Question

Let \( M \) be a \( 3 \times 3 \) real matrix such that \( M^2 = 2M + 3I \). If the determinant of \( M \) is \( -9 \), then the trace of \( M \) equals __________.

Hint:

To find the trace of a matrix from its characteristic equation, solve for the eigenvalues and sum them up. The trace is equal to the sum of the eigenvalues.

Answer 1

Correct Answer: 5

We are given the matrix equation \( M^2 = 2M + 3I \), where \( I \) is the identity matrix. We need to find the trace of the matrix \( M \). The trace of a matrix is the sum of its diagonal elements, and it is also equal to the sum of the eigenvalues of the matrix. Step 1: Eigenvalue equation. Let's assume that \( \lambda \) is an eigenvalue of the matrix \( M \). If \( M \) has an eigenvalue \( \lambda \), then for any eigenvector \( v \) corresponding to \( \lambda \), we have: \[ M v = \lambda v \] Substitute into the given equation \( M^2 = 2M + 3I \): \[ M^2 v = 2M v + 3I v \] \[ \lambda^2 v = 2 \lambda v + 3v \] Thus, the eigenvalue \( \lambda \) must satisfy the equation: \[ \lambda^2 = 2 \lambda + 3 \] Rearrange this to get: \[ \lambda^2 - 2\lambda - 3 = 0 \] This is a quadratic equation, and solving for \( \lambda \), we get: \[ \lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2} \] Thus, the eigenvalues are: \[ \lambda = \frac{2 + 4}{2} = 3 \quad \text{or} \quad \lambda = \frac{2 - 4}{2} = -1 \] Step 2: Use the determinant to find the eigenvalue multiplicities. We are given that the determinant of \( M \) is \( -9 \). The determinant of a matrix is the product of its eigenvalues. Let the multiplicities of the eigenvalues \( 3 \) and \( -1 \) be \( m_3 \) and \( m_{-1} \), respectively. Since \( M \) is a \( 3 \times 3 \) matrix, the sum of the multiplicities must be 3: \[ m_3 + m_{-1} = 3 \] The determinant of \( M \) is the product of the eigenvalues raised to their multiplicities: \[ \text{det}(M) = 3^{m_3} (-1)^{m_{-1}} = -9 \] Thus, we have: \[ 3^{m_3} (-1)^{m_{-1}} = -9 \] For \( m_3 = 2 \) and \( m_{-1} = 1 \), we get: \[ 3^2 (-1)^1 = 9 \times (-1) = -9 \] So, \( m_3 = 2 \) and \( m_{-1} = 1 \). Step 3: Compute the trace. The trace of \( M \) is the sum of its eigenvalues, weighted by their multiplicities: \[ \text{Tr}(M) = 3 \times 2 + (-1) \times 1 = 6 - 1 = 5 \] Final Answer: Thus, the trace of \( M \) is \( \boxed{5} \).