Let P1=I=100010001,P2=100001010,P3=010100001,P4=010001100,P5=001100010,P6=001010100 and X=∑k=16Pk213102321PkTWhere PkT denotes the transpose of the matrix Pk. Then which of the following options is/ are correct?

If X 1 1 1 = α 1 1 1 , then α = 30

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$\begin{array}{l} L e t P_{1} = I = [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}], P_{2} = [\begin{array}{l} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}], P_{3} = [\begin{array}{l} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}], P_{4} = [\begin{array}{l} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}], P_{5} = [\begin{array}{l} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}], P_{6} = [\begin{array}{l} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}] \\ and X = \sum_{k = 1}^{6} P_{k} [\begin{array}{l} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{array}] P_{k}^{T} \end{array}$ Where $P_{k}^{T}$ denotes the transpose of the matrix $P_{k}$ . Then which of the following options is/ are correct?

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X is a symmetric matrix

The sum of diagonal entries of X is 18

X – 30I is an invertible matrix

If $X [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = α [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}],$ then $α = 30$

answer is A.

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Detailed Solution

$\begin{array}{l} Let A = [\begin{array}{l} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{array}] \Rightarrow X = \sum_{k = 1}^{6} P_{k} A P_{k}^{T}, Here P_{k} P_{k}^{T} = P_{k}^{T} P_{k} = I for all k = 1, 2, 3, 4, 5, 6 . \\ P_{k}^{T} = P_{k}^{- 1} & A^{T} = A and X^{T} = \sum_{k = 1}^{6} {(P_{k} A P_{k}^{T})}^{T} = \sum_{k = 1}^{6} P_{k} A P_{k}^{T} = X \end{array}$

$\begin{array}{l} Let B = [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}] \Rightarrow X B = \sum_{K = 1}^{6} P_{K} A P_{K}^{T} B = \sum_{K = 1}^{6} P_{K} A B = (\sum_{K = 1}^{6} P_{K}) A B = [\begin{array}{l} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{array}] A B = [\begin{array}{l} 30 \\ 30 \\ 30 \end{array}] = 30 [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}] = 30 B \\ ∴ α = 30 and (X - 30 I) is singular, Tr (\sum_{k = 1}^{6} P_{k} A P_{k}^{T}) = 18 \end{array}$