Question

Let $P_1=I=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],P_2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right],P_3=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right],P_4=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$, $P_5=\left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right],P=\left[\begin{array}{ccc}1& 2& 4\\ 8& 16& 32\\ 64& 32& 1\end{array}\right]$ and $X=\sum _{K=1}^5{P}_{K}P{P}_K^T$ 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)

$\text{ We have }\mathrm{trace}(AB)=\mathrm{trace}(BA)$
$\Rightarrow \mathrm{Trace}(X)=\sum _{K=1}^5\mathrm{Trace}\left(P_{K}P{P}_K^T\right)$
$=\sum _{K=1}^5\mathrm{Trace}\left(P_K{P}_K^{T}P\right)$
$=\sum _{K=1}^5\mathrm{Trace}(P)=90$
As $\mathrm{Trace}(X)\ne 0$
$\Rightarrow X$ can not be skew-symmetric matrix.