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Given two 3×3 matrices P and Q. Such that P is not equal to Q, P³= Q³ and P²Q = Q²P. Find det(P²+Q²)? I tried converting the qiven condition to get a relation between matrices but was stuck by the fact that matrix multiplication is not commutative and was unable to use the first relation that P³ =Q³, any ideas to solve it?

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    $\begingroup$ $\det(P^2+Q^2) \det(P-Q) = \det(P^3+Q^2P-P^2Q-Q^3) = \det(P^3-Q^3) = 0$ $\endgroup$ – DHMO Apr 16 '17 at 16:34
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We have $(P^2+Q^2) (P-Q) = P^3+Q^2P-P^2Q-Q^3 = P^3-Q^3 = O$.

If $P^2+Q^2$ were invertible, then $(P-Q) = (P^2+Q^2)^{-1} O = O$, a contradiction.

Therefore, $P^2+Q^2$ is not invertible, so $\det(P^2+Q^2) = 0$.

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