JEE Advanced 2019 — Paper 2 — Question 26

$$P_1 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, P_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}, P_3 = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}, P_4 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}, P_5 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}, P_6 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$ $$X = \sum_{k=1}^{6} P_k \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{bmatrix} P_k^T$$ $$\text{Let } A = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{bmatrix}$$ $$A = A^T.$$ $$X^T = \left( \sum_{k=1}^{6} P_k A P_k^T \right)^T = \sum_{k=1}^{6} (P_k^T)^T A^T P_k^T = \sum_{k=1}^{6} P_k A P_k^T = X$$

So X is symmetric matrix

$$\text{Let } Q = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$ $$XQ = \sum_{k=1}^{6} P_k A P_k^T Q = \sum_{k=1}^{6} P_k A Q$$ $$AQ = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 6 \\ 3 \\ 6 \end{bmatrix}$$ $$XQ = (P_1 + P_2 + \ldots + P_6) AQ$$ $$P_1 + P_2 + P_3 + P_4 + P_5 + P_6 = \begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}$$ $$XQ = \begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix} \begin{bmatrix} 6 \\ 3 \\ 6 \end{bmatrix} = \begin{bmatrix} 12 + 6 + 12 \\ 12 + 6 + 12 \\ 12 + 6 + 12 \end{bmatrix} = \begin{bmatrix} 30 \\ 30 \\ 30 \end{bmatrix} = 30 Q$$ $$XQ = 30Q \Rightarrow (X - 30I) Q = 0$$

So, $(X - 30I) = 0,$ has non-trivial solution. So, not invertible. Where trace $(P_k A P_k^T) =$ trace $(A P_k^T P_k) =$trace $(A I) =$ trace $(A) = 2 + 0 + 1 = 3$ $\Rightarrow$ Trace $X = 3 \times 6 = 18.$