Question

Let $P_1=I=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}, P_2=\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix}, P_3=\begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}, P_4=\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}$, $P _5=\begin{bmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}, P =\begin{bmatrix}1 & 2 & 4 \\ 8 & 16 & 32 \\ 64 & 32 & 1\end{bmatrix}$ and $X =\displaystyle\sum_{ K =1}^5 P _{ K } PP _{ K }^{ T }$ where $P _{ K }^{ T }$ denotes the transpose of the matrix $P _{ K }$. Then which of the following options is correct?

• A. The sum of diagonal entries of $X$ is 18
• B. The sum of diagonal entries of $X$ is 90
• C. The sum of diagonal entries of $X$ is 36
• D. $X$ is skew symmetric matrix

Answer 1

Answer: (B)

$\text { We have } \operatorname{trace}( AB )=\operatorname{trace}( BA )$
$\Rightarrow \operatorname{Trace}( X )=\displaystyle\sum_{ K =1}^5 \operatorname{Trace}\left( P _{ K } PP _{ K }^{ T }\right)$
$=\displaystyle\sum_{ K =1}^5 \operatorname{Trace}\left( P _{ K } P _{ K }^{ T } P \right) $
$=\displaystyle\sum_{ K =1}^5 \operatorname{Trace}( P )=90$
As $\operatorname{Trace}( X ) \neq 0$
$\Rightarrow X$ can not be skew-symmetric matrix.