Daily Question 115

$\textbf{Q115.}$ Let

$P_1=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix},$ $P_2=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0\\ \end{bmatrix},$ $P_3=\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\\ \end{bmatrix},$ $P_4=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0\\ \end{bmatrix},$ $P_5=\begin{bmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0\\ \end{bmatrix},$ $P_6=\begin{bmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0\\ \end{bmatrix}$ and $X=\displaystyle\sum_{k=1}^{6}P_k\begin{bmatrix} 2 & 1 & 3\\ 1 & 0 & 2\\ 3 & 2 & 1\\ \end{bmatrix}P^{T}_{k}$

Where $P^{T}_{k}$ denotes the transpose of matrix $P_k.$ Then which of the following options is/are correct?

(A) $X$ is a symmetric matrix.

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

(C) If $X\begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix}=\alpha\begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix},$ then $\alpha=30.$

(D) $X-30I$ is an invertible matrix.

$\textbf{Ans.} (A), (B), (C)$

$\textbf{Sol.}$ Clearly form the given data,

$P_1=P_1^T=P_1^{-1}$
$P_2=P_2^T=P_2^{-1}$
$\vdots$
$P_6=P_6^T=P_6^{-1}$
Let $A=\begin{bmatrix} 2 & 1 & 3\\ 1 & 0 & 2\\ 3 & 2 & 1\\ \end{bmatrix}$

Here $A^{T}=A\implies A$ is symmetric matrix.$\qquad\ldots(1)$

Now, $X=P_{1}AP_{1}^{T}+P_{2}AP_{2}^{T}+\ldots+P_{6}AP_{6}^{T}$

So, $X^{T}=\left(P_{1}AP_{1}^{T}+P_{2}AP_{2}^{T}+\ldots+P_{6}AP_{6}^{T}\right)^{T}$

$\implies X^{T}=P_{1}A^{T}P_{1}^{T}+P_{2}A^{T}P_{2}^{T}+\ldots+P_{6}A^{T}P_{6}^{T}$

Using Eq. $(1)$

$\implies X^{T}=P_{1}AP_{1}^{T}+P_{2}AP_{2}^{T}+\ldots+P_{6}AP_{6}^{T}$

$\therefore X^{T}=X \implies X$ is symmetric.

Now Let $B=\begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix}$

$XB=P_{1}AP_{1}^{T}B+P_{2}AP_{2}^{T}B+\ldots+P_{6}AP_{6}^{T}B$

$\implies XB=P_{1}AB+P_{2}AB+\ldots+P_{6}AB$

$\implies XB=(P_{1}+P_{2}+\ldots+P_{6})\begin{bmatrix} 6\\ 3\\ 6\\ \end{bmatrix}$

$\implies XB=\begin{bmatrix} 2 & 2 & 2\\ 2 & 2 & 2\\ 2 & 2 & 2\\ \end{bmatrix}\begin{bmatrix} 6\\ 3\\ 6\\ \end{bmatrix}=\begin{bmatrix} 30\\ 30\\ 30\\ \end{bmatrix}=30B$

$\therefore\alpha=30$

$\implies (X-30I)B=0$ has non trivial solution $B=\begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix}$

$\implies |X-30I|=0$

$\text{trace}(X)=\text{tr}(P_1AP_1^{T})+\ldots+\text{tr}(P_6AP_6^{T})$

$\implies\text{trace}(X)=(2+0+1)+\ldots+(2+0+1)$

$\implies\text{trace}(X)=3\times 6=18$

Daily Question 114

Compiled Daily Questions 1-100