Expand the determinant: $\left| {\begin{array}{*{20}{c}}
2&{ - 2}&5 \\
4&6&{ - 2} \\
3&{ - 4}&1
\end{array}} \right|$ .
Answer
635.1k+ views
Hint-In this question, we use the concept of property of determinants. We apply some row and column operation to simplify determinants in easy form then after simplify we can easily expand the determinants.
Complete step-by-step answer:
We have a determinant $\left| {\begin{array}{*{20}{c}}
2&{ - 2}&5 \\
4&6&{ - 2} \\
3&{ - 4}&1
\end{array}} \right|$ and we have to expand it. So, before expanding we can simplify determinant by applying some row and column operation then we easily expand the determinant.
Now, we try to make zeros in row and column of determinant by using row and column operation and then we can easily expand the determinant.
We apply, ${C_2} \to {C_2} + {C_1}$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
2&0&5 \\
4&{10}&{ - 2} \\
3&{ - 1}&1
\end{array}} \right|$
Apply, ${C_3} \to {C_3} + {C_2}$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
2&0&5 \\
4&{10}&8 \\
3&{ - 1}&0
\end{array}} \right|$
Apply, ${R_2} \to {R_2} + \left( { - 2} \right){R_1}$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
2&0&5 \\
0&{10}&{ - 2} \\
3&{ - 1}&0
\end{array}} \right|$
Now we expand determinant,
$
\Rightarrow 2\left[ {10 \times 0 - \left( { - 2} \right) \times \left( { - 1} \right)} \right] - 0\left[ {0 - \left( { - 2} \right) \times 3} \right] + 5\left[ {0 \times \left( { - 1} \right) - 10 \times 3} \right] \\
\Rightarrow 2\left( { - 2} \right) - 0 + 5\left( { - 30} \right) \\
\Rightarrow - 4 - 150 \\
\Rightarrow - 154 \\
$
So, the answer of determinant $\left| {\begin{array}{*{20}{c}}
2&{ - 2}&5 \\
4&6&{ - 2} \\
3&{ - 4}&1
\end{array}} \right|$ is -154.
Note-In such types of problems we use some important points to solve questions in an easy way. First we try to make zeros in row and column of determinant by using row and column operation and then we can easily expand the determinant. In such types of problems if we make a maximum number of zeros in row and column so we can easily expand the determinant.
Complete step-by-step answer:
We have a determinant $\left| {\begin{array}{*{20}{c}}
2&{ - 2}&5 \\
4&6&{ - 2} \\
3&{ - 4}&1
\end{array}} \right|$ and we have to expand it. So, before expanding we can simplify determinant by applying some row and column operation then we easily expand the determinant.
Now, we try to make zeros in row and column of determinant by using row and column operation and then we can easily expand the determinant.
We apply, ${C_2} \to {C_2} + {C_1}$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
2&0&5 \\
4&{10}&{ - 2} \\
3&{ - 1}&1
\end{array}} \right|$
Apply, ${C_3} \to {C_3} + {C_2}$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
2&0&5 \\
4&{10}&8 \\
3&{ - 1}&0
\end{array}} \right|$
Apply, ${R_2} \to {R_2} + \left( { - 2} \right){R_1}$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
2&0&5 \\
0&{10}&{ - 2} \\
3&{ - 1}&0
\end{array}} \right|$
Now we expand determinant,
$
\Rightarrow 2\left[ {10 \times 0 - \left( { - 2} \right) \times \left( { - 1} \right)} \right] - 0\left[ {0 - \left( { - 2} \right) \times 3} \right] + 5\left[ {0 \times \left( { - 1} \right) - 10 \times 3} \right] \\
\Rightarrow 2\left( { - 2} \right) - 0 + 5\left( { - 30} \right) \\
\Rightarrow - 4 - 150 \\
\Rightarrow - 154 \\
$
So, the answer of determinant $\left| {\begin{array}{*{20}{c}}
2&{ - 2}&5 \\
4&6&{ - 2} \\
3&{ - 4}&1
\end{array}} \right|$ is -154.
Note-In such types of problems we use some important points to solve questions in an easy way. First we try to make zeros in row and column of determinant by using row and column operation and then we can easily expand the determinant. In such types of problems if we make a maximum number of zeros in row and column so we can easily expand the determinant.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What are the major means of transport Explain each class 12 social science CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

Why should a magnesium ribbon be cleaned before burning class 12 chemistry CBSE

