question:
Given $A,B$ be two square matrices (with real entries )of order $2$ where
$AB=BA$ $, \det(A)=\alpha>0$ , $\det(A+i\alpha B)=0$
then find value of L where ,
$L=\det(A^2-\alpha A B+\alpha^2B^2)$
my attempt:
we know for $\det(A+xB)=x^2\det B+mx+\det A$ where $m= \ tr((adj A) B)$
put $x=i\alpha$ we get,
$\det(A+i\alpha B)=-\alpha^2\det B+mi\alpha+\det A=0\implies m=0\\$ &$\ \ \ \alpha \det B=1$
i don't know how to proceed further .please help in solving to get final answer . thank you
I suspect you are supposed to assume $A$ and $B$ are real matrices (as the answer is not determined if $A$ and $B$ can have complex entries). Note that for real matrices, $\det(A+i\alpha B) = 0$ implies $\det(A-i\alpha B) = 0$, and thus $B^{-1} A$ has eigenvalues $\pm i \alpha$.