If A = begin{bmatrix} 1 & 5 & 2 4 & 1 & 3 2 | vedclass

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(B) First, we find the determinant of matrix $A$: $|A| = 1(1 	imes 3 - 3 	imes 6) - 5(4 	imes 3 - 3 	imes 2) + 2(4 	imes 6 - 1 	imes 2)$ $|A| =

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(B) First, we find the determinant of matrix $A$: $|A| = 1(1 	imes 3 - 3 	imes 6) - 5(4 	imes 3 - 3 	imes 2) + 2(4 	imes 6 - 1 	imes 2)$ $|A| = 1(3 - 18) - 5(12 - 6) + 2(24 - 2)$ $|A| = 1(-15) - 5(6) + 2(22)$ $|A| = -15 - 30 + 44 = -1$ We know that $|\operatorname{Adj} A| = |A|^{n-1}$, where $n$ is the order of the matrix. Here $n=3$, so $|\operatorname{Adj} A| = |A|^{3-1} = |A|^2 = (-1)^2 = 1$. We need to find $|(\operatorname{Adj} A)^{-1}|$. Using the property $|B^{-1}| = \frac{1}{|B|}$, we have $|(\operatorname{Adj} A)^{-1}| = \frac{1}{|\operatorname{Adj} A|} = \frac{1}{1} = 1$.

If $A = \begin{bmatrix} 1 & 5 & 2 \\ 4 & 1 & 3 \\ 2 & 6 & 3 \end{bmatrix}$, then $|(\operatorname{Adj} A)^{-1}| = $

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