Let $A=\left(\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right)$ and $10B=\left(\begin{array}{ccc}4& 2& 2\\ -5& 0& a\\ 1& -2& 3\end{array}\right).$ If $B$ is the inverse of matrix $A,$ then $a$ is

Since $B$ is the inverse of $A,$ $AB=I,$ which means $A(10B)=10I$.
$$\left(\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right)\left(\begin{array}{ccc}4& 2& 2\\ -5& 0& a\\ 1& -2& 3\end{array}\right)=\left(\begin{array}{ccc}10& 0& 0\\ 0& 10& 0\\ 0& 0& 10\end{array}\right)$$
.
Multiplying the first row of $A$ and the third column of $10B$:
$(1)(2)+(-1)(a)+(1)(3)=0$.
$2-a+3=0$
$5-a=0$
$a=5$.