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

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(C) Given,$A = \begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix}$.First,we calculate the determinant $|A|$:$|A| = 1(1 - (-3)) - (-1)

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$5$

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(C) Given,$A = \begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix}$.First,we calculate the determinant $|A|$:$|A| = 1(1 - (-3)) - (-1)(2 - (-3)) + 1(2 - 1)$$|A| = 1(4) + 1(5) + 1(1) = 4 + 5 + 1 = 10$.Next,we find the cofactor matrix $C$ of $A$:$C_{11} = (1 - (-3)) = 4, C_{12} = -(2 - (-3)) = -5, C_{13} = (2 - 1) = 1$$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$.Thus,$Adj(A) = C^T = \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3 \end{bmatrix}$.Since $B = A^{-1} = \frac{1}{|A|} Adj(A)$,we have $10 B = Adj(A)$.Comparing $10 B = \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3 \end{bmatrix}$ with $Adj(A) = \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3 \end{bmatrix}$,we get $\alpha = 5$.

If $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$,$10 B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$ and $B = A^{-1}$,then the value of $\alpha$ is:

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