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

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(B) First, we calculate the determinant of $A$: $|A| = 1(4-3) - 2(6-3) + 2(3-2) = 1(1) - 2(3) + 2(1) = 1 - 6 + 2 = -3$. Next, we find the adjoint of $

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$\frac{1}{3}$

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(B) First, we calculate the determinant of $A$: $|A| = 1(4-3) - 2(6-3) + 2(3-2) = 1(1) - 2(3) + 2(1) = 1 - 6 + 2 = -3$. Next, we find the adjoint of $A$, $adj(A)$. The matrix of cofactors is: $C_{11} = (4-3) = 1, C_{12} = -(6-3) = -3, C_{13} = (3-2) = 1$ $C_{21} = -(4-2) = -2, C_{22} = (2-2) = 0, C_{23} = -(1-2) = 1$ $C_{31} = (6-4) = 2, C_{32} = -(3-6) = 3, C_{33} = (2-6) = -4$ Thus, $adj(A) = \begin{bmatrix} 1 & -2 & 2 \ -3 & 0 & 3 \ 1 & 1 & -4 \end{bmatrix}^T = \begin{bmatrix} 1 & -3 & 1 \ -2 & 0 & 1 \ 2 & 3 & -4 \end{bmatrix}$. Then $A^{-1} = \frac{1}{|A|} adj(A) = -\frac{1}{3} \begin{bmatrix} 1 & -3 & 1 \ -2 & 0 & 1 \ 2 & 3 & -4 \end{bmatrix}$. The sum of all elements of $A^{-1}$ is: $-\frac{1}{3} (1 - 3 + 1 - 2 + 0 + 1 + 2 + 3 - 4) = -\frac{1}{3} (-1) = \frac{1}{3}$.

If $A=\begin{bmatrix} 1 & 2 & 2 \\ 3 & 2 & 3 \\ 1 & 1 & 2 \end{bmatrix}$ and $A^{-1}=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$, then $\sum_{1 \leq i, j \leq 3} a_{ij} =$

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