Given, $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$
and $\quad 10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$
Since, $B$ is the inverse of $A$,
i.e. $\quad B=A^{-1}$
So, $\quad 10 A^{-1}=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$
$\Rightarrow 10 A^{-1} A=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right] A$
$\Rightarrow \quad 10 I=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10\end{array}\right]=\left[\begin{array}{ccc}10 & 0 & 0 \\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0 & 0 & 10\end{array}\right]$
$\Rightarrow \quad-5+\alpha=0$
$\Rightarrow \quad \alpha=5$
Alternative $\quad A^{-1}=\frac{1}{|A|} \operatorname{adj}(A)$
$\operatorname{adj}(A)=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
$|A|=1[1+3]+1[2+3]+1[2-1]=0$
$\therefore \quad A^{-1}=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
$10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$
$$
\begin{aligned}
&=\left[\begin{array}{ccc}
4 & 2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right] \\
\Rightarrow \quad \alpha &=5
\end{aligned}
$$