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. $0$
• D. $-1$

Answer 1

The correct option is C $0$

On substracting, we get

$P^3-P^{2}Q=Q^3-Q^{2}P$

$\Rightarrow P^2(P-Q)+Q^2(P-Q)=0$

$\Rightarrow (P^2+Q^2)(P-Q)=0$

If $|P^2+Q^2|\ne 0$; then $P^2+Q^2$ is invertible

$\Rightarrow P-Q=0$, contradiction

$\therefore |P^2+Q^2|=0$