Question

Let \( \alpha, \beta \) and \( \gamma \) be the eigenvalues of \[ M = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 3 & 3 \\ -1 & 2 & 2 \end{bmatrix}. \] If \( \gamma = 1 \) and \( \alpha > \beta \), then the value of \( 2\alpha + 3\beta \) is .............

Hint:

For a 3×3 matrix, use the trace (sum of eigenvalues) and determinant (product of eigenvalues) to check consistency after finding roots.

Answer 1

Correct Answer: 7

Step 1: Write the characteristic equation.
We find the eigenvalues from \[ |M - \lambda I| = 0. \] So, \[ \begin{vmatrix} -\lambda & 1 & 0 \\ 1 & 3 - \lambda & 3 \\ -1 & 2 & 2 - \lambda \end{vmatrix} = 0. \]
Step 2: Expand the determinant.
Expanding along the first row: \[ (-\lambda) \begin{vmatrix} 3 - \lambda & 3 \\ 2 & 2 - \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 3 \\ -1 & 2 - \lambda \end{vmatrix} + 0 = 0. \] Compute each term: \[ (-\lambda)[(3 - \lambda)(2 - \lambda) - 6] - [1(2 - \lambda) - (-3)] = 0. \] Simplify: \[ (-\lambda)[\lambda^2 - 5\lambda] - [(2 - \lambda) + 3] = 0, \] \[ -\lambda^3 + 5\lambda^2 - (5 - \lambda) = 0, \] \[ -\lambda^3 + 5\lambda^2 + \lambda - 5 = 0. \] Multiply by \(-1\): \[ \lambda^3 - 5\lambda^2 - \lambda + 5 = 0. \]
Step 3: Use the given eigenvalue.
Since \(\gamma = 1\) is an eigenvalue, substitute \(\lambda = 1\): \[ 1 - 5 - 1 + 5 = 0. \] Thus, divide the polynomial by \((\lambda - 1)\).
Step 4: Perform synthetic division.
Coefficients: \(1, -5, -1, 5\) \[ \begin{array}{r|rrrr} 1 & 1 & -5 & -1 & 5 \\ & & 1 & -4 & -5 \\ \hline & 1 & -4 & -5 & 0 \end{array} \] The quotient is \(\lambda^2 - 4\lambda - 5 = 0\). Hence, the other roots are: \[ \lambda = 5, \ -1. \]
Step 5: Identify eigenvalues.
Eigenvalues: \(\alpha = 5, \beta = -1, \gamma = 1\). Given \(\alpha > \beta\), we use these.
Step 6: Compute required value.
\[ 2\alpha + 3\beta = 2(5) + 3(-1) = 10 - 3 = 7. \] Final Answer: \[ \boxed{7} \]