Let $P_1 = I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, $P_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}$, $P_3 = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, $P_4 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$, $P_5 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$, $P_6 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$ and $X = \sum_{k=1}^{6} P_k \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{bmatrix} P_k^T$ Where $P_k^T$ denotes the transpose of the matrix $P_k$. Then which of the following options is/are correct?

Let P₁ = I = 1 0 0 0 1 0 0 0 1 , P₂ = 1 0 0 0 0 1 0 1 1 , P₃ = 0 1 0 1 0 0 0 0 1 , P₄ = 0 1 0 0 0 1 1 0 0 , P₅ = 0 0 1 1 0 0 0 1 0 , P₆ = 0 0 1 0 1 0 1 0 0 and X = ∑ₖ₌₁⁶ Pₖ 2 1 3 1 0 2 A=2 1 3 1 0 2 3 2 1

$X=\sum_{k=1}^{6} P_k A P_k^{T}$

$P_1,\dots,P_6:\ 3\times 3 \text{ matrices (as given)}$

$(1)\ X \text{ symmetric};\qquad (2)\ \operatorname{tr}(X)=18$

$(3)\ X-30I \text{ invertible};\qquad (4)\ X\mathbf{1}=\alpha\mathbf{1}\Rightarrow \alpha=30$

$P_1AP_1^{T}=A$

$P_3,P_4,P_5,P_6 \text{ permutation} \Rightarrow P_k^{T}P_k=I$

$P_2^{T}P_2=P_2^{2}\neq I \Rightarrow \text{handle separately}$

P₁AP₁^T=A=2 1 3 1 0 2 3 2 1 P₂A=2 1 3 3 2 1 4 2 3 P₂AP₂^T=2 3 4 3 1 3 4 3 5

$\operatorname{tr}(P_2AP_2^{T})=2+1+5=8$

P₃AP₃^T=0 1 2 1 2 3 2 3 1, P₄AP₄^T=0 2 1 2 1 3 1 3 2 P₅AP₅^T=1 3 2 3 2 1 2 1 0, P₆AP₆^T=1 2 3 2 0 1 3 1 2 X=6 12 15 12 6 13 15 13 11

$X^{T}=X \Rightarrow (1) \text{ true}$

$\operatorname{tr}(X)=6+6+11=23$

$23\neq 18 \Rightarrow (2) \text{ false}$

X-30I=-24 12 15 12 -24 13 15 13 -19

$\det(X-30I)=5928\neq 0 \Rightarrow (3) \text{ true}$

X1 1 1=33 31 39≠ α1 1 1

$\Rightarrow (4) \text{ false}$

$\text{Correct options: } (1) \text{ and } (3)$