If B=left[begin{array}{ll}1 & 3 1 & alphaend | vedclass

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(B) We know that for any square matrix $A$,the property relating the adjoint and the determinant is $|\operatorname{adj}(A)| = |A|^{n-1}$,where $n$ is

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$5$

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(B) We know that for any square matrix $A$,the property relating the adjoint and the determinant is $|\operatorname{adj}(A)| = |A|^{n-1}$,where $n$ is the order of the matrix.Here,$n=2$ and $|A|=2$.So,$|\operatorname{adj}(A)| = |A|^{2-1} = |A|^1 = 2$.Given $\operatorname{adj}(A) = B = \left[\begin{array}{ll}1 & 3 \ 1 & \alpha\end{array}ight]$,we calculate the determinant of $B$:$|B| = (1)(\alpha) - (3)(1) = \alpha - 3$.Since $|\operatorname{adj}(A)| = |B|$,we have $\alpha - 3 = 2$.Therefore,$\alpha = 2 + 3 = 5$.

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

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