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

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

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$-42$

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(B) First,we calculate the determinant of matrix $A$: $|A| = 1((-1)(1) - (3)(2)) - (-3)((-2)(-1) - (3)(3)) + 2((-2)(2) - (1)(3))$ $|A| = 1(-1 - 6) + 3(2 - 9) + 2(-4 - 3)$ $|A| = 1(-7) + 3(-7) + 2(-7) = -7 - 21 - 14 = -42$. We know the property $A \cdot \operatorname{Adj} A = |A| I$,where $I$ is the identity matrix. Therefore,$A^2 \operatorname{Adj} A = A(A \operatorname{Adj} A) = A(|A| I) = |A| A$. Substituting the value of $|A|$,we get $A^2 \operatorname{Adj} A = -42 A$.

If $A = \begin{bmatrix} 1 & -3 & 2 \\ -2 & 1 & 3 \\ 3 & 2 & -1 \end{bmatrix}$,then $A^2 \operatorname{Adj} A = $ (in $I$)

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