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Let $\quad P_1=I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_2=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right], \quad P_3=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_4=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right], \quad P_5=\left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right]$,
$P_6=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$ and $X=\sum_{k=1}^6 P_k\left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right] P_k^{\top}$
where $P _{ K }^{ T }$ denotes the transpose of the matrix $P _{ K }$. Then which of the following options is/are correct?
$(1)$ $X -30 I$ is an invertible matrix
$(2)$ The sum of diagonal entries of $X$ is 18
$(3)$ If $X \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\alpha\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$, then $\alpha=30$
$(4)$ $X$ is a symmetric matrix
$1,2,3$
$2,3,4$
$1,2,4$
$2,4$
Solution
Let $Q=\left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right]$
$X=\sum_{k=1}^6\left(P_k Q P_k^T\right)$
$X^T=\sum_{k=1}^6\left(P_k Q P_k^T\right)^T=X .0$
$X$ is symmetric
Let $R=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$
$X R=\sum_{1=1}^6 P_k Q P_k^{\top} R \cdot\left[\because P_k{ }^{\top} R=R\right]$
$=\sum_{k=1}^6 P_k Q R \cdot=\left(\sum_{k=1}^6 P_k\right) Q R$
$\begin{array}{l}\sum_{ K =1}^6 P _{ x }=\left[\begin{array}{lll}2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2\end{array}\right] \quad QR =\left[\begin{array}{l}6 \\ 3 \\ 6\end{array}\right] \\ \Rightarrow XR =\left[\begin{array}{lll}2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2\end{array}\right]\left[\begin{array}{l}6 \\ 3 \\ 6\end{array}\right]=\left[\begin{array}{l}30 \\ 30 \\ 30\end{array}\right]=30 R \end{array}$
$\Rightarrow \alpha=30 \text {. }$
Trace $X =\operatorname{Trace}\left(\sum_{ K =1}^{\circ} P _{ k } Q F _{ K }^{ T }\right)$
$\begin{array}{l}=\sum_{ K =1}^6 \operatorname{Trace}\left( P _{ x } QP _{ k }^{ T }\right)=6(\text { Irace } Q )=18 \\ x \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=30\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] \\ \Rightarrow( X -30 I )\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]= O \Rightarrow| X -30 I |=0 \\ \Rightarrow X -301 \text { is non-imvertible } \\\end{array}$
