Question

firstly go through the attachment. now answer that which of the following options is/are correct.

(1) X is a symmetric matrix.
(2) if $\begin{gathered}\sf x \left[\begin{array}{c} \sf 1 \\\sf 1 \\\sf 1 \end{array}\right] = \alpha \left[\begin{array}{c} \sf 1 \\\sf 1 \\\sf 1\end{array}\right] \end{gathered}​$
, then α = 30
(3) sum of diagonal entries of X is 18.
(4) X-30 I is an invertible matrix.​

Answer 1

Answer:

Let $\sf A = \left[ \begin{array}{ccc} \sf 2&\sf 1&\sf 3\\\sf 1&\sf 0&\sf 2\\\sf 3&\sf 2&\sf 1 \end{array}\right]$

$\rm$

then $\sf A^T = \left[ \begin{array}{ccc} \sf 2&\sf 1&\sf 3\\\sf 1&\sf 0&\sf 2\\\sf 3&\sf 2&\sf 1 \end{array}\right]$

$\rm$

$\sf \implies A = A^T$

$\rm$

Now, $\sf X^T = (P_1 AP_1^T + P_2 AP_2^T + \dots + P_6 AP_6^T)^T = X$

so, X is symmetric matrix.

let $\sf Q = \left[\begin{array}{c} \sf 1\\\sf 1 \\\sf 1 \end{array}\right]$

$\rm$

$\sf XQ = P_1 AP_1^T Q + P_2 AP_2^T Q + \dots P_6 AP_6^T Q$

$\rm$

$\sf = P_2 AQ + P_2 AQ + \dots + P_6 AQ \left[ \because P_1^T Q = \left[\begin{array}{ccc}\sf 1&0&0\\\sf 0&\sf 1&\sf 0\\\sf 0&\sf 0&\sf 1\end{array}\right] \left[\begin{array}{c} \sf 1 \\\sf 1 \\\sf 1 \end{array}\right] = Q \right]$

$\rm$

$\sf = (P_1 + P_2 + \dots + P_6) AQ_1$

$\rm$

$\sf = \left[\begin{array}{ccc} \sf 2&\sf 2&\sf 2\\\sf 2 &\sf 2 &\sf 2\\\sf 2&\sf 2 &\sf 2\end{array}\right] \left[\begin{array}{c}\sf 6\\\sf 3\\\sf 6 \end{array}\right] = \left[\begin{array}{c}\sf 30\\\sf 30\\\sf 30 \end{array}\right]$

$\rm$

$\sf = 30Q \because \left[ AQ = \left[\begin{array}{ccc} \sf 2&\sf 1 &\sf 3 \\\sf 1 & \sf 0 &\sf 2\\\sf 3 & \sf 3 & \sf 1\end{array}\right] \left[\begin{array}{c} \sf 1\\\sf 1 \\\sf 1 \end{array}\right] \right]$

$\rm$

$\sf \implies XQ = 30 Q \implies (X-30I) Q = 0$

so, $\sf |X-30I| =0$ has non trivial solution

so $\sf |X-30I|$ is non invertible.

$\rm \implies$ Trace $\sf X = 3\times 6 = 18$

therefore options 1,2,3 are correct.