$\mathrm{Let}\mathrm{Q}=\left[\begin{array}{ccc}2& 1& 3\\ 1& 0& 2\\ 3& 2& 1\end{array}\right]$

$\mathrm{X}=\sum _{\mathrm{k}=1}^6\left({\mathrm{P}}_{\mathrm{k}}{\mathrm{QP}}_{\mathrm{K}}^{\mathrm{T}}\right)$

${\mathrm{X}}^{\mathrm{T}}=\sum _{\mathrm{k}=1}^6{\left({\mathrm{P}}_{\mathrm{k}}{\mathrm{QP}}_{\mathrm{K}}^{\mathrm{T}}\right)}^{\mathrm{T}}=\mathrm{X}$

X is symmetric

$\mathrm{Let}\mathrm{R}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

$\mathrm{XR}=\sum _{\mathrm{k}=1}^6{\mathrm{P}}_{\mathrm{K}}{\mathrm{QP}}_{\mathrm{K}}^{\mathrm{T}}\mathrm{R}[\because {\mathrm{P}}_{\mathrm{K}}^{\mathrm{T}}\mathrm{R}=\mathrm{R}]$

$=\sum _{\mathrm{K}=1}^6{\mathrm{P}}_{\mathrm{K}}\mathrm{QR}=\left(\sum _{\mathrm{K}=1}^6{\mathrm{P}}_{\mathrm{K}}\right)\mathrm{QR}$

$\sum _{\mathrm{K}=1}^6{\mathrm{P}}_{\mathrm{K}}=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right];\mathrm{QR}=\left[\begin{array}{c}6\\ 3\\ 6\end{array}\right]$

$\Rightarrow \mathrm{XR}=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]\left[\begin{array}{c}6\\ 3\\ 6\end{array}\right]=\left[\begin{array}{c}30\\ 30\\ 30\end{array}\right]=30\mathrm{R}$

⇒ α = 30

Trace X = Trace $\left(\sum _{\mathrm{k}=1}^6{\mathrm{P}}_{\mathrm{K}}{\mathrm{QP}}_{\mathrm{K}}^{\mathrm{T}}\right)=\sum _{\mathrm{k}=1}^6\mathrm{Trace}\left({\mathrm{P}}_{\mathrm{K}}{\mathrm{QP}}_{\mathrm{K}}^{\mathrm{T}}\right)$ = 6 (Trace Q) = 18

$\mathrm{X}\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=30\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

⇒ (X – 30I) $\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$= O ⇒ | X – 30I | = 0

⇒ X – 30I is non‑invertible.