Matrices & Determinants
Permutation Matrices
Grade Class 12
Question:
Let $P_1 = I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, P_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}, P_3 = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}, P_4 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}, P_5 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}, P_6 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$ and $X = \sum_{k=1}^6 P_k \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{bmatrix} P_k^T$ where $P_k^T$ denotes the transpose of the matrix $P_k$. Then which of the following options is/are correct?<br>(A) $X - 30I$ is an invertible matrix<br>(B) The sum of diagonal entries of $X$ is 18<br>(C) If $X \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \alpha \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$, then $\alpha = 30$<br>(D) $X$ is a symmetric matrix
(A) $X - 30I$ is an invertible 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) $X$ is a symmetric matrix
Step-by-Step Solution
Key Concept: The matrices Pk are permutation matrices. The sum X = sum Pk A Pk^T represents the sum of all possible permutations of the rows and columns of matrix A. Since A is symmetric, X will also be symmetric. The sum of diagonal entries of X is 6 times the sum of diagonal entries of A. The vector [1, 1, 1]^T is an eigenvector of X.
The matrix $A = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{bmatrix}$ is symmetric. $X = \sum_{k=1}^6 P_k A P_k^T$. Since $P_k$ are permutation matrices, $X$ is a symmetric matrix (D is correct). The sum of diagonal entries of $X$ is $6 \times (2+0+1) = 18$ (B is correct). For $v = [1, 1, 1]^T$, $P_k v = v$ and $P_k^T v = v$, so $Xv = \sum P_k A P_k^T v = \sum P_k A v = \sum P_k [6, 3, 6]^T$. Summing over all permutations gives $30v$, so $\alpha = 30$ (C is correct). $X-30I$ has eigenvalue 0, so it is not invertible (A is incorrect).
Correct Answer: 1, 3, 4