Question

Let $P$ and $Q$ be $3\times 3$matrices, $P\ne Q$.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

$0$

Explanation for the correct option:

Finding the determinant of $(P^2+Q^2)$:

Given

$$\begin{gathered} P^3=Q^3\dots \dots \left(1\right) \\ {P}^{2}Q=Q^{2}P\dots \dots \left(2\right) \end{gathered}$$

On subtracting equations $\left(1\right)$ and $\left(2\right)$, we get

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

Solving the equation, we get

$$\begin{gathered} \Rightarrow P^2(P-Q)+Q^2(P-Q)=0 \\ \Rightarrow (P^2+Q^2)(P-Q)=0 \end{gathered}$$

This shows that either $\left(P^2+Q^2\right)=0\text{or}\left(P-Q\right)\text{=0}$

If

$$\begin{gathered} \left(P-Q\right)\text{=0} \\ \Rightarrow P=Q \end{gathered}$$

But it is given that $P\ne Q$.

Hence, $\left(P-Q\right)\text{=0}$ is not satisfied.

If

$\begin{array}{rcl}{P}^2+Q^2& =& 0\\ & \Rightarrow & \left|P^2+Q^2\right|=0\end{array}$

Thus, $\left(P^2+Q^2\right)=0$

Therefore, the correct answer is option (C).