Let P _ 1 = I = [ cases 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 cases ], P _...

$∵$\begin{cases}{P}_{1} = {P}_{1}^{T} = {P}_{1}^{- 1}, & {P}_{3} = {P}_{3}^{T} = {P}_{3}^{- 1}, & {P}_{5} = {P}_{5}^{T} = {P}_{5}^{T}, \\ {P}_{2} = {P}_{2}^{T} = {P}_{2}^{- 1}, & {P}_{4} = {P}_{4}^{T} = {P}_{4}^{T}, & {P}_{6} = {P}_{6}^{T} = {P}_{6}^{- 1}\end{cases}<br />Now<br />Let $Q = \left[$\begin{cases}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{cases}$\right]$<br />$X = \mathop{∑}_{K = 1}^{6}{P}_{K}Q{P}_{K}^{T}$<br />$⇒ $\begin{cases}Trace\left(X\right) = \mathop{∑}_{K = 1}^{6}{T}_{r}\left(\left({P}_{K}Q\right){P}_{K}^{T}\right) & \left(∵ {T}_{r}\left(AB\right) = {T}_{r}\left(BA\right)\right)\end{cases}
$\begin{cases}= \mathop{∑}_{K = 1}^{6}{T}_{r}\left({P}_{K}^{T}{P}_{K}Q\right) & \left(∵ {P}_{K}^{T}{P}_{K} = I\right)\end{cases}$
$= \mathop{∑}_{K = 1}^{6}{T}_{r}\left(Q\right)$
$= 6{T}_{race}\left(Q\right)$
$= 6 \times 3$
$= 18$
$⇒ X = \mathop{∑}_{K = 1}^{6}\left({P}_{K}Q{P}_{K}^{T}\right)$
${X}^{T} = \mathop{∑}_{K = 1}^{6}{\left({P}_{K}Q{P}_{K}^{T}\right)}^{T}$
$\begin{cases}{X}^{T} = \mathop{∑}_{K = 1}^{6}{\left({P}_{K}^{T}\right)}^{T}{Q}^{T}{\left({P}_{K}\right)}^{T} & \left(∵ {\left({A}^{T}\right)}^{T} = A\right)\end{cases}$
$\begin{cases}{X}^{T} = \mathop{∑}_{K = 1}^{6}{P}_{K}Q{P}_{K}^{T} & \left(∵ {Q}^{T} = Q\right)\end{cases}$
${X}^{T} = X$
$⇒$ Let R = \left[$\begin{array}{ll}1 \\ 1 \\ 1\end{array}$\right] then $XR = \mathop{∑}_{K = 1}^{6}{P}_{K}Q{P}_{K}^{T}R$
$\begin{cases}XR = \mathop{∑}_{K = 1}^{6}{P}_{K}Q{P}_{K}R & \left(∵ {P}_{K}^{T} = {P}_{K}\right)\end{cases}$
$\begin{cases}XR = \mathop{∑}_{K = 1}^{6}{P}_{K}QR & \left(∵ {P}_{K}R = R\right)\end{cases}$
$$\begin{array}{ll}XR = \mathop{∑}_{K = 1}^{6}{P}_{K} 6 \\ 3 \\ 6\end{array}$$\left(∵ QR = \left[$$\begin{array}{ll}6 \\ 3 \\ 6\end{array}$$\right]\right)$<br />$$\begin{array}{ll}XR = 2 2 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2\end{array}$\left[$\begin{array}{ll}6 \\ 3 \\ 6\end{array}$\right] \left(∵ \mathop{∑}_{K = 1}^{6}{P}_{K} = \left[\begin{cases}2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2\end{cases}\right]\right)
XR = \left[$\begin{array}{ll}30 \\ 30 \\ 30\end{array}$\right] ⇒ XR = 30\left[$\begin{array}{ll}1 \\ 1 \\ 1\end{array}$\right] ⇒ \alpha = 30
⇒ XR = 30\left[$\begin{array}{ll}1 \\ 1 \\ 1\end{array}$\right]
$XR = 30R$
\left(X - 30I\right)R = 0$<br />Now \left|x - 30I\right|$ must be zero else $R = 0$
If \left|x - 30I\right| \ne 0$<br />then${\left(x - 30I\right)}^{- 1}\left(x - 30I\right)R = 0$<br />$R = 0$not possible<br />Hence, \left|x - 30I\right| = 0$
$X - 30I$ is not invertible.