Let quad P_1=I=left[begin{array}{lll}1 & 0 & | vedclass

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(B) Let $Q = \left[\begin{array}{lll}2 & 1 & 3 \ 1 & 0 & 2 \ 3 & 2 & 1\end{array}ight]$.$X = \sum_{k=1}^6 (P_k Q P_k^T)$.$X^T = \sum_{k=1}^6 (P_k

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$2, 3, 4$

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(B) Let $Q = \left[\begin{array}{lll}2 & 1 & 3 \ 1 & 0 & 2 \ 3 & 2 & 1\end{array}ight]$.$X = \sum_{k=1}^6 (P_k Q P_k^T)$.$X^T = \sum_{k=1}^6 (P_k Q P_k^T)^T = \sum_{k=1}^6 (P_k Q^T P_k^T) = \sum_{k=1}^6 (P_k Q P_k^T) = X$ (since $Q$ is symmetric).Thus,$X$ is a symmetric matrix. Option $(4)$ is correct.Let $R = \left[\begin{array}{l}1 \ 1 \ 1\end{array}ight]$. Note that for all permutation matrices $P_k$,$P_k R = R$ and $P_k^T R = R$.$X R = \sum_{k=1}^6 P_k Q P_k^T R = \sum_{k=1}^6 P_k Q R = (\sum_{k=1}^6 P_k) Q R$.$\sum_{k=1}^6 P_k = \left[\begin{array}{lll}2 & 2 & 2 \ 2 & 2 & 2 \ 2 & 2 & 2\end{array}ight]$ and $Q R = \left[\begin{array}{lll}2 & 1 & 3 \ 1 & 0 & 2 \ 3 & 2 & 1\end{array}ight] \left[\begin{array}{l}1 \ 1 \ 1\end{array}ight] = \left[\begin{array}{l}6 \ 3 \ 6\end{array}ight]$.$X R = \left[\begin{array}{lll}2 & 2 & 2 \ 2 & 2 & 2 \ 2 & 2 & 2\end{array}ight] \left[\begin{array}{l}6 \ 3 \ 6\end{array}ight] = \left[\begin{array}{l}30 \ 30 \ 30\end{array}ight] = 30 R$. Thus,$\alpha = 30$. Option $(3)$ is correct.Since $X R = 30 R$,$(X - 30I) R = 0$,which implies $X - 30I$ is non-invertible. Option $(1)$ is incorrect.$	ext{Trace}(X) = \sum_{k=1}^6 	ext{Trace}(P_k Q P_k^T) = \sum_{k=1}^6 	ext{Trace}(P_k^T P_k Q) = \sum_{k=1}^6 	ext{Trace}(I Q) = 6 	imes 	ext{Trace}(Q) = 6 	imes (2+0+1) = 18$. Option $(2)$ is correct.Therefore,options $(2), (3), (4)$ are correct.

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