If \(A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right], 10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]\) and \(B\) is the inverse of \(A\), then the value of \(\alpha\) is

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 \(\quad A^{-1}=\frac{1}{|A|} \operatorname{adj}(A)\)
\(\operatorname{adj}(A)=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\)
\(|A|=1[1+3]+1[2+3]+1[2-1]=0\)
\(\therefore \quad A^{-1}=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\)
\(10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]\)
\[
\begin{aligned}
&=\left[\begin{array}{ccc}
4 & 2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right] \\
\Rightarrow \quad \alpha &=5
\end{aligned}
\]