Question

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

• A. $-2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

The correct option is C $0$

Given that $P^3=Q^3$
$\Rightarrow P^3-P^{2}Q=Q^3-Q^{2}P[\because P^{2}Q=Q^{2}P]$
$\Rightarrow P^2(P-Q)=Q^2(Q-P)$
$\Rightarrow P^2(P-Q)+Q^2(P-Q)=0$
$\Rightarrow (P^2+Q^2)(P-Q)=0$
$\Rightarrow P^2+Q^2=0[\because P\ne Q]$
$\Rightarrow \mid P^2+Q^2\mid =0$