Let P and Q be 3 times 3 matrices such that P | vedclass

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(C) Given that $P$ and $Q$ are $3 	imes 3$ matrices such that $P 
eq Q$. We are given the equations: $1) P^3 = Q^3$ $2) P^2Q = Q^2P$ Subtracting the

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(C) Given that $P$ and $Q$ are $3 	imes 3$ matrices such that $P 
eq Q$. We are given the equations: $1) P^3 = Q^3$ $2) P^2Q = Q^2P$ Subtracting the second equation from the first,we get: $P^3 - P^2Q = Q^3 - Q^2P$ Factoring the terms,we have: $P^2(P - Q) = Q^2(Q - P)$ $P^2(P - Q) = -Q^2(P - Q)$ $(P^2 + Q^2)(P - Q) = 0$ Since $P 
eq Q$,the matrix $(P - Q)$ is not necessarily a zero matrix,but the product $(P^2 + Q^2)(P - Q) = 0$ implies that the matrix $(P^2 + Q^2)$ must be singular if $(P - Q)$ is non-zero and the system holds. Specifically,if $(P^2 + Q^2)(P - Q) = 0$ and $P - Q 
eq 0$,it implies that the determinant of the product is zero: $\det((P^2 + Q^2)(P - Q)) = \det(0) = 0$ $\det(P^2 + Q^2) \cdot \det(P - Q) = 0$ In the context of such matrix problems,this leads to $\det(P^2 + Q^2) = 0$.

Let $P$ and $Q$ be $3 \times 3$ matrices such that $P \neq Q$. If $P^3 = Q^3$ and $P^2Q = Q^2P$,then the determinant $\det(P^2 + Q^2)$ is equal to:

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