Given, $A=\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right]$

Then, $\left|A\right|=1\left(1+3\right)+1\left(2+3\right)+1\left(2-1\right)$

$\Rightarrow \left|A\right|=10$ 

Also, $C_{11}=4, C_{12}=-5, C_{13}=1, C_{21}=2, C_{22}=0, C_{23}=-2, C_{31}=2, C_{32}=5, C_{33}=3$

And $\mathrm{Adj}\left(\mathrm{A}\right)=\left[\begin{array}{ccc}{\mathrm{C}}_{11}& {\mathrm{C}}_{21}& {\mathrm{C}}_{31}\\ {\mathrm{C}}_{12}& {\mathrm{C}}_{22}& {\mathrm{C}}_{32}\\ {\mathrm{C}}_{13}& {\mathrm{C}}_{23}& {\mathrm{C}}_{33}\end{array}\right]=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]$

We know that, $A^{-1}=\frac{1}{\left|A\right|}·\mathrm{Adj}\left(\mathrm{A}\right)$

$\Rightarrow A^{-1}=\frac{1}{10}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]$

Also given that, $B=A^{-1}$ and $10B=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]$

$\Rightarrow \alpha =5$