Question

Let \( M = (m_{ij}) \) be a \( 3 \times 3 \) real, invertible matrix and \( \sigma \in S_3 \) be the permutation defined by \( \sigma(1) = 2, \sigma(2) = 3 \) and \( \sigma(3) = 1 \). The matrix \( M_\sigma \) is defined by \( n_{ij} = m_{i\sigma(j)} \) for all \( i,j \in \{1, 2, 3\} \). Then, which one of the following is TRUE?

• A. \( \det(M) = \det(M_\sigma) \), and the nullity of the matrix \( M - M_\sigma \) is 0
• B. \( \det(M) = -\det(M_\sigma) \), and the nullity of the matrix \( M - M_\sigma \) is 1
• C. \( \det(M) = \det(M_\sigma) \), and the nullity of the matrix \( M - M_\sigma \) is 1
• D. \( \det(M) = -\det(M_\sigma) \), and the nullity of the matrix \( M - M_\sigma \) is 0

Hint:

When dealing with permutations of rows in a matrix, remember that the determinant remains invariant. The nullity can be calculated by checking the rank of the matrix.

Answer 1

The Correct Option is C

Step 1: Property of Permutation Matrices.
The matrix \( M_\sigma \) is obtained by permuting the rows of \( M \), so \( \det(M_\sigma) = \det(M) \) because the determinant is invariant under row permutations. 
Step 2: Nullity of \( M - M_\sigma \).
Since \( M \) is invertible and \( M_\sigma \) is simply a permutation of the rows, the matrix \( M - M_\sigma \) has nullity 1, meaning it has a nontrivial null space. 
Final Answer: \[ \boxed{\det(M) = \det(M_\sigma), \text{ and the nullity of the matrix } M - M_\sigma \text{ is } 1.} \]