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(D) We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix of order $3 	imes 3$. Given $A = \begin{bmatrix} 0 & 1 & 2 \ 1 & 2 & 3 \ 3 &

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$5$

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(D) We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix of order $3 	imes 3$. Given $A = \begin{bmatrix} 0 & 1 & 2 \ 1 & 2 & 3 \ 3 & 1 & 1 \end{bmatrix}$ and $A^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & 1 & -1 \ 8 & -6 & 2 \ -x & 3 & -1 \end{bmatrix}$. Let $B = \begin{bmatrix} -1 & 1 & -1 \ 8 & -6 & 2 \ -x & 3 & -1 \end{bmatrix}$. Then $A^{-1} = -\frac{1}{2} B$. Thus,$A \cdot (-\frac{1}{2} B) = I$,which implies $A \cdot B = -2I = \begin{bmatrix} -2 & 0 & 0 \ 0 & -2 & 0 \ 0 & 0 & -2 \end{bmatrix}$. Multiplying the third row of $A$ with the first column of $B$ to find the element at position $(3, 1)$ of the product matrix: $(3 	imes -1) + (1 	imes 8) + (1 	imes -x) = -2$. $-3 + 8 - x = -2$. $5 - x = -2$. $x = 5 + 2 = 7$. Wait,let us re-calculate the determinant $|A|$: $|A| = 0(2-3) - 1(1-9) + 2(1-6) = 0 - 1(-8) + 2(-5) = 8 - 10 = -2$. Since $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$,the matrix $B$ must be $	ext{adj}(A)$. The element at $(3, 1)$ of $	ext{adj}(A)$ is the cofactor $C_{13}$ of matrix $A$. $C_{13} = (-1)^{1+3} \begin{vmatrix} 1 & 2 \ 3 & 1 \end{vmatrix} = 1(1 - 6) = -5$. Comparing this with the matrix $B$,the element at $(3, 1)$ is $-x$. So,$-x = -5$,which means $x = 5$. Therefore,the correct option is $D$.

If matrix $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$ and the inverse of matrix $A$ is $A^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -x & 3 & -1 \end{bmatrix}$,then find the value of $x$.

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