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(C) Given $B = A^{-1}$,we have $AB = I$,where $I$ is the identity matrix. Given $10B = \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3 \end

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$5$

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(C) Given $B = A^{-1}$,we have $AB = I$,where $I$ is the identity matrix. Given $10B = \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3 \end{bmatrix}$,so $B = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3 \end{bmatrix}$. Since $AB = I$,we have $A(10B) = 10I$. $\begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 10 & 0 & 0 \ 0 & 10 & 0 \ 0 & 0 & 10 \end{bmatrix}$. We need to find $\alpha$. Let us multiply the second row of $A$ with the third column of $10B$: $(2)(2) + (1)(\alpha) + (-3)(3) = 0$. $4 + \alpha - 9 = 0$. $\alpha - 5 = 0$. $\alpha = 5$.

If $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$,$10B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$ and $B$ is the inverse of matrix $A$,then $\alpha$ is equal to . . . . . . .

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