If \(B=\left[\begin{array}{ll}1 & 3 \\ 1 & \alpha\end{array}\right]\) be the adjoint of a matrix \(A\) and \(|A|=2\), then the value of \(\alpha\) is

For a \(2 \times 2\) matrix \(A\),
\(\operatorname{adj}(A)=|A| \cdot A^{-1} \Rightarrow A^{-1}=\frac{1}{|A|} \cdot \operatorname{adj}(A)\)
So:
\(A^{-1}=\frac{1}{2}\left[\begin{array}{ll}
1 & 3 \\
1 & \alpha
\end{array}\right]\)
Now, recall for a \(2 \times 2\) matrix:
\(A^{-1}=\frac{1}{\operatorname{det}(A)}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]\)
So, comparing:
\(\frac{1}{2}\left[\begin{array}{ll}
1 & 3 \\
1 & \alpha
\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right] \Rightarrow \operatorname{adj}(A)=\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]\)
Matching entries:
- \(d=1\)
- \(-b=3 \Rightarrow b=-3\)
- \(-c=1 \Rightarrow c=-1\)
- \(a=\alpha\)
Now compute:
\(|A|=a d-b c=\alpha(1)-(-3)(-1)=\alpha-3\)
Given \(|A|=2\), so:
\(\alpha-3=2 \Rightarrow \alpha=5\)