if $2$ is root of equation $|A-xI|=0$ (where $A$ is non-singular matrix), $\frac{|A|}{2}$ root of equation $|B-xI|=0$ then $B$ can be

Solution 1:

$$ \begin{align} &\ \det(A-2I) = 0 \\ \implies &\ \det\left(\frac{A}{2} - I\right) = 0 \\ \implies &\ \det\left(\frac{A\det(A)}{2} - \det(A) I\right) = 0 ~~(\because \det(A) \ne 0)\\ \implies &\ \det \left(\frac{A\det(A)}{2} - A ~\text{adj} (A)\right) = 0 \\ \implies &\ \det(A)\det\left(\frac{\det(A)}{2}I - \text{adj} (A)\right) = 0 \\ \implies &\ \det\left(\frac{\det(A)}{2}I - \text{adj} (A)\right) = 0 ~~(\because \det(A) \ne 0) \\ \implies &\ \det\left(\text{adj} (A)-\frac{\det(A)}{2}I\right) = 0 \\ \end{align}$$ Therefore, $\frac{\det(A)}{2}$ is a root of the equation $\det(B-xI) = 0$ where $B = \text{adj}(A)$