Answer
Verified
396.9k+ views
Hint: We prove the question using the formula for determinant of order $3 \times 3$ \[\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right| = a\left| {\begin{array}{*{20}{c}}
e&f \\
h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
d&f \\
g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
d&e \\
g&h
\end{array}} \right|.\]
Then use the following formula to calculate the determinant of order $2 \times 2.$
\[\left| {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right| = ad - bc.\]
Foil method: The multiplication of $(x + y)(u + v)$ is given by $xu + xv + yu + yv.$
Complete step-by-step answer:
We are given\[\left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right|.\]
Using the formula\[\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right| = a\left| {\begin{array}{*{20}{c}}
e&f \\
h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
d&f \\
g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
d&e \\
g&h
\end{array}} \right|\] expand the given determinant.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right| = (1 + a)\left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right|\] $...(1)$
Now we will calculate values of each determinant of order $2 \times 2$ separately and then substitute it back in equation (1).
Calculate the value of \[\left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right|\] using \[\left| {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right| = ad - bc\] formula.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right| = (1 + b)(1 + c) - (1)(1)\]
Find the multiplication of \[(1 + b)(1 + c)\] by using foil method.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right| = (1)(1) + (1)(c) + b(1) + bc - 1\]
Simplify by doing multiplication.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right| = 1 + c + b + bc - 1\]
Simplify by adding like terms.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right| = c + b + bc\] $...(2)$
Calculate the value of \[\left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right|\] using \[\left| {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right| = ad - bc\] formula.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right| = (1)(1 + c) - (1)(1)\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right| = 1 + c - 1\]
Simplify by adding like terms.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right| = c\] $...(3)$
Calculate the value of \[\left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right|\] using \[\left| {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right| = ad - bc\] formula.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right| = (1)(1) - (1 + b)(1)\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right| = 1 - (1 + b)\]
Simplify by distributing negative signs over parenthesis.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right| = 1 - 1 - b\]
Simplify by adding like terms.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right| = - b\] $...(4)$
Step 5: Substitute values of \[\left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right|,\left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right|\] and \[\left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right|\] from equation$(2), (3)$ and $(4)$ in equation \[\left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right| = (1 + a)\left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right|\].
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right| = (1 + a)(c + b + bc) - 1(c) + 1( - b)\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right| = (1)(c + b + bc) + a(c + b + bc) - c - b\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right| = c + b + bc + ac + ab + abc - c - b\]
Simplify adding like terms.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right| = ab + bc + ca + abc\]
Hence, proved.
Additional Information: * Any matrix with a row having all elements as 0 will have its determinant is 0.
* The determinant of a diagonal matrix is the product of its diagonal entries.
* The determinant of a matrix is 0 if and only if its rows are linearly dependent, which means elements of a row can be written as a linear combination of elements of another row. If the rows are linearly independent, then the determinant is non-zero.
Note: Students can easily get confused while calculating the determinant if they don’t break it into three parts of $2 \times 2$ matrices. Also, many students make the mistake of not writing negative signs along with the second term of the determinant. Keep in mind we move in an alternate sign way, we take first value positive, then second negative and then third value again positive.
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right| = a\left| {\begin{array}{*{20}{c}}
e&f \\
h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
d&f \\
g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
d&e \\
g&h
\end{array}} \right|.\]
Then use the following formula to calculate the determinant of order $2 \times 2.$
\[\left| {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right| = ad - bc.\]
Foil method: The multiplication of $(x + y)(u + v)$ is given by $xu + xv + yu + yv.$
Complete step-by-step answer:
We are given\[\left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right|.\]
Using the formula\[\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right| = a\left| {\begin{array}{*{20}{c}}
e&f \\
h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
d&f \\
g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
d&e \\
g&h
\end{array}} \right|\] expand the given determinant.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right| = (1 + a)\left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right|\] $...(1)$
Now we will calculate values of each determinant of order $2 \times 2$ separately and then substitute it back in equation (1).
Calculate the value of \[\left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right|\] using \[\left| {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right| = ad - bc\] formula.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right| = (1 + b)(1 + c) - (1)(1)\]
Find the multiplication of \[(1 + b)(1 + c)\] by using foil method.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right| = (1)(1) + (1)(c) + b(1) + bc - 1\]
Simplify by doing multiplication.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right| = 1 + c + b + bc - 1\]
Simplify by adding like terms.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right| = c + b + bc\] $...(2)$
Calculate the value of \[\left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right|\] using \[\left| {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right| = ad - bc\] formula.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right| = (1)(1 + c) - (1)(1)\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right| = 1 + c - 1\]
Simplify by adding like terms.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right| = c\] $...(3)$
Calculate the value of \[\left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right|\] using \[\left| {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right| = ad - bc\] formula.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right| = (1)(1) - (1 + b)(1)\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right| = 1 - (1 + b)\]
Simplify by distributing negative signs over parenthesis.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right| = 1 - 1 - b\]
Simplify by adding like terms.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right| = - b\] $...(4)$
Step 5: Substitute values of \[\left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right|,\left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right|\] and \[\left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right|\] from equation$(2), (3)$ and $(4)$ in equation \[\left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right| = (1 + a)\left| {\begin{array}{*{20}{c}}
{1 + b}&1 \\
1&{1 + c}
\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}
1&1 \\
1&{1 + c}
\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}
1&{1 + b} \\
1&1
\end{array}} \right|\].
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right| = (1 + a)(c + b + bc) - 1(c) + 1( - b)\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right| = (1)(c + b + bc) + a(c + b + bc) - c - b\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right| = c + b + bc + ac + ab + abc - c - b\]
Simplify adding like terms.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + a}&1&1 \\
1&{1 + b}&1 \\
1&1&{1 + c}
\end{array}} \right| = ab + bc + ca + abc\]
Hence, proved.
Additional Information: * Any matrix with a row having all elements as 0 will have its determinant is 0.
* The determinant of a diagonal matrix is the product of its diagonal entries.
* The determinant of a matrix is 0 if and only if its rows are linearly dependent, which means elements of a row can be written as a linear combination of elements of another row. If the rows are linearly independent, then the determinant is non-zero.
Note: Students can easily get confused while calculating the determinant if they don’t break it into three parts of $2 \times 2$ matrices. Also, many students make the mistake of not writing negative signs along with the second term of the determinant. Keep in mind we move in an alternate sign way, we take first value positive, then second negative and then third value again positive.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE