Answer
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417.3k+ 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.
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