Question

Let \[ A = \begin{pmatrix} 2 & -1 & 3 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{pmatrix}. \] Then the largest eigenvalue of \(A\) is _________.

Hint:

When a matrix has identical rows (or columns), it immediately implies one eigenvalue is zero. Simplify determinant algebraically to find remaining eigenvalues efficiently.

Answer 1

Correct Answer: 4

Step 1: Observe structure of the matrix.
The first two rows of \(A\) are identical. Hence, the matrix is rank-deficient, and one eigenvalue must be \(0.\)
Step 2: Simplify using linear dependence.
Let’s find the characteristic polynomial: \[ \det(A - \lambda I) = \begin{vmatrix} 2-\lambda & -1 & 3 \\ 2 & -1-\lambda & 3 \\ 3 & 2 & -1-\lambda \end{vmatrix}. \]
Step 3: Expand determinant.
Using expansion along the first row: \[ (2-\lambda) \begin{vmatrix} -1-\lambda & 3 \\ 2 & -1-\lambda \end{vmatrix} + 1 \begin{vmatrix} 2 & 3 \\ 3 & -1-\lambda \end{vmatrix} + 3 \begin{vmatrix} 2 & -1-\lambda \\ 3 & 2 \end{vmatrix}. \] Simplifying gives: \[ (2-\lambda)((-1-\lambda)^2 - 6) + (2(-1-\lambda) - 9) + 3(4 + 3(1+\lambda)). \] \[ (2-\lambda)(\lambda^2 + 2\lambda - 5) + (-2 - 2\lambda - 9) + (12 + 9 + 9\lambda). \] \[ (2-\lambda)(\lambda^2 + 2\lambda - 5) + (-11 - 2\lambda + 21 + 9\lambda). \] \[ (2-\lambda)(\lambda^2 + 2\lambda - 5) + (10 + 7\lambda). \] Expanding: \[ 2\lambda^2 + 4\lambda - 10 - \lambda^3 - 2\lambda^2 + 5\lambda + 10 + 7\lambda. \] Simplify: \[ -\lambda^3 + 16\lambda. \]
Step 4: Solve for eigenvalues.
\[ \lambda(-\lambda^2 + 16) = 0 \Rightarrow \lambda = 0,\; \pm 4. \]
Step 5: Conclusion.
Largest eigenvalue \(= \boxed{4}.\)