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(A) Let $A = \begin{bmatrix} 0 & 1 & 2 \ 1 & 2 & 3 \ 3 & 1 & 1 \end{bmatrix}$.First,we find the determinant $|A|$:$|A| = 0(2-3) - 1(1-9) + 2(1-6) =

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$\begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \ -4 & 3 & -1 \ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}$

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(A) Let $A = \begin{bmatrix} 0 & 1 & 2 \ 1 & 2 & 3 \ 3 & 1 & 1 \end{bmatrix}$.First,we find the determinant $|A|$:$|A| = 0(2-3) - 1(1-9) + 2(1-6) = 0 - 1(-8) + 2(-5) = 8 - 10 = -2$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.Next,we find the matrix of cofactors $C_{ij}$:$C_{11} = +(2-3) = -1, C_{12} = -(1-9) = 8, C_{13} = +(1-6) = -5$$C_{21} = -(1-2) = 1, C_{22} = +(0-6) = -6, C_{23} = -(0-3) = 3$$C_{31} = +(3-4) = -1, C_{32} = -(0-2) = 2, C_{33} = +(0-1) = -1$The adjoint matrix $adj(A)$ is the transpose of the cofactor matrix:$adj(A) = \begin{bmatrix} -1 & 1 & -1 \ 8 & -6 & 2 \ -5 & 3 & -1 \end{bmatrix}$.Finally,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-2} \begin{bmatrix} -1 & 1 & -1 \ 8 & -6 & 2 \ -5 & 3 & -1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \ -4 & 3 & -1 \ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}$.

The inverse matrix of $\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$ is

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