Question

Let
$P_1=\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|,$$P_2=\left|\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right|,$$P_3=\left|\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right|,$$P_4=\left|\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right|,$$P_5=\left|\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right|,$$P_6=\left|\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right|,$ and $X=\sum _{k=1}^6{P}_k\left[\begin{array}{ccc}2& 1& 3\\ 1& 0& 2\\ 3& 2& 1\end{array}\right]P_k^T,$ where $P_k^T$ denotes the transpose of the matrix $P_k.$ Then which of the following options is/are correct?

• A. $X-30I$ is an invertible matrix
• B. The sum of diagonal entries of $X$ is $18$
• C. $X$ is a symmetric matrix
• D. $\text{If}X\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\alpha \left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],\text{then}\alpha =30$

Answer 1

Answer: (B), (C), (D)

$\text{If}X\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\alpha \left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],\text{then}\alpha =30$

$\text{Let}A=\left[\begin{array}{ccc}2& 1& 3\\ 1& 0& 2\\ 3& 2& 1\end{array}\right]$
$X=\sum _{k=1}^6(P_{k}A{P}_k^T)=(P_{1}A{P}_1^T+P_{2}A{P}_2^T+\cdots +P_{6}A{P}_6^T)$
$X^T=\sum _{k=1}^6(P_{k}A{P}_k^T{)}^T=X[\because A=A^T]$
$\Rightarrow X$ is a symmetric matrix.

$\text{Let}Q=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

$P_1^{T}Q=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=Q,$

Similarly, $P_k^{T}Q=Q$
$\therefore XQ=\sum _{k=1}^6{P}_{k}A{P}_k^{T}Q$
$=\sum _{k=1}^6{P}_{k}AQ$
$=\left(\sum _{k=1}^6{P}_k\right)AQ$
$\sum _{k=1}^6{P}_k=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]\text{and}AQ=\left[\begin{array}{c}6\\ 3\\ 6\end{array}\right]$
$\Rightarrow XQ=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]\left[\begin{array}{c}6\\ 3\\ 6\end{array}\right]=\left[\begin{array}{c}30\\ 30\\ 30\end{array}\right]$
$\Rightarrow XQ=30\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=30Q=\alpha Q$
$\therefore \alpha =30$
$\text{Also,}X\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=30\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]\Rightarrow (X-30I)\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=0$
$\therefore (X-30I)Q=0$
$\therefore$ The system has a non-trivial solution.
So, $|X-30I|=0\Rightarrow X-30I$ is a non-invertible matrix.

$\text{Trace}\left(X\right)=\text{Trace}(\sum _{k=1}^6{P}_{k}A{P}_k^T)=6\text{Trace}\left(A\right)=6\times 3=18[\because \text{Trace}(A)=2+0+1=3]$