Given the matrix A = \( \begin{pmatrix} 1 & -2 & 0 \\ -2 & 1 & -2 \\ 0 & -2 & 1 \end{pmatrix} \), find the characteristic equation.
Answer
Step 3: Therefore, the characteristic equation of matrix A is \(\lambda^3 - 3\lambda^2 + 4\lambda - 1 = 0\).
Step 1 ：Step 1: We start by finding the determinant of the matrix \(A - \lambda I\), where \(I\) is the identity matrix and \(\lambda\) is a scalar. So, \(A - \lambda I = \begin{pmatrix} 1-\lambda & -2 & 0 \\ -2 & 1-\lambda & -2 \\ 0 & -2 & 1-\lambda \end{pmatrix}\).
Step 2 ：Step 2: The determinant of \(A - \lambda I\) is \((1-\lambda)[(1-\lambda)^2 - 4] + 2 * 2 * 2 - 0\), which simplifies to \(\lambda^3 - 3\lambda^2 + 4\lambda - 1\).
Step 3 ：Step 3: Therefore, the characteristic equation of matrix A is \(\lambda^3 - 3\lambda^2 + 4\lambda - 1 = 0\).
