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

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

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$5$

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(D) Given $B = A^{-1}$,we have $AB = I$,where $I$ is the identity matrix.$B = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & \alpha \ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 0.4 & 0.2 & 0.2 \ -0.5 & 0 & \alpha/10 \ 0.1 & -0.2 & 0.3 \end{bmatrix}$.For $AB = I$,the product of the second row of $A$ and the third column of $B$ must be $0$ (since it is the element at position $(2,3)$ of the identity matrix).The second row of $A$ is $(2, 1, -3)$ and the third column of $B$ is $\begin{bmatrix} 0.2 \ \alpha/10 \ 0.3 \end{bmatrix}$.Thus,$(2)(0.2) + (1)(\alpha/10) + (-3)(0.3) = 0$.$0.4 + \frac{\alpha}{10} - 0.9 = 0$.$\frac{\alpha}{10} - 0.5 = 0$.$\frac{\alpha}{10} = 0.5$.$\alpha = 5$.

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

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