Question

Find the determinant:--
$\red{\left|\begin{array}{ccc}6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2\end{array}\right|}$
Answer with proper explanation.

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Answer 1

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given determinant is

$\rm :\longmapsto\:{\left|\begin{array}{ccc}6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2\end{array}\right|}$

$\red{\rm :\longmapsto\:\boxed{\tt{ OP \: R_1 \: \to \: R_1 - 3R_2 \: }}}$

So, we get

$\rm \:  =  \: {\left|\begin{array}{ccc}0 & 0 & - 4 \\ 2 & -1 & 2 \\ -10 & 5 & 2\end{array}\right|}$

$\red{\rm :\longmapsto\:\boxed{\tt{ OP \: R_3 \: \to \: R_3 + 5R_2 \: }}}$

So, we get

$\rm \:  =  \: {\left|\begin{array}{ccc}0 & 0 & - 4 \\ 2 & -1 & 2 \\ 0 & 0 & 12\end{array}\right|}$

$\red{\rm :\longmapsto\:\boxed{\tt{ OP \: R_3 \: \to \: R_3 + 4R_1 \: }}}$

$\rm \:  =  \: {\left|\begin{array}{ccc}0 & 0 & - 4 \\ 2 & -1 & 2 \\ 0 & 0 & 0\end{array}\right|}$

Since, elements of third row, all are 0, so determinant value is 0.

$\rm \:  =  \: 0$

Hence,

$\rm \implies\:\boxed{\tt{ \: \: \: \: {\left|\begin{array}{ccc}6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2\end{array}\right|} = 0 \: \: \: }}$

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Additional Information

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.