value of $\det(P^2+Q^2)$

Question

If $P$ and $Q$ be $3\times 3$ matrices and $P\neq Q. $

If $P^3=Q^3$ and $P^2Q=Q^2P.$ Then $\det(P^2+Q^2)$ is

Try: From

$P^3-Q^3=O\Rightarrow (P-Q)(P^2+PQ+Q^2)=O$

So either $P=Q$ or either $P^2+PQ+Q^2=O$

So $P^2+Q^2=-PQ$, Now i did not understand how

i use $P^2Q=Q^2P$ and find $\det(P^2+Q^2)$

Could some help me how to find it, thanks

Answer 1

Some hints:

• Establish the equality $\left(P^2+Q^2\right)\left(P-Q\right)=0$.
• Using $\det(AB)=\det(A)\det(B)$, we derive that either $\det\left(P^2+Q^2\right)=0$ or $\det\left(P-Q\right)=0$.
• Suppose that $\det\left(P^2+Q^2\right)\neq 0$. Then $P^2+Q^2$ is invertible: left-multiply by $\left(P^2+Q^2\right)^{-1}$ in equality $\left(P^2+Q^2\right)\left(P-Q\right)=0$ to reach a contradiction.