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# 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

$(a)\;5\qquad(b)\;-1\qquad(c)\;2\qquad(d)\;-2$

$10B=\begin{bmatrix}4 &2&3\\-5&0&\alpha\\1&-2&3\end{bmatrix}$
$\Rightarrow B=\large\frac{1}{10}$$\begin{bmatrix}4 &2&2\\-5&0&\alpha\\1&-2&3\end{bmatrix} Also since B=A^{-1} \Rightarrow AB=I \large\frac{1}{10}$$\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}$
$\Rightarrow \begin{bmatrix}1 &0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow \large\frac{5-\alpha}{10}$$=0$
$\Rightarrow \alpha=5$
Hence (a) is the correct answer.