Question

Let P and Q be $3\times 3$ matrices such that $P\ne Q.$ If $P^3=Q^{3}and{P}^{2}Q=Q^{2}P,$ then determinant of $(P^2+Q^2)$ is equal to:

• A. -2
• B. 1
• C. -1

Answer 1

We have $P^3=Q^{3}and{P}^{2}Q=P{Q}^2$
Substracting, we get
$P^3-P^{2}Q=Q^3-Q^{2}P{P}^2(P-Q)+Q^2(P-Q)=0(P^2+Q^2)(P-Q)=0$
$If|P^2+Q^2|\ne 0then{P}^2+Q^2$ is invertible
$\Rightarrow P-Q=0$ (contradiction)
Hence $|P^2+Q^2|=0$