First slide

Applications of determinant

Question

Let P and Q be 3x3 matrices with $P \neq Q$ If $P^{3} = Q^{3}$ and $P^{2} Q = Q^{2} P$ then determinant of $(P^{2} + Q^{2})$ is equal to

Moderate

A

1
B

0
C

-1
D

-2

Solution

$P^{3} = Q^{3}$ and $P^{2} Q = Q^{2} P$ gives
$\begin{array}{l} P^{3} - P^{2} Q = Q^{3} - Q^{2} P \\ \Rightarrow P^{2} (P - Q) = - Q^{2} (P - Q) \\ \Rightarrow (P^{2} + Q^{2}) (P - Q) = 0 \end{array}$
If det $(P^{2} + Q^{2}) \neq 0$ then $P^{2} + Q^{2}$ is invertible and hence P $= Q$ . Therefore, $\det (P^{2} + Q^{2}) = 0$ .

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