Evaluate the determinant: left|begin{array}{r | vedclass

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(C) Let $A = \left|\begin{array}{rrr}3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0\end{array}ight|$.It can be observed that in the second row,two entries

Vedclass · Accepted Answer

$-12$

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(C) Let $A = \left|\begin{array}{rrr}3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0\end{array}ight|$.It can be observed that in the second row,two entries are zero. Thus,we expand along the second row for easier calculation.The expansion along the second row is given by:$|A| = -a_{21} M_{21} + a_{22} M_{22} - a_{23} M_{23}$$|A| = -0 \left|\begin{array}{cc}-1 & -2 \ -5 & 0\end{array}ight| + 0 \left|\begin{array}{cc}3 & -2 \ 3 & 0\end{array}ight| - (-1) \left|\begin{array}{cc}3 & -1 \ 3 & -5\end{array}ight|$$|A| = 0 + 0 + 1 	imes ((3 	imes -5) - (-1 	imes 3))$$|A| = 1 	imes (-15 - (-3))$$|A| = 1 	imes (-15 + 3)$$|A| = -12$

Evaluate the determinant: $\left|\begin{array}{rrr}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$

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