\left| {\begin{array}{*{20}{c}}  { {x}}&{{ {sin\theta }}}&{{ {cos\theta }}} \\   {{ { - sin\theta }}}&{{ { - x}}}&{ {1}} \\   {{ {cos\theta }}}&{ {1}}&{ {x}}  \end{array}} \right|\]

\[\begin{align}  { { =  x - }}\left( {{{ {x}}^{ {2}}}{ { - 1}}} \right){ { - sin\theta ( - xsin\theta  - cos\theta ) + cos\theta ( - sin\theta  + xcos\theta )}} \hfill \\  { { =   - }}{{ {x}}^{ {3}}}{ { - x + xsi}}{{ {n}}^{ {2}}}{ {\theta  + sin\theta cos\theta  - cos\theta sin\theta  + xco}}{{ {s}}^{ {2}}}{ {\theta }} \hfill \\  { { =   - }}{{ {x}}^{ {3}}}{ { - x + x}}\left( {{ {si}}{{ {n}}^{ {2}}}{ {\theta  + co}}{{ {s}}^{ {2}}}{ {\theta }}} \right) \hfill \\  { { =   - }}{{ {x}}^{ {3}}}{ { - x + x}} \hfill \\  { { =   - }}{{ {x}}^{ {2}}} \hfill \\  \end{align} \]

जो \[{ {\theta }}\] से स्वतंत्र है। 

2. सारणिक का प्रसरण किए बिना सिद्ध कीजिए की \[\left| {\begin{array}{*{20}{l}}  { {a}}&{{{ {a}}^{ {2}}}}&{{ {bc}}} \\   { {b}}&{{{ {b}}^{ {2}}}}&{{ {bc}}} \\   { {c}}&{{{ {c}}^{ {2}}}}&{{ {ab}}}  \end{array}} \right|{ {  =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{{{ {a}}^{ {2}}}}&{{{ {a}}^{ {3}}}} \\   { {1}}&{{{ {b}}^{ {2}}}}&{{{ {b}}^{ {3}}}} \\   { {1}}&{{{ {c}}^{ {2}}}}&{{{ {c}}^{ {3}}}}  \end{array}} \right|\]

उत्तर: \[\left| {\begin{array}{*{20}{l}}  { {a}}&{{{ {a}}^{ {2}}}}&{{ {bc}}} \\   { {b}}&{{{ {b}}^{ {2}}}}&{{ {ca}}} \\   { {c}}&{{{ {c}}^{ {2}}}}&{{ {ab}}} \end{array}} \right|{ {  =  }}\dfrac{{ {1}}}{{{ {abc}}}}\left| {\begin{array}{*{20}{l}}  {{{ {a}}^{ {2}}}}&{{{ {a}}^{ {3}}}}&{{ {abc}}} \\   {{{ {b}}^{ {2}}}}&{{{ {b}}^{ {3}}}}&{{ {abc}}} \\   {{{ {c}}^{ {2}}}}&{{{ {c}}^{ {3}}}}&{{ {abc}}}  \end{array}} \right|\quad \;\;\;\;\;{ {[}}{{ {R}}_1} \to { {a}}{{ {R}}_1}{ {, }}{{ {R}}_2} \to { {b}}{{ {R}}_2}{ {, }}{{ {R}}_3} \to { {c}}{{ {R}}_3}{ {]}}\]

\[{ { =  }}\dfrac{{{ {abc}}}}{{{ {abc}}}}\left| {\begin{array}{*{20}{l}}  {{{ {a}}^{ {2}}}}&{{{ {a}}^{ {3}}}}&{ {1}} \\   {{{ {b}}^{ {2}}}}&{{{ {b}}^{ {3}}}}&{ {1}} \\   {{{ {c}}^{ {2}}}}&{{{ {c}}^{ {3}}}}&{ {1}}  \end{array}} \right|\quad \]      [\[{{ {C}}_3}\] से \[{ {abc}}\] आम लेना]

\[\begin{align}  { { =  ( - 1}}{{ {)}}^{ {2}}}\left| {\begin{array}{*{20}{l}}  { {1}}&{{{ {a}}^{ {3}}}}&{{{ {a}}^{ {2}}}} \\   { {1}}&{{{ {b}}^{ {3}}}}&{{{ {b}}^{ {2}}}} \\   { {1}}&{{{ {c}}^{ {3}}}}&{{{ {c}}^{ {2}}}}  \end{array}} \right|\,\;\;\,{ {   }}\,\,\,{ {   [}}{{ {C}}_1} \leftrightarrow {{ {C}}_3}{ {]}} \hfill \\  { { =  ( - 1}}{{ {)}}^{ {2}}}\left| {\begin{array}{*{20}{l}}  { {1}}&{{{ {a}}^{ {2}}}}&{{{ {a}}^{ {3}}}} \\   { {1}}&{{{ {b}}^{ {2}}}}&{{{ {b}}^{ {3}}}} \\   { {1}}&{{{ {c}}^{ {2}}}}&{{{ {c}}^{ {3}}}}  \end{array}} \right|\quad \;\;\;\;\;\;{ {[}}{{ {C}}_2} \leftrightarrow {{ {C}}_3}{ {]}} \hfill \\  \end{align} \]

\[{ { =  }}\left| {\begin{array}{*{20}{l}}  { {1}}&{{{ {a}}^{ {2}}}}&{{{ {a}}^{ {3}}}} \\   { {1}}&{{{ {b}}^{ {2}}}}&{{{ {b}}^{ {3}}}} \\   { {1}}&{{{ {c}}^{ {2}}}}&{{{ {c}}^{ {3}}}}  \end{array}} \right|\]

3. \[\left| {\begin{array}{*{20}{c}}  {{ {cos\alpha cos\beta }}}&{{ {cos\alpha sin\beta }}}&{{ { - sin\alpha }}} \\   {{ { - sin\beta }}}&{{ {cos\beta }}}&{ {0}} \\   {{ {sin\alpha cos\beta }}}&{{ {sin\alpha sin\beta }}}&{{ {cos\alpha }}}  \end{array}} \right|\] का मान ज्ञात कीजिए। 

उत्तर: \[{ { =   - sin\alpha }}\left( {{ { - sin\alpha  si}}{{ {n}}^{ {2}}}{ {\beta  - sin\alpha  co}}{{ {s}}^{ {2}}}{ {\beta }}} \right){ { - 0(cos\alpha  cos\beta  sin\alpha  sin\beta  - cos\alpha  sin\beta  sin\alpha  cos\beta )}}\]

\[{ { + cos\alpha }}\left( {{ {cos\alpha  co}}{{ {s}}^{ {2}}}{ {\beta  + cos\alpha  co}}{{ {s}}^{ {2}}}{ {\beta }}} \right)\]

\[\begin{align}  { { =  si}}{{ {n}}^{ {2}}}{ {\alpha }}\left( {{ {si}}{{ {n}}^{ {2}}}{ {\beta  + co}}{{ {s}}^{ {2}}}{ {\beta }}} \right){ { + co}}{{ {s}}^{ {2}}}{ {\alpha }}\left( {{ {co}}{{ {s}}^{ {2}}}{ {\beta  + si}}{{ {n}}^{ {2}}}{ {\beta }}} \right) \hfill \\  { { =  si}}{{ {n}}^{ {2}}}{ {\alpha  + co}}{{ {s}}^{ {2}}}{ {\alpha }} \hfill \\  { { =  1}} \hfill \\  \end{align} \]

4. यदि \[{ {a, b, c}}\] वास्तविक संखयाए हो और सारणिक \[{ {\Delta   =  }}\left| {\begin{array}{*{20}{l}}  {{ {b + c}}}&{{ {c + a}}}&{{ {a + b}}} \\   {{ {c + a}}}&{{ {a + b}}}&{{ {b + c}}} \\   {{ {a + b}}}&{{ {b + c}}}&{{ {c + a}}}  \end{array}} \right|{ {  =  0}}\] हो तो दर्शाइए की या तो \[{ {a + b + c  =  0}}\] या \[{ {a  =  b  =  c}}\] है। 

उत्तर: \[\left| {\begin{array}{*{20}{l}}  {{ {b + c}}}&{{ {c + a}}}&{{ {a + b}}} \\   {{ {c + a}}}&{{ {a + b}}}&{{ {b + c}}} \\   {{ {a + b}}}&{{ {b + c}}}&{{ {c + a}}}  \end{array}} \right|{ {  =  0}}\]

\[{ { =  }}\left| {\begin{array}{*{20}{c}}  {{ {2(a + b + c)}}}&{{ {2(a + b + c)}}}&{{ {2(a + b + c)}}} \\   {{ {c + a}}}&{{ {a + b}}}&{{ {b + c}}} \\   {{ {a + b}}}&{{ {b + c}}}&{{ {c + a}}}  \end{array}} \right|{ {  =  0}}\;\,\quad { {[}}{{ {R}}_1} \to {{ {R}}_1}{ { + }}{{ {R}}_2}{ { + }}{{ {R}}_3}{ {]}}\]

\[{ { =  2(a + b + c)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   {{ {c + a}}}&{{ {a + b}}}&{{ {b + c}}} \\   {{ {a + b}}}&{{ {b + c}}}&{{ {c + a}}}  \end{array}} \right|{ {  =  0}}\]                

[\[{{ {R}}_1}\] के सामान्य रूप से \[{ {2(a + b + c)}}\] ]

\[{ { =  2(a + b + c)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {0}}&{ {0}} \\   {{ {c + a}}}&{{ {a + b}}}&{{ {b + c}}} \\   {{ {a + b}}}&{{ {b + c}}}&{{ {c + a}}}  \end{array}} \right|{ {  =  0}}\;\;\,\,\;\;\;{ {  [}}{{ {C}}_2} \to {{ {C}}_2}{ { - C1, }}{{ {C}}_3} \to {{ {C}}_3}{ { - }}{{ {C}}_1}]\]

\[\begin{align}  \therefore \;{ {2(a + b + c)[(b - c)(c - b) - (b - a)(c - a)]  =  0}} \hfill \\  { {2(a + b + c)}}\left[ {{ {bc - }}{{ {b}}^{ {2}}}{ { - }}{{ {c}}^{ {2}}}{ { + bc - }}\left( {{ {bc - ab - ac + }}{{ {a}}^{ {2}}}} \right)} \right]{ {  =  0}} \hfill \\  \end{align} \]

\[\begin{align}  { {2(a + b + c)}}\left[ {{ {bc - }}{{ {b}}^{ {2}}}{ { - }}{{ {c}}^{ {2}}}{ { + bc - bc + ab + ac - }}{{ {a}}^{ {2}}}} \right]{ {  =  0}} \hfill \\  { { - (a + b + c)}}\left[ {{ {2}}{{ {a}}^{ {2}}}{ { + 2}}{{ {b}}^{ {2}}}{ { + 2}}{{ {c}}^{ {2}}}{ { - 2ab - 2bc - 2ca}}} \right]{ {  =  0}} \hfill \\   \end{align} \]

\[\begin{align}  { { - (a + b + c)}}\left[ {\left( {{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}{ { + 2ab}}} \right){ { + }}\left( {{{ {b}}^{ {2}}}{ { + }}{{ {c}}^{ {2}}}{ { - 2bc}}} \right)} \right.\left. {{ { + }}\left( {{{ {c}}^{ {2}}}{ { + }}{{ {a}}^{ {2}}}{ { - 2ca}}} \right)} \right]{ {  =  0}} \hfill \\  { { - (a + b + c)}}\left[ {{{{ {(a - b)}}}^{ {2}}}{ { + (b - c}}{{ {)}}^{ {2}}}{ { + (c - a}}{{ {)}}^{ {2}}}} \right]{ {  =  0}} \hfill \\  \end{align} \]

\[{ {a + b + c  =  0 ; (a - b}}{{ {)}}^{ {2}}}{ {  =  0, (b - c}}{{ {)}}^{ {2}}}{ {  =  0, (c - a}}{{ {)}}^{ {2}}}{ {  =  0}}\]

\[\begin{align}  { {a + b + c  =  0 ; (a - b)  =  0, (b - c)  =  0, (c - a)  =  0}} \hfill \\  { {a + b + c  =  0 ; a  =  b, b  =  c, c  =  a}} \hfill \\  { {a + b + c  =  0 ; a  =  b  =  c}} \hfill \\  \end{align} \]

5. यदि \[{ {a }} \ne { { 0}}\] हो तो समीकरण \[\left| {\begin{array}{*{20}{c}}  {{ {x + a}}}&{ {x}}&{ {x}} \\   { {x}}&{{ {x + a}}}&{ {x}} \\   { {x}}&{ {x}}&{{ {x + a}}}   \end{array}} \right|{ {  =  0}}\] को हल कीजिए। 

उत्तर: \[\left| {\begin{array}{*{20}{c}}  {{ {x + a}}}&{ {x}}&{ {x}} \\   { {x}}&{{ {x + a}}}&{ {x}} \\   { {x}}&{ {x}}&{{ {x + a}}}   \end{array}} \right|{ {  =  0}}\]

\[{ { =  }}\left| {\begin{array}{*{20}{c}}  {{ {3x + a}}}&{{ {3x + a}}}&{{ {3x + a}}} \\   { {x}}&{{ {x + 2}}}&{ {x}} \\   { {x}}&{ {x}}&{{ {x + a}}}  \end{array}} \right|{ {  =  0}}\quad \;\;\;\;{ {[}}{{ {R}}_1} \to {{ {R}}_1}{ { + }}{{ {R}}_2}{ { + }}{{ {R}}_3}{ {]}}\]

\[{ { =  (3x + a)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {1}}&{ {1}} \\   { {x}}&{{ {x + 2}}}&{ {x}} \\   { {x}}&{ {x}}&{{ {x + a}}}  \end{array}} \right|{ {  =  0}}\]    [\[{{ {R}}_1}\] से सामान्य रूप मे \[{ {(3x + a)}}\] ले]

\[\begin{align}  { { =  (3x + a)}}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {0}}&{ {0}} \\   { {x}}&{ {a}}&{ {0}} \\   { {x}}&{ {0}}&{ {a}}  \end{array}} \right|{ {  =  0      }}\quad { {[}}{{ {C}}_2} \to {{ {C}}_2}{ { - }}{{ {C}}_1}{ {, }}{{ {C}}_3} \to {{ {C}}_3}{ { - }}{{ {C}}_1}{ {]}} \hfill \\  { { =  (3x + a)}}\left[ {{{ {a}}^{ {2}}}{ { - 0}}} \right] \hfill \\  { { =  0}} \hfill \\  \end{align} \]

\[\begin{align}  { { =  }}{{ {a}}^{ {2}}}{ {(3x + a)  =  0}} \hfill \\   \Rightarrow { { (3x + a)  =  0      }}\quad { {[}}\because { { a}} \ne { {0]}} \hfill \\   \Rightarrow { { x  =   - }}\dfrac{{ {a}}}{{ {3}}} \hfill \\   \end{align} \]

6. सिद्ध कीजिए की \[\left| {\begin{array}{*{20}{c}}  {{{ {a}}^{ {2}}}}&{{ {bc}}}&{{ {ac + }}{{ {c}}^{ {2}}}} \\   {{{ {a}}^{ {2}}}{ { + ab}}}&{{{ {b}}^{ {2}}}}&{{ {ac}}} \\   {{ {ab}}}&{{{ {b}}^{ {2}}}{ { + bc}}}&{{{ {c}}^{ {2}}}}  \end{array}} \right|{ {  =  4}}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}}\]

उत्तर: \[\left| {\begin{array}{*{20}{c}}  {{{ {a}}^{ {2}}}}&{{ {bc}}}&{{ {ac + }}{{ {c}}^{ {2}}}} \\   {{{ {a}}^{ {2}}}{ { + ab}}}&{{{ {b}}^{ {2}}}}&{{ {ac}}} \\   {{ {ab}}}&{{{ {b}}^{ {2}}}{ { + bc}}}&{{{ {c}}^{ {2}}}}  \end{array}} \right|{ {  =  }}\left| {\begin{array}{*{20}{c}}  {{ {2}}{{ {a}}^{ {2}}}}&{ {0}}&{{ {2ac}}} \\   {{{ {a}}^{ {2}}}{ { + ab}}}&{{{ {b}}^{ {2}}}}&{{ {ac}}} \\   {{ {ab}}}&{{{ {b}}^{ {2}}}{ { + bc}}}&{{{ {c}}^{ {2}}}}   \end{array}} \right|\]

\[{ { =  abc}}\left| {\begin{array}{*{20}{c}}  {{ {2a}}}&{ {0}}&{{ {2a}}} \\   {{ {a + b}}}&{ {b}}&{ {a}} \\   { {b}}&{{ {b + c}}}&{ {c}}  \end{array}} \right|{ {           [}}{{ {R}}_1} \to {{ {R}}_2}{ { + }}{{ {R}}_2}{ { - }}{{ {R}}_3}{ {]}}\]

\[\begin{align}  { { =  abc}}\left| {\begin{array}{*{20}{c}}  {{ {2a}}}&{ {0}}&{ {0}} \\   {{ {a + b}}}&{ {b}}&{{ { - b}}} \\   { {b}}&{{ {b + c}}}&{{ {c - b}}}  \end{array}} \right| \hfill \\  { { =  (abc)2a[b(c - a) - b(b + c)]}} \hfill \\  { { =  2}}{{ {a}}^{ {2}}}{ {bc}}\left[ {{ {bc - }}{{ {b}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}{ { + bc}}} \right] \hfill \\  { { =  2}}{{ {a}}^{ {2}}}{ {bc[2bc]}} \hfill \\  { { =  4}}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}} \hfill \\   \end{align} \]

7. \[{{ {A}}^{{ { - 1}}}}{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{{ { - 1}}}&{ {1}} \\   {{ { - 15}}}&{ {6}}&{{ { - 5}}} \\   { {5}}&{{ { - 2}}}&{ {2}}  \end{array}} \right]\] और \[{ {B  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {2}}&{{ { - 2}}} \\   {{ { - 1}}}&{ {3}}&{ {0}} \\   { {0}}&{{ { - 2}}}&{ {1}}  \end{array}} \right]\] हो तो \[{{ {(AB)}}^{{ { - 1}}}}\] का मान ज्ञात कीजिए। 

उत्तर: \[{ {B  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{ {2}}&{{ { - 2}}} \\   {{ { - 1}}}&{ {3}}&{ {0}} \\   { {0}}&{{ { - 2}}}&{ {1}}  \end{array}} \right]\]

\[\begin{array}{*{20}{c}}  {{{ {B}}_{11}}{ {  =  3}}}&{{{ {B}}_{12}}{ {  =  1}}}&{{{ {B}}_{13}}{ {  =  2}}} \\   {{{ {B}}_{21}}{ {  =  2}}}&{{{ {B}}_{22}}{ {  =  1}}}&{{{ {B}}_{23}}{ {  =  2}}} \\   {{{ {B}}_{31}}{ {  =  6}}}&{{{ {B}}_{32}}{ {  =  2}}}&{{{ {B}}_{33}}{ {  =  5}}}   \end{array}\]

\[\begin{align}  {{ {B}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|B|}}}}{ { adj B}} \hfill \\  { { =  }}\dfrac{{ {1}}}{{ {1}}}\left[ {\begin{array}{*{20}{l}}  {{{ {B}}_{{ {11}}}}}&{{{ {B}}_{{ {12}}}}}&{{{ {B}}_{{ {13}}}}} \\   {{{ {B}}_{{ {21}}}}}&{{{ {B}}_{{ {22}}}}}&{{{ {B}}_{{ {23}}}}} \\   {{{ {B}}_{{ {31}}}}}&{{{ {B}}_{{ {32}}}}}&{{{ {B}}_{{ {33}}}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{ {2}}&{ {6}} \\   { {1}}&{ {1}}&{ {2}} \\   { {2}}&{ {2}}&{ {5}}  \end{array}} \right] \hfill \\  \therefore \;{{ {(AB)}}^{{ { - 1}}}}{ {  =  }}{{ {B}}^{{ { - 1}}}}{{ {A}}^{{ { - 1}}}} \hfill \\  {{ {(AB)}}^{{ { - 1}}}}{ {  =  }}{{ {B}}^{{ { - 1}}}}{{ {A}}^{{ { - 1}}}}{ {  =  }}\left[ {\begin{array}{*{20}{l}}  { {3}}&{ {2}}&{ {6}} \\   { {1}}&{ {1}}&{ {2}} \\   { {2}}&{ {2}}&{ {5}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {3}}&{{ { - 1}}}&{ {1}} \\   {{ { - 15}}}&{ {6}}&{{ { - 5}}} \\   { {5}}&{{ { - 2}}}&{ {2}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  {{ {9 - 30 + 30}}}&{{ { - 3 + 12 - 12}}}&{{ {3 - 10 + 12}}} \\   {{ {3 - 15 + 10}}}&{{ { - 1 + 6 - 4}}}&{{ {1 - 5 + 4}}} \\   {{ {6 - 30 + 25}}}&{{ { - 2 + 12 - 10}}}&{{ {2 - 10 + 10}}}  \end{array}} \right] \hfill \\  { { =  }}\left[ {\begin{array}{*{20}{c}}  { {9}}&{{ { - 3}}}&{ {5}} \\   {{ { - 2}}}&{ {1}}&{ {0}} \\   { {1}}&{ {0}}&{ {2}}  \end{array}} \right] \hfill \\  \end{align} \]

8. मान लीजिए \[\left[ {\begin{array}{*{20}{c}}  { {1}}&{{ { - 2}}}&{ {1}} \\   {{ { - 2}}}&{ {3}}&{ {1}} \\   { {1}}&{ {1}}&{ {5}}  \end{array}} \right]\] हो तो सत्यापित कीजिए की 

(i) \[{{ {[adjA]}}^{{ { - 1}}}}{ {  =  adj}}\left( {{{ {A}}^{{ { - 1}}}}} \right)\]

उत्तर: \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{{ { - 2}}}&{ {1}} \\   {{ { - 2}}}&{ {3}}&{ {1}} \\   { {1}}&{ {1}}&{ {5}}  \end{array}} \right]\]

\[\left| { {A}} \right|{ {  =  1(15 - 1) + 2( - 10 - 1) + 1( - 2 - 3)  =   - 13 }} \ne { { 0 }} \Rightarrow { { }}{{ {A}}^{{ { - 1}}}}\]

\[\begin{array}{*{20}{c}}  {{{ {A}}_{11}}{ {  =  14}}}&{{{ {A}}_{12}}{ {  =  11}}}&{{{ {A}}_{13}}{ {  =   - 5}}} \\   {{{ {A}}_{21}}{ {  =  11}}}&{{{ {A}}_{22}}{ {  =  4}}}&{{{ {A}}_{23}}{ {  =   - 3}}} \\   {{{ {A}}_{31}}{ {  =   - 5}}}&{{{ {A}}_{32}}{ {  =   - 3}}}&{{{ {A}}_{33}}{ {  =   - 1}}}   \end{array}\]

\[{{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}{ { adj A  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {13}}}}} \\   {{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {23}}}}} \\   {{{ {A}}_{{ {31}}}}}&{{{ {A}}_{{ {32}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ { - 13}}}}\left[ {\begin{array}{*{20}{c}}  {{ {14}}}&{{ {11}}}&{{ { - 5}}} \\   {{ {11}}}&{ {4}}&{{ { - 3}}} \\   {{ { - 5}}}&{{ { - 3}}}&{{ { - 1}}}   \end{array}} \right]\]  ………………..(1)

\[\begin{align}  { {B  =  adjA}} \hfill \\  { {B  =  }}\left[ {\begin{array}{*{20}{c}}  {{ {14}}}&{{ {11}}}&{{ { - 5}}} \\   {{ {11}}}&{ {4}}&{{ { - 3}}} \\   {{ { - 5}}}&{{ { - 3}}}&{{ { - 1}}}  \end{array}} \right] \hfill \\  \left| { {B}} \right|{ {  =  14( - 4 - 9) - 11( - 11 - 15) - 5( - 33 + 20)  =   - 182 + 286 + 65  =  169 }} \ne { { 0}}\; \Rightarrow \;{{ {B}}^{{ { - 1}}}} \hfill \\  \end{align} \]

\[\begin{array}{*{20}{c}}  {{{ {B}}_{{ {11}}}}{ {  =   - 13}}}&{{{ {B}}_{{ {12}}}}{ {  =  26}}}&{{{ {B}}_{{ {13}}}}{ {  =   - 13}}} \\   {{{ {B}}_{{ {21}}}}{ {  =  26}}}&{{{ {B}}_{{ {22}}}}{ {  =   - 39}}}&{{{ {B}}_{{ {23}}}}{ {  =   - 13}}} \\   {{{ {B}}_{{ {31}}}}{ {  =   - 13}}}&{{{ {B}}_{{ {32}}}}{ {  =   - 13}}}&{{{ {B}}_{{ {33}}}}{ {  =   - 65}}}  \end{array}\]

\[{{ {B}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|B|}}}}{ {adjB  =  }}\dfrac{{ {1}}}{{{ {|B|}}}}\left[ {\begin{array}{*{20}{l}}  {{{ {B}}_{{ {11}}}}}&{{{ {B}}_{{ {12}}}}}&{{{ {B}}_{{ {13}}}}} \\   {{{ {B}}_{{ {21}}}}}&{{{ {B}}_{{ {22}}}}}&{{{ {B}}_{{ {23}}}}} \\   {{{ {B}}_{{ {31}}}}}&{{{ {B}}_{{ {32}}}}}&{{{ {B}}_{{ {33}}}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {169}}}}\left[ {\begin{array}{*{20}{c}}  {{ { - 13}}}&{{ {26}}}&{{ { - 13}}} \\   {{ {26}}}&{{ { - 39}}}&{{ { - 13}}} \\   {{ { - 13}}}&{{ { - 13}}}&{{ { - 65}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {13}}}}\left[ {\begin{array}{*{20}{c}}  {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\   { {2}}&{{ { - 3}}}&{{ { - 1}}} \\   {{ { - 1}}}&{{ { - 1}}}&{{ { - 5}}}  \end{array}} \right]\]

\[{ {adj}}{{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {13}}}}\left[ {\begin{array}{*{20}{c}}  {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\   { {2}}&{{ { - 3}}}&{{ { - 1}}} \\   {{ { - 1}}}&{{ { - 1}}}&{{ { - 5}}}   \end{array}} \right]\] …………………..(2)

\[\begin{align}  { {C  =  }}{{ {A}}^{{ { - 1}}}} \hfill \\  { {C  =  }}\dfrac{{ {1}}}{{{ { - 13}}}}\left[ {\begin{array}{*{20}{c}}  {{ {14}}}&{{ {11}}}&{{ { - 5}}} \\   {{ {11}}}&{ {4}}&{{ { - 3}}} \\   {{ { - 5}}}&{{ { - 3}}}&{{ { - 1}}}  \end{array}} \right] \hfill \\   \end{align} \]

\[\begin{array}{*{20}{c}}  {{{ {C}}_{{ {11}}}}{ {  =  3}}}&{{{ {C}}_{{ {12}}}}{ {  =  1}}}&{{{ {C}}_{{ {13}}}}{ {  =  2}}} \\   {{{ {C}}_{21}}{ {  =  2}}}&{{{ {C}}_{22}}{ {  =  1}}}&{{{ {C}}_{23}}{ {  =  2}}} \\   {{{ {C}}_{31}}{ {  =  6}}}&{{{ {C}}_{32}}{ {  =  2}}}&{{{ {C}}_{33}}{ {  =  5}}}   \end{array}\]

\[{ {adjC  =  }}\dfrac{{ {1}}}{{ {1}}}\left[ {\begin{array}{*{20}{l}}  {{{ {C}}_{{ {11}}}}}&{{{ {C}}_{{ {12}}}}}&{{{ {C}}_{{ {13}}}}} \\   {{{ {C}}_{{ {21}}}}}&{{{ {C}}_{{ {22}}}}}&{{{ {C}}_{{ {23}}}}} \\   {{{ {C}}_{{ {31}}}}}&{{{ {C}}_{{ {32}}}}}&{{{ {C}}_{{ {33}}}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}&{\dfrac{{ {2}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}} \\   {\dfrac{{ {2}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {3}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}} \\   {{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {5}}}{{{ {13}}}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {13}}}}\left[ {\begin{array}{*{20}{c}}  {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\   { {2}}&{{ { - 3}}}&{{ { - 1}}} \\   {{ { - 1}}}&{{ { - 1}}}&{{ { - 5}}}  \end{array}} \right]\]

\[{ {adjC  =  adj}}\left( {{{ {A}}^{{ { - 1}}}}} \right){ {  =  }}\dfrac{{ {1}}}{{{ {13}}}}\left[ {\begin{array}{*{20}{c}}  {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\   { {2}}&{{ { - 3}}}&{{ { - 1}}} \\   {{ { - 1}}}&{{ { - 1}}}&{{ { - 5}}}   \end{array}} \right]\] ………………..(3)

समीकरण (2) और (3) से हमारे पास है \[{{ {(adjA)}}^{{ { - 1}}}}{ {  =  adj}}\left( {{{ {A}}^{{ { - 1}}}}} \right)\]

(ii) \[{\left( {{{ {A}}^{{ { - 1}}}}} \right)^{{ { - 1}}}}{ {  =  A}}\]

उत्तर: समीकरण (1) से हमारे पास 

\[\begin{align}  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ { - 13}}}}\left[ {\begin{array}{*{20}{c}}  {{ {14}}}&{{ {11}}}&{{ { - 5}}} \\   {{ {11}}}&{ {4}}&{{ { - 3}}} \\   {{ { - 5}}}&{{ { - 3}}}&{{ { - 1}}}  \end{array}} \right] \hfill \\  { {D  =  }}{{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ { - 13}}}}\left[ {\begin{array}{*{20}{c}}  {{ {14}}}&{{ {11}}}&{{ { - 5}}} \\   {{ {11}}}&{ {4}}&{{ { - 3}}} \\   {{ { - 5}}}&{{ { - 3}}}&{{ { - 1}}}   \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {\dfrac{{{ {14}}}}{{{ {13}}}}}&{\dfrac{{{ {11}}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {5}}}{{{ {13}}}}} \\   {\dfrac{{{ {11}}}}{{{ {13}}}}}&{\dfrac{{ {4}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {3}}}{{{ {13}}}}} \\   {{ { - }}\dfrac{{ {5}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {3}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}  \end{array}} \right] \hfill \\  \left| { {D}} \right|{ {  =   - }}{\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)^{ {3}}}{ {[14( - 4 - 9) - 11( - 11 - 15) - 5( - 33 + 20)]}} \hfill \\  { { =   - }}{\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)^{ {3}}}{ {( - 182 + 286 + 65)  =   - }}{\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)^{ {3}}}{ {169  =   - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right){ { }} \ne { { 0 }} \Rightarrow { { }}{{ {D}}^{{ { - 1}}}} \hfill \\   \end{align} \]

\[\begin{array}{*{20}{l}}  {{{ {D}}_{11}}{ {  =   - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}&{{{ {D}}_{12}}{ {  =   - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}&{{{ {D}}_{13}}{ {  =   - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)} \\   {{{ {D}}_{21}}{ {  =   - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}&{{{ {D}}_{22}}{ {  =   - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}&{{{ {D}}_{23}}{ {  =   - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)} \\   {{{ {D}}_{31}}{ {  =   - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}&{{{ {D}}_{23}}{ {  =   - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}&{{{ {D}}_{33}}{ {  =   - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}  \end{array}\]

\[\begin{align}  {{ {D}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|D|}}}}{ {adjD  =  }}\dfrac{{ {1}}}{{{ {|D|}}}}\left[ {\begin{array}{*{20}{l}}  {{{ {D}}_{{ {11}}}}}&{{{ {D}}_{{ {12}}}}}&{{{ {D}}_{{ {13}}}}} \\   {{{ {D}}_{{ {21}}}}}&{{{ {D}}_{{ {22}}}}}&{{{ {D}}_{{ {23}}}}} \\   {{{ {D}}_{{ {31}}}}}&{{{ {D}}_{{ {32}}}}}&{{{ {D}}_{{ {33}}}}}  \end{array}} \right] \hfill \\  { { =  }}\dfrac{{ {1}}}{{{ { - 1}}}}\left[ {\begin{array}{*{20}{c}}  {{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}&{\dfrac{{ {2}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}} \\   {\dfrac{{ {2}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {3}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}} \\   {{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {5}}}{{{ {13}}}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{{ { - 2}}}&{ {1}} \\   {{ { - 2}}}&{ {3}}&{ {1}} \\   { {1}}&{ {1}}&{ {5}}   \end{array}} \right] \hfill \\  {{ {D}}^{{ { - 1}}}}{ {  =  }}{\left( {{{ {A}}^{{ { - 1}}}}} \right)^{{ { - 1}}}}{ {  =  }}\left| {\begin{array}{*{20}{c}}  { {1}}&{{ { - 2}}}&{ {1}} \\   {{ { - 2}}}&{ {3}}&{ {1}} \\   { {1}}&{ {1}}&{ {5}}  \end{array}} \right|{ {  =  A}} \hfill \\  \end{align} \]

9. \[\left| {\begin{array}{*{20}{c}}  { {x}}&{ {y}}&{{ {x + y}}} \\   { {y}}&{{ {x + y}}}&{ {x}} \\   {{ {x + y}}}&{ {x}}&{ {y}}   \end{array}} \right|\] का मान ज्ञात कीजिए। 

उत्तर: \[{ { =  }}\left| {\begin{array}{*{20}{c}}  {{ {2(x + y)}}}&{ {y}}&{{ {x + y}}} \\   {{ {2(x + y)}}}&{{ {x + y}}}&{ {x}} \\   {{ {2(x + y)}}}&{ {x}}&{ {y}}  \end{array}} \right|{ {       [}}{{ {C}}_1} \to {{ {C}}_1}{ { + }}{{ {C}}_2}{ { + }}{{ {C}}_3}{ {]}}\]

\[{ { =  2(x + y)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{ {y}}&{{ {x + y}}} \\   { {1}}&{{ {x + y}}}&{ {x}} \\   { {1}}&{ {x}}&{ {y}}  \end{array}} \right|\]         [\[{{ {C}}_{ {1}}}\] से सामान्य के रूप मे \[{ {2(x + y)}}\] लेना]

\[\begin{align}  { { =  2(x + y)}}\left| {\begin{array}{*{20}{c}}  { {0}}&{{ { - x}}}&{ {y}} \\   { {0}}&{ {y}}&{{ {x - y}}} \\   { {1}}&{ {x}}&{ {y}}  \end{array}} \right|\;\;\;\;\;\;\,\;\;\;\;\;\;\left[ {{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { - }}{{ {R}}_{ {2}}}{ {, }}{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {3}}}} \right.{ {]}} \hfill \\  { { =  2(x + y)[( - x)(x - y) - y \times y]}} \hfill \\  { { =  2(x + y)}}\left( {{ { - }}{{ {x}}^{ {2}}}{ { + xy - }}{{ {y}}^{ {2}}}} \right) \hfill \\  { { =   - 2(x + y)}}\left( {{{ {x}}^{ {2}}}{ { - xy + }}{{ {y}}^{ {2}}}} \right) \hfill \\  { { =   - 2}}\left( {{{ {x}}^{ {2}}}{ { + }}{{ {y}}^{ {2}}}} \right) \hfill \\  \end{align} \]

10. \[\left| {\begin{array}{*{20}{c}}  { {1}}&{ {x}}&{{ {x + y}}} \\   { {1}}&{{ {x + y}}}&{ {x}} \\   { {1}}&{ {x}}&{ {y}}  \end{array}} \right|\] का मान ज्ञात कीजिए। 

उत्तर: \[{ { =  }}\left| {\begin{array}{*{20}{c}}  { {0}}&{{ { - y}}}&{ {0}} \\   { {0}}&{ {y}}&{{ { - x}}} \\   { {1}}&{ {x}}&{{ {x + y}}}  \end{array}} \right|{ {                [}}{{ {R}}_1} \to {{ {R}}_1}{ { - }}{{ {R}}_2}{ {, }}{{ {R}}_2} \to {{ {R}}_2}{ { - }}{{ {R}}_3}{ {]}}\]

\[\begin{align}  { { =  [( - y)( - x) - y \times 0]}} \hfill \\  { { =  xy}} \hfill \\  \end{align} \]

सारणिकों के गुणधर्मों का प्रयोग करके निम्नलिखित 11 से 15 तक प्रश्नों को सिद्ध कीजिए: 

11. \[\left| {\begin{array}{*{20}{l}}  { {\alpha }}&{{{ {\alpha }}^{ {2}}}}&{{ {\beta  + \gamma }}} \\   { {\beta }}&{{{ {\beta }}^{ {2}}}}&{{ {\gamma  + \alpha }}} \\   { {\gamma }}&{{{ {\gamma }}^{ {2}}}}&{{ {\alpha  + \beta }}}  \end{array}} \right|{ {  =  (\beta  - \gamma )(\gamma  - \alpha )(\alpha  - \beta )(\alpha  + \beta  + \gamma )}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{l}}  { {\alpha }}&{{{ {\alpha }}^{ {2}}}}&{{ {\beta  + \gamma }}} \\   { {\beta }}&{{{ {\beta }}^{ {2}}}}&{{ {\gamma  + \alpha }}} \\   { {\gamma }}&{{{ {\gamma }}^{ {2}}}}&{{ {\alpha  + \beta }}}  \end{array}} \right|\]

\[{ { =  }}\left| {\begin{array}{*{20}{l}}  { {\alpha }}&{{{ {\alpha }}^{ {2}}}}&{{ {\alpha  + \beta  + \gamma }}} \\   { {\beta }}&{{{ {\beta }}^{ {2}}}}&{{ {\alpha  + \beta  + \gamma }}} \\   { {\gamma }}&{{{ {\gamma }}^{ {2}}}}&{{ {\alpha  + \beta  + \gamma }}}  \end{array}} \right|{ {               [}}{{ {C}}_3} \to {{ {C}}_3}{ { + }}{{ {C}}_2}{ {]}}\]

\[{ { =  (\alpha  + \beta  + \gamma )}}\left| {\begin{array}{*{20}{l}}  { {\alpha }}&{{{ {\alpha }}^{ {2}}}}&{ {1}} \\   { {\beta }}&{{{ {\beta }}^{ {2}}}}&{ {1}} \\   { {\gamma }}&{{{ {\gamma }}^{ {2}}}}&{ {1}}  \end{array}} \right|\]       [\[{{ {C}}_3}\] से सामान्य के रूप मे \[{ {(\alpha  + \beta  + \gamma )}}\] लेना]

\[{ { =  (\alpha  + \beta  + \gamma )}}\left| {\begin{array}{*{20}{c}}  {{ {\alpha  - \beta }}}&{{{ {\alpha }}^{ {2}}}{ { - }}{{ {\beta }}^{ {2}}}}&{ {0}} \\   {{ {\beta  - \gamma }}}&{{{ {\beta }}^{ {2}}}{ { - }}{{ {\gamma }}^{ {2}}}}&{ {0}} \\   { {\gamma }}&{{{ {\gamma }}^{ {2}}}}&{ {1}}  \end{array}} \right|\quad \;\;\;\;\;\;\;\,\;{ {[}}{{ {R}}_1} \to {{ {R}}_1}{ { - }}{{ {R}}_2}{ {, }}{{ {R}}_2} \to {{ {R}}_2}{ { - }}{{ {R}}_3}{ {]}}\]

\[{ { =  (\alpha  + \beta  + \gamma )(\alpha  - \beta )(\beta  - \gamma )}}\left| {\begin{array}{*{20}{c}}  { {1}}&{{ {\alpha  + \beta }}}&{ {0}} \\   { {1}}&{{ {\beta  + \gamma }}}&{ {0}} \\   { {\gamma }}&{{{ {\gamma }}^{ {2}}}}&{ {1}}  \end{array}} \right|\]         [\[{{ {R}}_{ {1}}}\] से सामान्य के रूप मे \[{ {(\alpha  - \beta )}}\] लेना और \[{{ {R}}_{ {2}}}\] से सामान्य               रूप से \[{ {(\beta  - \gamma )}}\] लेना]

\[\begin{align}  { { =  (\alpha  + \beta  + \gamma )(\alpha  - \beta )(\beta  - \gamma )[(\beta  + \gamma ) - (\alpha  + \beta )]}} \hfill \\  { { =  (\alpha  + \beta  + \gamma )(\alpha  - \beta )(\beta  - \gamma )(\gamma  - \alpha )}} \hfill \\  \end{align} \]

= RHS

12. \[\left| {\begin{array}{*{20}{l}}  { {x}}&{{{ {x}}^{ {2}}}}&{{ {1 + p}}{{ {x}}^{ {2}}}} \\   { {y}}&{{{ {y}}^{ {2}}}}&{{ {1 + p}}{{ {y}}^{ {2}}}} \\   { {z}}&{{{ {z}}^{ {2}}}}&{{ {1 + p}}{{ {z}}^{ {2}}}}  \end{array}} \right|{ {  =  (1 + pxyz)(x - y)(y - z)(z - x)}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{l}}  { {x}}&{{{ {x}}^{ {2}}}}&{{ {1 + p}}{{ {x}}^{ {2}}}} \\   { {y}}&{{{ {y}}^{ {2}}}}&{{ {1 + p}}{{ {y}}^{ {2}}}} \\   { {z}}&{{{ {z}}^{ {2}}}}&{{ {1 + p}}{{ {z}}^{ {2}}}}  \end{array}} \right|\]

\[{ { =  }}\left| {\begin{array}{*{20}{l}}  { {x}}&{{{ {x}}^{ {2}}}}&{ {1}} \\   { {y}}&{{{ {y}}^{ {2}}}}&{ {1}} \\   { {z}}&{{{ {z}}^{ {2}}}}&{ {1}}  \end{array}} \right|{ { + }}\left| {\begin{array}{*{20}{l}}  { {x}}&{{{ {x}}^{ {2}}}}&{{ {p}}{{ {x}}^{ {2}}}} \\   { {y}}&{{{ {y}}^{ {2}}}}&{{ {p}}{{ {y}}^{ {2}}}} \\   { {z}}&{{{ {z}}^{ {2}}}}&{{ {p}}{{ {z}}^{ {2}}}}  \end{array}} \right|\]

\[{ { =  ( - 1}}{{ {)}}^{ {1}}}\left| {\begin{array}{*{20}{l}}  { {1}}&{{{ {x}}^{ {2}}}}&{ {x}} \\   { {1}}&{{{ {y}}^{ {2}}}}&{ {y}} \\   { {1}}&{{{ {z}}^{ {2}}}}&{ {z}}  \end{array}} \right|{ { + p}}\left| {\begin{array}{*{20}{l}}  { {x}}&{{{ {x}}^{ {2}}}}&{{{ {x}}^{ {3}}}} \\   { {y}}&{{{ {y}}^{ {2}}}}&{{{ {y}}^{ {3}}}} \\   { {z}}&{{{ {z}}^{ {2}}}}&{{{ {z}}^{ {3}}}}  \end{array}} \right|\]                   

[\[{{ {C}}_3}\] से सामान्य के रूप मे \[{ {p}}\] लेना]

\[{ { =  ( - 1}}{{ {)}}^{ {2}}}\left| {\begin{array}{*{20}{l}}  { {1}}&{{{ {x}}^{ {2}}}}&{ {x}} \\   { {1}}&{{{ {y}}^{ {2}}}}&{ {y}} \\   { {1}}&{{{ {z}}^{ {2}}}}&{ {z}}  \end{array}} \right|{ { + pxyz}}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {x}}&{{{ {x}}^{ {2}}}} \\   { {1}}&{ {y}}&{{{ {y}}^{ {2}}}} \\   { {1}}&{ {z}}&{{{ {z}}^{ {2}}}}  \end{array}} \right|\]          [\[{{ {R}}_1}{ {,}}{{ {R}}_2}{ {,}}{{ {R}}_3}\] से सामान्य के रूप मे \[{ {x,y,z}}\] लेना]

\[{ { =  (1 + pxyz)}}\left| {\begin{array}{*{20}{l}}  { {1}}&{ {x}}&{{{ {x}}^{ {2}}}} \\   { {1}}&{ {y}}&{{{ {y}}^{ {2}}}} \\   { {1}}&{ {z}}&{{{ {z}}^{ {2}}}}   \end{array}} \right|\]                     

[\[\,\left| {\begin{array}{*{20}{l}}  { {1}}&{ {x}}&{{{ {x}}^{ {2}}}} \\   { {1}}&{ {y}}&{{{ {y}}^{ {2}}}} \\   { {1}}&{ {z}}&{{{ {z}}^{ {2}}}}  \end{array}} \right|\] सामान्य के रूप मे लेना]

\[{ { =  (1 + pxyz)}}\left| {\begin{array}{*{20}{c}}  { {0}}&{{ {x - y}}}&{{{ {x}}^{ {2}}}{ { - }}{{ {y}}^{ {2}}}} \\   { {0}}&{{ {y - z}}}&{{{ {y}}^{ {2}}}{ { - }}{{ {z}}^{ {2}}}} \\   { {1}}&{ {z}}&{{{ {z}}^{ {2}}}}  \end{array}} \right|\;\;\;\,\,\,{ {    }}\,\;{ {[}}{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { - }}{{ {R}}_{ {2}}}{ {, }}{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {3}}}]\]

\[{ { =  (1 + pxyz)(x - y)(y - z)}}\left| {\begin{array}{*{20}{c}}  { {0}}&{ {1}}&{{ {x + y}}} \\   { {0}}&{ {1}}&{{ {y + z}}} \\   { {1}}&{ {z}}&{{{ {z}}^{ {2}}}}  \end{array}} \right|\]       [\[{{ {R}}_1}\] से सामान्य के रूप मे \[{ {(x - y)}}\] लेना और \[{{ {R}}_2}\] से सामान्य के रूप मे \[{ {y - z}}\] लेना]

\[\begin{align}  { { =  (1 + pxyz)(x - y)(y - z)[(y + z) - (x + y)]}} \hfill \\  { { =  (1 + pxyz)(x - y)(y - z)(z - x)}} \hfill \\  \end{align} \] 

= RHS

13. \[\left| {\begin{array}{*{20}{c}}  {{ {3a}}}&{{ { - a + b}}}&{{ { - a + c}}} \\   {{ { - b + a}}}&{{ {3b}}}&{{ { - b + c}}} \\   {{ { - c + a}}}&{{ { - c + b}}}&{{ {3c}}}  \end{array}} \right|{ {  =  3(a + b + c)(ab + bc + ca)}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}}  {{ {3a}}}&{{ { - a + b}}}&{{ { - a + c}}} \\   {{ { - b + a}}}&{{ {3b}}}&{{ { - b + c}}} \\   {{ { - c + a}}}&{{ { - c + b}}}&{{ {3c}}}  \end{array}} \right|\]

\[{ { =  }}\left| {\begin{array}{*{20}{c}}  {{ {a + b + c}}}&{{ { - a + b}}}&{{ { - a + c}}} \\   {{ {a + b + c}}}&{{ {3b}}}&{{ { - b + c}}} \\   {{ {a + b + c}}}&{{ { - c + b}}}&{{ {3c}}}  \end{array}} \right|{ {              [}}{{ {C}}_1} \to {{ {C}}_1}{ { + }}{{ {C}}_2}{ { + }}{{ {C}}_3}{ {]}}\]

\[{ { =  (a + b + c)}}\left| {\begin{array}{*{20}{c}}  { {1}}&{{ { - a + b}}}&{{ { - a + c}}} \\   { {1}}&{{ {3b}}}&{{ { - b + c}}} \\   { {1}}&{{ { - c + b}}}&{{ {3c}}}  \end{array}} \right|\]      [\[{{ {C}}_{ {1}}}\] से सामान्य के रूप मे \[{ {(a + b + c)}}\] लेना]

\[{ { =  (a + b + c)}}\left| {\begin{array}{*{20}{c}}  { {0}}&{{ { - a - 2b}}}&{{ { - a + c}}} \\   { {0}}&{{ {2b + c}}}&{{ { - b - 2c}}} \\   { {1}}&{{ { - c + b}}}&{{ {3c}}}  \end{array}} \right|{ {             [}}{{ {R}}_1} \to {{ {R}}_1}{ { - }}{{ {R}}_2}{ {, }}{{ {R}}_2} \to {{ {R}}_2}{ { - }}{{ {R}}_3}{ {]}}\]

\[\begin{align}  { { =  (a + b + c)[( - a - 2b)( - b - 2c) - (2b + c)( - a + b)]}} \hfill \\  { { =  (a + b + c)}}\left( {{ {ab + 2ac + 2}}{{ {b}}^{ {2}}}{ { + 4bc}}} \right.\left. {{ { - }}\left( {{ { - 2ab + 2}}{{ {b}}^{ {2}}}{ {  =   - ac + bc}}} \right)} \right) \hfill \\  { { =  (a + b + c)(3ab + 3bc + 3ca)}} \hfill \\  { { =  3(a + b + c)(ab + bc + ca)}} \hfill \\  \end{align} \]

= RHS

14. \[\left| {\begin{array}{*{20}{c}}  { {1}}&{{ {1 + p}}}&{{ {1 + p + q}}} \\   { {2}}&{{ {3 + 2p}}}&{{ {4 + 3p + 2q}}} \\   { {3}}&{{ {6 + 3p}}}&{{ {10 + 6p + 3q}}}  \end{array}} \right|{ {  =  1}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}}  { {1}}&{{ {1 + p}}}&{{ {1 + p + q}}} \\   { {2}}&{{ {3 + 2p}}}&{{ {4 + 3p + 2q}}} \\   { {3}}&{{ {6 + 3p}}}&{{ {10 + 6p + 3q}}}   \end{array}} \right|\]

\[\begin{align}  { { =  }}\left| {\begin{array}{*{20}{c}}  { {1}}&{{ {1 + p}}}&{{ {1 + p + q}}} \\   { {0}}&{ {1}}&{{ {2 + p}}} \\   { {0}}&{ {3}}&{{ {7 + 3p}}}  \end{array}} \right|{ {                [}}{{ {R}}_2} \to {{ {R}}_2}{ { - 2}}{{ {R}}_1}{ {, }}{{ {R}}_3} \to {{ {R}}_3}{ { - 3}}{{ {R}}_1}{ {]}} \hfill \\  { { =  1(1(7 + 3p) - (3)(2 + p))}} \hfill \\  { { =  7 + 3p - 6 - 3p}} \hfill \\  { { =  1}} \hfill \\  \end{align} \]

= RHS

15. \[\left| {\begin{array}{*{20}{c}}  {{ {sin\alpha }}}&{{ {cos\alpha }}}&{{ {cos(\alpha  + \delta )}}} \\   {{ {sin\beta }}}&{{ {cos\beta }}}&{{ {cos(\beta  + \delta )}}} \\   {{ {sin\gamma }}}&{{ {cos\gamma }}}&{{ {cos(\gamma  + \delta )}}}  \end{array}} \right|{ {  =  0}}\]

उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}}  {{ {sin\alpha }}}&{{ {cos\alpha }}}&{{ {cos(\alpha  + \delta )}}} \\   {{ {sin\beta }}}&{{ {cos\beta }}}&{{ {cos(\beta  + \delta )}}} \\   {{ {sin\gamma }}}&{{ {cos\gamma }}}&{{ {cos(\gamma  + \delta )}}}  \end{array}} \right|\]

\[\begin{align}  { { =  }}\left| {\begin{array}{*{20}{l}}  {{ {sin\alpha }}}&{{ {cos\alpha cos\delta  - sin\alpha sin\delta }}}&{{ {cos(\alpha  + \delta )}}} \\   {{ {sin\beta }}}&{{ {cos\beta cos\delta  - sin\beta sin\delta }}}&{{ {cos(\beta  + \delta )}}} \\   {{ {sin\gamma }}}&{{ {cos\gamma cos\delta  - sin\gamma sin\delta }}}&{{ {cos(\gamma  + \delta )}}}  \end{array}} \right|{ {               [}}{{ {C}}_2} \to { {cos\delta  }}{{ {C}}_2}{ { - sin\delta  }}{{ {C}}_1}{ {]}} \hfill \\  { { =  }}\left| {\begin{array}{*{20}{l}}  {{ {sin\alpha }}}&{{ {cos(\alpha  + \delta )}}}&{{ {cos(\alpha  + \delta )}}} \\   {{ {sin\beta }}}&{{ {cos(\beta  + \delta )}}}&{{ {cos(\beta  + \delta )}}} \\   {{ {sin\gamma }}}&{{ {cos(\gamma  + \delta )}}}&{{ {cos(\gamma  + \delta )}}}  \end{array}} \right|\quad \;\;\,\;\;\,\,\,\left[ {\because \;{{ {C}}_{ {2}}}{ {  =  }}{{ {C}}_{ {3}}}} \right] \hfill \\  { { =  0}} \hfill \\  \end{align} \]

= RHS

16. निम्नलिखित समीकरण निकाय को हल कीजिए: 

\[\dfrac{{ {2}}}{{ {x}}}{ { + }}\dfrac{{ {3}}}{{ {y}}}{ { + }}\dfrac{{{ {10}}}}{{ {z}}}{ {  =  4, }}\dfrac{{ {4}}}{{ {x}}}{ { - }}\dfrac{{ {6}}}{{ {y}}}{ { + }}\dfrac{{ {5}}}{{ {z}}}{ {  =  1, }}\dfrac{{ {6}}}{{ {x}}}{ { + }}\dfrac{{ {9}}}{{ {y}}}{ { - }}\dfrac{{{ {20}}}}{{ {z}}}{ {  =  2}}\]

उत्तर: समीकरणों की इस प्रणाली को \[{ {AX}}\,{ { = }}\;{ {B}}\] के रूप मे लिखा जा सकता है:

\[\begin{align}  { {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {2}}&{ {3}}&{{ {10}}} \\   { {4}}&{{ { - 6}}}&{ {5}} \\   { {6}}&{ {9}}&{{ { - 7}}}   \end{array}} \right]{ {, X  =  }}\left[ {\begin{array}{*{20}{l}}  {\dfrac{{ {1}}}{{ {x}}}} \\   {\dfrac{{ {1}}}{{ {y}}}} \\   {\dfrac{{ {1}}}{{ {z}}}}  \end{array}} \right]{ {, B  =  }}\left[ {\begin{array}{*{20}{l}}  { {4}} \\   { {1}} \\   { {2}}  \end{array}} \right] \hfill \\  { {|A|  =  2(120 - 45) - 3( - 80 - 30) + 10(36 + 36)}} \hfill \\  { { =  150 + 330 + 720  =  1200 }} \ne { { 0 }} \Rightarrow { { }}{{ {A}}^{{ { - 1}}}} \hfill \\  \end{align} \]

\[\begin{array}{*{20}{c}}  {{{ {A}}_{11}}{ {  =  75}}}&{{{ {A}}_{12}}{ {  =  110}}}&{{{ {A}}_{13}}{ {  =  72}}} \\   {{{ {A}}_{21}}{ {  =  150}}}&{{{ {A}}_{22}}{ {  =   - 100}}}&{{{ {A}}_{23}}{ {  =  0}}} \\   {{{ {A}}_{31}}{ {  =  75}}}&{{{ {A}}_{32}}{ {  =  30}}}&{{{ {A}}_{33}}{ {  =   - 24}}}  \end{array}\]

\[{{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {adjA  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {13}}}}} \\   {{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {23}}}}} \\   {{{ {A}}_{{ {31}}}}}&{{{ {A}}_{{ {32}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {1200}}}}\left[ {\begin{array}{*{20}{c}}  {{ {75}}}&{{ {150}}}&{{ {75}}} \\   {{ {110}}}&{{ { - 100}}}&{{ {30}}} \\   {{ {72}}}&{ {0}}&{{ { - 24}}}  \end{array}} \right]\]

\[\begin{align}  { {X  =  }}{{ {A}}^{{ { - 1}}}}{ {B}} \hfill \\  \left[ {\begin{array}{*{20}{l}}  {\dfrac{{ {1}}}{{ {x}}}} \\   {\dfrac{{ {1}}}{{ {y}}}} \\   {\dfrac{{ {1}}}{{ {z}}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {1200}}}}\left[ {\begin{array}{*{20}{c}}  {{ {75}}}&{{ {150}}}&{{ {75}}} \\   {{ {110}}}&{{ { - 100}}}&{{ {30}}} \\   {{ {72}}}&{ {0}}&{{ { - 24}}}  \end{array}} \right]\left[ {\begin{array}{*{20}{l}}  { {4}} \\   { {1}} \\   { {2}}  \end{array}} \right] \hfill \\  \end{align} \]

\[\begin{align}  \left[ {\begin{array}{*{20}{l}}  {\dfrac{{ {1}}}{{ {x}}}} \\   {\dfrac{{ {1}}}{{ {y}}}} \\   {\dfrac{{ {1}}}{{ {z}}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {1200}}}}\left[ {\begin{array}{*{20}{c}}  {{ {300 + 150 + 150}}} \\   {{ {440 - 100 + 60}}} \\   {{ {288 + 0 - 48}}}  \end{array}} \right] \hfill \\  \left[ {\begin{array}{*{20}{l}}  {\dfrac{{ {1}}}{{ {x}}}} \\   {\dfrac{{ {1}}}{{ {y}}}} \\   {\dfrac{{ {1}}}{{ {z}}}}  \end{array}} \right]{ {  =  }}\dfrac{{ {1}}}{{{ {1200}}}}\left[ {\begin{array}{*{20}{l}}  {{ {600}}} \\   {{ {400}}} \\   {{ {240}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{c}}  {\dfrac{{ {1}}}{{ {2}}}} \\   { {1}} \\   { {3}} \\   { {1}} \\   { {5}}  \end{array}} \right] \hfill \\  \end{align} \]

\[\begin{align}   \Rightarrow \;\dfrac{{ {1}}}{{ {x}}}{ {  =  }}\dfrac{{ {1}}}{{ {2}}}{ {, }}\dfrac{{ {1}}}{{ {y}}}{ {  =  }}\dfrac{{ {1}}}{{ {3}}}{ {, }}\dfrac{{ {1}}}{{ {z}}}{ {  =  }}\dfrac{{ {1}}}{{ {5}}} \hfill \\   \Rightarrow \;{ {x  =  2, y  =  3, z  =  5}} \hfill \\  \end{align} \]

निम्नलिखित प्रश्नों 17 से 19 मे सही उत्तर का चुनाव कीजिए। 

17. यदि \[{ {a, b, c}}\] समातर श्रहई मे हो तो सारणिक \[\left| {\begin{array}{*{20}{c}}  {{ {x + 2}}}&{{ {x + 3}}}&{{ {x + 2a}}} \\   {{ {x + 3}}}&{{ {x + 4}}}&{{ {x + 2b}}} \\   {{ {x + 4}}}&{{ {x + 5}}}&{{ {x + 2c}}}  \end{array}} \right|\] का मान होगा:

(a) \[{ {0}}\]

(b) \[{ {1}}\]

(c) \[{ {x}}\]

(d) \[{ {2x}}\]

उत्तर: \[\left| {\begin{array}{*{20}{c}}  {{ {x + 2}}}&{{ {x + 3}}}&{{ {x + 2a}}} \\   {{ {x + 3}}}&{{ {x + 4}}}&{{ {x + 2b}}} \\   {{ {x + 4}}}&{{ {x + 5}}}&{{ {x + 2c}}}  \end{array}} \right|\]

\[{ { =  }}\left| {\begin{array}{*{20}{c}}  {{ {x + 2}}}&{{ {x + 3}}}&{{ {x + 2a}}} \\   { {0}}&{ {0}}&{{ {2(2b - a - c)}}} \\   {{ {x + 4}}}&{{ {x + 5}}}&{{ {x + 2c}}}  \end{array}} \right|\,\,\,\,\;\;\,\quad \left[ {{{ {R}}_{ {2}}} \to { {2}}{{ {R}}_{ {2}}}{ { - }}} \right.\left( {{{ {R}}_{ {1}}}{ { + }}{{ {R}}_{ {3}}}} \right){ {]}}\]

\[{ { =  }}\left| {\begin{array}{*{20}{c}}  {{ {x + 2}}}&{{ {x + 3}}}&{{ {x + 2a}}} \\   { {0}}&{ {0}}&{ {0}} \\   {{ {x + 4}}}&{{ {x + 5}}}&{{ {x + 2c}}}  \end{array}} \right|\]          [\[{ {a,b,c}}\] समांतर श्रेणी मे है]

\[{ { =  0}}\]

अतः विकल्प (a) सही है। 

18. यदि \[{ {x, y, z}}\] शुनएतर वास्तविक संखयाए हो तो आव्यूह \[{ {A  =  }}\left| {\begin{array}{*{20}{l}}  { {x}}&{ {0}}&{ {0}} \\   { {0}}&{ {y}}&{ {0}} \\   { {0}}&{ {0}}&{ {z}}  \end{array}} \right|\] का वयुक्तकर्म है: 

(a) \[\left[ {\begin{array}{*{20}{c}}  {{{ {x}}^{{ { - 1}}}}}&{ {0}}&{ {0}} \\   { {0}}&{{{ {y}}^{{ { - 1}}}}}&{ {0}} \\   { {0}}&{ {0}}&{{{ {z}}^{{ { - 1}}}}}  \end{array}} \right]\]

(b) \[{ {xyz}}\left[ {\begin{array}{*{20}{c}}  {{{ {x}}^{{ { - 1}}}}}&{ {0}}&{ {0}} \\   { {0}}&{{{ {y}}^{{ { - 1}}}}}&{ {0}} \\   { {0}}&{ {0}}&{{{ {z}}^{{ { - 1}}}}}  \end{array}} \right]\]

(c) \[\dfrac{{ {1}}}{{{ {xyz}}}}\left[ {\begin{array}{*{20}{l}}  { {x}}&{ {0}}&{ {0}} \\   { {0}}&{ {y}}&{ {0}} \\   { {0}}&{ {0}}&{ {z}}  \end{array}} \right]\]

(d) \[\dfrac{{ {1}}}{{{ {xyz}}}}\left[ {\begin{array}{*{20}{l}}  { {1}}&{ {0}}&{ {0}} \\   { {0}}&{ {1}}&{ {0}} \\   { {0}}&{ {0}}&{ {1}}  \end{array}} \right]\]

उत्तर: \[{ {A  =  }}\left[ {\begin{array}{*{20}{l}}  { {x}}&{ {0}}&{ {0}} \\   { {0}}&{ {y}}&{ {0}} \\   { {0}}&{ {0}}&{ {z}}  \end{array}} \right]\]

\[\begin{align}  { {|A|  =  x(yz - 0) - 0(0 - 0) + 0(0 + 0)}} \hfill \\  { { =  xyz }} \ne { { 0 }} \Rightarrow { { }}{{ {A}}^{{ { - 1}}}} \hfill \\  \end{align} \]

\[\begin{array}{*{20}{c}}  {{{ {A}}_{{ {11}}}}{ {  =  yz }}}&{{{ {A}}_{{ {12}}}}{ {  =  0 }}}&{{{ {A}}_{{ {13}}}}{ {  =  0 }}} \\   {{{ {A}}_{{ {21}}}}{ {  =  0 }}}&{{{ {A}}_{{ {22}}}}{ {  =  xz  }}}&{{{ {A}}_{{ {23}}}}{ {  =  0}}} \\   {{{ {A}}_{{ {31}}}}{ {  =  0 }}}&{{{ {A}}_{{ {32}}}}{ {  =  0 }}}&{{{ {A}}_{{ {33}}}}{ {  =  xy }}}  \end{array}\]

\[\begin{align}  {{ {A}}^{{ { - 1}}}}{ {  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {adjA  =  }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}}  {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {13}}}}} \\   {{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {23}}}}} \\   {{{ {A}}_{{ {31}}}}}&{{{ {A}}_{{ {32}}}}}&{{{ {A}}_{{ {33}}}}}  \end{array}} \right] \hfill \\  { { =  }}\dfrac{{ {1}}}{{{ {xyZ}}}}\left[ {\begin{array}{*{20}{c}}  {{ {yz}}}&{ {0}}&{ {0}} \\   { {0}}&{{ {xz}}}&{ {0}} \\   { {0}}&{ {0}}&{{ {xy}}}  \end{array}} \right]{ {  =  }}\left[ {\begin{array}{*{20}{l}}  {\dfrac{{ {1}}}{{ {x}}}}&{ {0}}&{ {0}} \\   { {0}}&{\dfrac{{ {1}}}{{ {y}}}}&{ {0}} \\   { {0}}&{ {0}}&{\dfrac{{ {1}}}{{ {z}}}}  \end{array}} \right] \hfill \\   \end{align} \]

\[{ { =  }}\left[ {\begin{array}{*{20}{c}}  {{{ {x}}^{{ { - 1}}}}}&{ {0}}&{ {0}} \\   { {0}}&{{{ {y}}^{{ { - 1}}}}}&{ {0}} \\   { {0}}&{ {0}}&{{{ {z}}^{{ { - 1}}}}}  \end{array}} \right]\]

अतः विकल्प (a) सही है। 

19. यदि \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{{ {sin\theta }}}&{ {1}} \\   {{ { - sin\theta }}}&{ {1}}&{{ {sin\theta }}} \\   {{ { - 1}}}&{{ { - sin\theta }}}&{ {1}}  \end{array}} \right]\] जहा \[{ {0 }} \leqslant { { \theta  }} \leqslant { { 2\pi }}\] हो तो: 

(a) \[{ {det(A)  =  0}}\]

(b) \[{ {det(A)}} \in { {(2,}}\infty { {)}}\]

(c) \[{ {det(A)}} \in { {(2,4)}}\]

(d) \[{ {det(A)}} \in { {[2,4]}}\]

उत्तर: \[{ {A  =  }}\left[ {\begin{array}{*{20}{c}}  { {1}}&{{ {sin\theta }}}&{ {1}} \\   {{ { - sin\theta }}}&{ {1}}&{{ {sin\theta }}} \\   {{ { - 1}}}&{{ { - sin\theta }}}&{ {1}}  \end{array}} \right]\]

\[\begin{align}  { { =  1}}\left( {{ {1 + si}}{{ {n}}^{ {2}}}{ {\theta }}} \right){ { + sin\theta ( - sin\theta  + sin\theta ) + 1}}\left( {{ {si}}{{ {n}}^{ {2}}}{ {\theta  + 1}}} \right) \hfill \\  { { =  2}}\left( {{ {1 + si}}{{ {n}}^{ {2}}}{ {\theta }}} \right) \hfill \\  \end{align} \] 

अब यह देखते हुए की \[{ {0}} \leqslant { {\theta }} \leqslant { {2\pi }}\]

\[ \Rightarrow { { 0}} \leqslant { {sin\theta }} \leqslant { {1}}\]

\[\begin{align}   \Rightarrow \;{ {0}} \leqslant { {si}}{{ {n}}^{ {2}}}{ {\theta }} \leqslant { {1}} \hfill \\   \Rightarrow \;{ {1}} \leqslant { {1 + si}}{{ {n}}^{ {2}}}{ {\theta }} \leqslant { {2}} \hfill \\   \Rightarrow \;{ {2}} \leqslant { {2}}\left( {{ {1 + si}}{{ {n}}^{ {2}}}{ {\theta }}} \right) \leqslant { {4}} \hfill \\   \Rightarrow \;{ {det(A)}} \in { {[2,4]}} \hfill \\  \end{align} \]

अतः विकल्प (d) सही है। 

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