If Q is the inverse of A,where A = begin{bmat | vedclass

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(D) Given $Q = A^{-1}$.First,calculate the determinant of $A$:$|A| = 1(1 - (-3)) - (-1)(2 - (-3)) + 1(2 - 1)$$|A| = 1(4) + 1(5) + 1(1) = 4 + 5 + 1 = 1

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$5$

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(D) Given $Q = A^{-1}$.First,calculate the determinant of $A$:$|A| = 1(1 - (-3)) - (-1)(2 - (-3)) + 1(2 - 1)$$|A| = 1(4) + 1(5) + 1(1) = 4 + 5 + 1 = 10$.Next,find the matrix of cofactors $C_{ij}$:$C_{11} = +(1+3) = 4, C_{12} = -(2+3) = -5, C_{13} = +(2-1) = 1$$C_{21} = -(-1-1) = 2, C_{22} = +(1-1) = 0, C_{23} = -(1+1) = -2$$C_{31} = +(3-1) = 2, C_{32} = -(-3-2) = 5, C_{33} = +(1+2) = 3$Thus,$	ext{Adj } A = \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3 \end{bmatrix}$.Since $A^{-1} = \frac{1}{|A|} 	ext{Adj } A$,we have:$Q = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3 \end{bmatrix}$.Multiplying by $10$,we get $10Q = \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3 \end{bmatrix}$.Comparing this with the given $10Q = \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & x \ 1 & -2 & 3 \end{bmatrix}$,we find $x = 5$.Therefore,option $(d)$ is correct.

If $Q$ is the inverse of $A$,where $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$ and $10Q = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & x \\ 1 & -2 & 3 \end{bmatrix}$,find $x$.

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