For any 3 times 3 matrix M,let | M | denote t | vedclass

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(C) First,calculate $PEP$:$PEP = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & 4 \ 8 & 13 &

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$A, B, D$

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(C) First,calculate $PEP$:$PEP = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & 4 \ 8 & 13 & 18 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \ 8 & 13 & 18 \ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 2 \ 8 & 18 & 13 \ 2 & 4 & 3 \end{bmatrix} = F$.Also,$P^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$. Thus,$(A)$ is $TRUE$.For $(B)$,note that $|E| = 0$ and $|F| = 0$. Since $|EQ| = |E||Q| = 0$ and $|PFQ^{-1}| = |P||F||Q|^{-1} = 0$,the equation $|EQ + PFQ^{-1}| = 0 + 0 = 0$ holds. Thus,$(B)$ is $TRUE$.For $(C)$,$|EF| = |E||F| = 0 	imes 0 = 0$. Thus,$|(EF)^3| = 0$ and $|EF|^2 = 0$. The inequality $0 > 0$ is $FALSE$.For $(D)$,since $P^2 = I$,$P^{-1} = P$. Then $P^{-1}FP = PFP = P(PEP)P = P^2EP^2 = I E I = E$. Thus,$E + P^{-1}FP = 2E$. Also,$P^{-1}EP + F = PEP + F = F + F = 2F$. The sum of diagonal entries (Trace) of $2E$ is $2(1+3+18) = 44$,while the trace of $2F$ is $2(1+18+3) = 44$. Thus,$(D)$ is $TRUE$.

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