Mathematics - Matrices Question with Solution | TestHub

$\because \mathrm{ }\begin{array}{ccc}{\mathrm{P}}_1=P_1^T=P_1^{-1},& P_3=P_3^T=P_3^{-1},& {\mathrm{P}}_5=P_5^T=P_5^T,\\ {\mathrm{P}}_2=P_2^T=P_2^{-1},& {\mathrm{P}}_4=P_4^T=P_4^T,& {\mathrm{P}}_6=P_6^T=P_6^{-1}\end{array}$
Now
Let$Q=\left[\begin{array}{ccc}2& 1& 3\\ 1& 0& 2\\ 3& 2& 1\end{array}\right]$
$X=\sum _{K=1}^6{P}_{K}Q{P}_K^T$
$\Rightarrow \begin{array}{cc}Trace\left(X\right)=\sum _{K=1}^6{T}_r\left(\left(P_{K}Q\right)P_K^T\right)& \left(\because T_r\left(AB\right)=T_r\left(BA\right)\right)\end{array}$
$\begin{array}{cc}=\sum _{K=1}^6{T}_r\left(P_K^T{P}_{K}Q\right)& \left(\because P_K^T P_K=I\right)\end{array}$
$=\sum _{K=1}^6{T}_r\left(Q\right)$
$=6 T_{\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{e}}\left(Q\right)$
$=6\times 3$
$=18$
$\Rightarrow X=\sum _{K=1}^6\left(P_{K}Q{P}_K^T\right)$
$X^T=\sum _{K=1}^6{\left(P_{K}Q{P}_K^T\right)}^T$
$\begin{array}{cc}{X}^T=\sum _{K=1}^6{\left(P_K^T\right)}^T{Q}^T{\left(P_K\right)}^T& \left(\because {\left(A^T\right)}^T=A\right)\end{array}$
$\begin{array}{cc}{X}^T=\sum _{K=1}^6{P}_{K}Q{P}_K^T& \left(\because Q^T=Q\right)\end{array}$
$X^T=X$
$\Rightarrow$Let$R=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$then$XR=\sum _{K=1}^6{P}_{K}Q{P}_K^{T}R$
$\begin{array}{cc}XR=\sum _{K=1}^6{P}_{K}Q{P}_{K}R& \left(\because P_K^T=P_K\right)\end{array}$
$\begin{array}{cc}XR=\sum _{K=1}^6{P}_{K}QR& \left(\because P_{K}R=R\right)\end{array}$
$\begin{array}{cc}XR=\sum _{K=1}^6{P}_K\left[\begin{array}{c}6\\ 3\\ 6\end{array}\right]& \left(\because QR=\left[\begin{array}{c}6\\ 3\\ 6\end{array}\right]\right)\end{array}$
$\begin{array}{cc}XR=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]\left[\begin{array}{c}6\\ 3\\ 6\end{array}\right]& \left(\because \sum _{K=1}^6{P}_K=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]\right)\end{array}$
$XR=\left[\begin{array}{c}30\\ 30\\ 30\end{array}\right]\Rightarrow XR=30\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]\Rightarrow \alpha =30$
$\Rightarrow XR=30\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$
$XR=30 \mathrm{R}$
$\left(X-30I\right)R=0$
Now$\left|x-30\mathrm{I}\right|$must be zero else$R=0$
If$\left|x-30\mathrm{I}\right|\ne 0$
then${\left(x-30\mathrm{I}\right)}^{-1}\left(x-30\mathrm{I}\right)R=0$
$R=0$not possible
Hence,$\left|x-30\mathrm{I}\right|=0$
$X-30\mathrm{I}$is not invertible.