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

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(C) Given $A = \begin{bmatrix} 1 & 2 & 3 \ 1 & 3 & 5 \ 2 & 1 & 6 \end{bmatrix}$.First, calculate the determinant $|A| = 1(18-5) - 2(6-10) + 3(1-6) =

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$\frac{1}{36} \begin{bmatrix} 13 & -9 & 1 \ 4 & 0 & -2 \ -5 & 3 & 1 \end{bmatrix}$

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(C) Given $A = \begin{bmatrix} 1 & 2 & 3 \ 1 & 3 & 5 \ 2 & 1 & 6 \end{bmatrix}$.First, calculate the determinant $|A| = 1(18-5) - 2(6-10) + 3(1-6) = 13 + 8 - 15 = 6$.We know the property $\operatorname{Adj}(\operatorname{Adj} A) = |A|^{n-2} A$, where $n$ is the order of the matrix.Here $n = 3$, so $\operatorname{Adj}(\operatorname{Adj} A) = |A|^{3-2} A = |A| A = 6A$.Therefore, $(\operatorname{Adj}(\operatorname{Adj} A))^{-1} = (6A)^{-1} = \frac{1}{6} A^{-1}$.Since $A^{-1} = \frac{1}{|A|} \operatorname{Adj} A$, we first find $\operatorname{Adj} A$:$\operatorname{Adj} A = \begin{bmatrix} 13 & -9 & 1 \ 4 & 0 & -2 \ -5 & 3 & 1 \end{bmatrix}$.Thus, $A^{-1} = \frac{1}{6} \begin{bmatrix} 13 & -9 & 1 \ 4 & 0 & -2 \ -5 & 3 & 1 \end{bmatrix}$.Finally, $(\operatorname{Adj}(\operatorname{Adj} A))^{-1} = \frac{1}{6} A^{-1} = \frac{1}{6} \left( \frac{1}{6} \begin{bmatrix} 13 & -9 & 1 \ 4 & 0 & -2 \ -5 & 3 & 1 \end{bmatrix} ight) = \frac{1}{36} \begin{bmatrix} 13 & -9 & 1 \ 4 & 0 & -2 \ -5 & 3 & 1 \end{bmatrix}$.

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

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