(B)
We know,
$\begin{array}{l}
\mathrm{A} \cdot \mathrm{A}^{-1}=\mathrm{I} \\
{\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\left[\begin{array}{ccc}
3 & -1 & 1 \\
\alpha & 6 & -5 \\
\beta & -2 & 2
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]} \\
{\left[\begin{array}{ccc}
6+0-\beta & -2+0+2 & 2+0-2 \\
15+\alpha+0 & -5+6+0 & 5-5+0 \\
0+\alpha+3 \beta & 0+6-6 & 0-5+6
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]} \\
{\left[\begin{array}{ccc}
6-\beta & 0 & 0 \\
15+\alpha & 1 & 0 \\
\alpha+3 \beta & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]} \\
6-\beta=1 \Rightarrow \beta=5 \text { and } 15+\alpha=0 \Rightarrow \alpha=-15
\end{array}$
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 $A$.