If \(A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]\) and \(A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2\end{array}\right]\), then the values of \(\alpha\) and \(\beta\) are. respectively

(B)
We know,
\(A \cdot A ^{-1}= I\)
\(\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)
\(\left[\begin{array}{ccc}6+0-\beta & -2+0+2 & 2+0-2 \\ 15+\alpha+0 & -5+6+0 & 5-5+0 \\ 0+\alpha+3 \beta & 0+6-6 & 0-5+6\end{array}\right]\) \(=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)
\(\left[\begin{array}{ccc}6-\beta & 0 & 0 \\ 15+\alpha & 1 & 0 \\ \alpha+3 \beta & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)
\(6-\beta=1 \Rightarrow \beta=5\) and \(15+\alpha=0 \Rightarrow \alpha=-15\)
This problem can also be solved as follows :
We have \(A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right] \Rightarrow|A|=2(3)-(5)=1 \alpha\) is \(a_{21}\) in \(A^{-1}\). So we will find cofactor of \(a_{12}\) in \(A\).