Question

If $\alpha$ is a characteristic root of a non-singular matrix $A$, then characteristic root of $adjA$ is

Answer 1

Since $\alpha$ is a characteristic root of a non-singular matrix, therefore $\alpha \ne 0.$
Also $\alpha$ is a characteristic root of $A$ implies that there exists a non-zero vector $X$ such that $AX=\alpha X$
$\Rightarrow \left(adjA\right)\left(AX\right)=\left(adjA\right)\left(\alpha X\right)\Rightarrow \left[\right(adjA\left)A\right]X=\alpha \left(adjA\right)X\Rightarrow |A|IX=\alpha \left(adjA\right)X[\because (adjA)A=|A|I]$ $\Rightarrow |A|X=\alpha \left(adjA\right)X\Rightarrow \frac{|A|}{\alpha }X=\left(adjA\right)X\Rightarrow \left(adjA\right)X=\frac{|A|}{\alpha }X$
Since $X$ is a non-zero vector, therefore $\left|\frac{A}{\alpha }\right|$ is the characteristic root of the matrix $adjA$.