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

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(B) Let $A = \left|\begin{array}{ccc}2 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0\end{array}ight|$.Expanding along the first row,we have:$|A| = 2 \left|\

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$5$

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(B) Let $A = \left|\begin{array}{ccc}2 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0\end{array}ight|$.Expanding along the first row,we have:$|A| = 2 \left|\begin{array}{cc} 2 & -1 \ -5 & 0 \end{array}ight| - (-1) \left|\begin{array}{cc} 0 & -1 \ 3 & 0 \end{array}ight| + (-2) \left|\begin{array}{cc} 0 & 2 \ 3 & -5 \end{array}ight|$.Calculating the $2 	imes 2$ determinants:$|A| = 2(2 	imes 0 - (-1) 	imes (-5)) + 1(0 	imes 0 - (-1) 	imes 3) - 2(0 	imes (-5) - 2 	imes 3)$.$|A| = 2(0 - 5) + 1(0 + 3) - 2(0 - 6)$.$|A| = 2(-5) + 1(3) - 2(-6)$.$|A| = -10 + 3 + 12$.$|A| = 5$.

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

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