Question

Let
$P_{1} = I = \begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix},\quad P_{2} = \begin{bmatrix}1&0&0\\ 0&0&1\\ 0&1&0\end{bmatrix},\quad P_{3} = \begin{bmatrix}0&1&0\\ 1&0&0\\ 0&0&1\end{bmatrix},
P_{4} = \begin{bmatrix}0&1&0\\ 0&0&1\\ 1&0&0\end{bmatrix}, \quad P_{5} = \begin{bmatrix}0&0&1\\ 1&0&0\\ 0&1&0\end{bmatrix},\quad P_{6} = \begin{bmatrix}0&0&1\\ 0&1&0\\ 1&0&0\end{bmatrix}$
and $x = \displaystyle\sum^{6}_{K = 1}P_{K}\begin{bmatrix}2&1&3\\ 1&0&2\\ 3&2&1\end{bmatrix}P^{T}_{K}$
where $P^{T}_{K}$ denotes the transpose of the matrix $P_{K}$. Then which of the following options is/are correct?

• A. If $X\begin{bmatrix}1\\ 1\\ 1\end{bmatrix} = \alpha\begin{bmatrix}1\\ 1\\ 1\end{bmatrix},$ then $\alpha$ = 30
• B. X is a symmetric matrix
• C. The sum of diagonal entries of X is 18
• D. X — 30I is an invertible matrix

Answer 1

Answer: (A), (B), (C)

Let $Q=\begin{bmatrix}2&1&3\\ 1&0&2\\ 3&2&1\end{bmatrix}$
$X=\displaystyle \sum_{k=1}^6$ $\left(P_{k}QP^{T}_{K}\right)$
$X^{T}=$ $=\displaystyle \sum_{k=1}^6$ $\left(P_{k}QP^{T}_{K}\right)^{T}=X.$
$X$ is symmetric
Let $R=\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}$
$XR=$ $\displaystyle \sum_{k=1}^6$$P_{K}QP_{K}^{T}R.\left[\because P_{K}\,{}^{T}R=R\right]$
$=\displaystyle \sum_{K=1}^6$$P_{K}=\begin{bmatrix}2&2&2\\ 2&2&2\\ 2&2&2\end{bmatrix} QR=\begin{bmatrix}6\\ 3\\ 6\end{bmatrix}$
$\Rightarrow XR=\begin{bmatrix}2&2&2\\ 2&2&2\\ 2&2&2\end{bmatrix}\begin{bmatrix}6\\ 3\\ 6\end{bmatrix}=\begin{bmatrix}30\\ 30\\ 30\end{bmatrix}=30R$
$\Rightarrow \alpha =30.$
Trace $X =$ Trace $\left(\displaystyle \sum_{K=1}^6P_{K}QP_{K}^{T}\right)$
$=\displaystyle \sum_{K=1}^6$ Trace $\left(P_{K}QP_{K}^{T}\right)=6$ (Trace$Q$) $=18$
$X\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}=30\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}$
$\Rightarrow \left(X-30\,I\right)\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}=O \Rightarrow \left|X-30\,I\right|=0$
$\Rightarrow X-30\,I$ is non-invertible