Question

Let $P_1=I=\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$, $P_2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$, $P_3=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$, $P_4=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$, $P_5=\left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]$, $P_6=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]$ 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. If $X\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\alpha \left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ then $\alpha =30$
• D. $X$ is a symmetric matrix

Answer 1

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

$X$ is a symmetric matrix

Explanation for correct options:

Option (D): $X$ is a symmetric matrix

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

$$\begin{gathered} X=\sum _{k=1}^6\left(P_{k}Q{P_k}^T\right) \\ \Rightarrow X^T=\sum _{k=1}^6{\left(P_{k}Q{P_k}^T\right)}^T \\ \Rightarrow X^T=X \end{gathered}$$

So, $X$ is a symmetric matrix.

Option (C): If $X\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\alpha \left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ then $\alpha =30$

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

$$\begin{gathered} XR=\sum _{k=1}^6\left(P_{k}Q{P_k}^T\right)R \\ \Rightarrow XR=\sum _{k=1}^6\left(P_{k}QR\right)\left[\because {P_k}^{T}R=R\right] \\ \Rightarrow XR=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]QR \\ \Rightarrow XR=\left[\begin{array}{ccc}2& 2& 2\\ 2& 2& 2\\ 2& 2& 2\end{array}\right]\left[\begin{array}{c}6\\ 3\\ 6\end{array}\right]\left[\because QR=\left[\begin{array}{c}6\\ 3\\ 6\end{array}\right]\right] \\ \Rightarrow XR=\left[\begin{array}{c}30\\ 30\\ 30\end{array}\right] \\ \Rightarrow XR=30\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right] \end{gathered}$$

It implies that $\alpha =30$. Hence, if $X\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\alpha \left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ then $\alpha =30$.

Option (B): The sum of diagonal entries of $X$ is $18$

Sum of diagonal elements is known as trace. Now,

$$\begin{gathered} Trace\left(X\right)=Trace\left(\sum _{k=1}^6{P}_{k}Q{P_k}^T\right) \\ =\sum _{k=1}^{6}Trace\left(P_{k}Q{P_k}^T\right) \\ =6\left(Trace\left(Q\right)\right) \\ =6\times 3 \\ =18 \end{gathered}$$

So, sum of diagonal elements of $X$ is equal to $18$.

Explanation for incorrect option

Option (A): $X-30I$ is an invertible matrix

We know that,

$$\begin{gathered} X\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=30\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right] \\ \Rightarrow \left(X-30I\right)\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=0 \\ \Rightarrow \left|X-30I\right|=0 \end{gathered}$$

It implies that $X-30I$ is non-invertible matrix.

Hence, the correct answer is option (B), (C) and (D).