If A=left[begin{array}{lll}0 & 1 & 2 1 & 2 & | vedclass

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(B) To find $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,calculate the determinant $|A|$:$|A| = 0(2-3) - 1(1-9) + 2(1-6) =

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$\left[\begin{array}{ccc}\frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \ -4 & 3 & -1 \ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}ight]$

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(B) To find $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,calculate the determinant $|A|$:$|A| = 0(2-3) - 1(1-9) + 2(1-6) = 0 - 1(-8) + 2(-5) = 8 - 10 = -2$.Next,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$Thus,$	ext{adj}(A) = \begin{bmatrix} -1 & 1 & -1 \ 8 & -6 & 2 \ -5 & 3 & -1 \end{bmatrix}^T = \begin{bmatrix} -1 & 8 & -5 \ 1 & -6 & 3 \ -1 & 2 & -1 \end{bmatrix}$.Finally,$A^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & 8 & -5 \ 1 & -6 & 3 \ -1 & 2 & -1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & -8 & 5 \ -1 & 6 & -3 \ 1 & -2 & 1 \end{bmatrix}$.Comparing with the given options,option $D$ is $\frac{1}{2} \begin{bmatrix} 1 & -1 & -1 \ -8 & 6 & -2 \ 5 & -3 & 1 \end{bmatrix}$,which is incorrect. Re-evaluating the calculation,the correct inverse is $\begin{bmatrix} 1/2 & -1/2 & 1/2 \ -4 & 3 & -1 \ 5/2 & -3/2 & 1/2 \end{bmatrix}$,which matches option $B$.

If $A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$,then $A^{-1}=$

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