Types of matrices

Solution

Option (c) is correct.

Explanation:

Given the condition \( M^3 = M \), we can factor it as follows: \[ M^3 - M = 0 \] This can be factored into: \[ M(M^2 - I) = 0 \] where \( I \) is the identity matrix. Thus, we have two cases: either \( M = 0 \) or \( M^2 = I \). The eigenvalues of \( M \), denoted by \( \lambda \), must satisfy the equation \( \lambda^3 - \lambda = 0 \). Factoring gives: \[ \lambda(\lambda^2 - 1) = 0 \implies \lambda(\lambda - 1)(\lambda + 1) = 0 \] This yields the eigenvalues \( \lambda = 0, 1, -1 \). The determinant of \( M \) is the product of its eigenvalues. Possible combinations of eigenvalues for a \( n \times n \) matrix yield different determinants: - If all eigenvalues are \( 0 \): \( \text{det}(M) = 0 \) - If all eigenvalues are \( 1 \): \( \text{det}(M) = 1 \) - If one eigenvalue is \( 0 \) and others can be \( 1 \) or \( -1\): - \( \text{det}(M) = 0 \) - \( \text{det}(M) = 1 \) - \( \text{det}(M) = -1 \) Thus, the possible values of \( |M| \) are \( 0, 1, -1 \). Therefore, there are three unique values possible for \( |M| \).