Question

Consider the matrix: \[ A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \] The eigenvalues of the matrix are 0.27 and __________ (rounded off to 2 decimal places).

Hint:

To find the eigenvalues of a matrix, solve the characteristic equation \( {det}(A - \lambda I) = 0 \). The roots of the resulting quadratic equation are the eigenvalues.

Answer 1

Correct Answer: 3.68

The eigenvalues of a matrix \( A \) are found by solving the characteristic equation: \[ \det(A - \lambda I) = 0 \] where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix. For the given matrix \( A \): \[ A - \lambda I = \begin{bmatrix} 2 - \lambda & 3 \\ 1 & 2 - \lambda \end{bmatrix} \] Now, calculate the determinant: \[ \det(A - \lambda I) = (2 - \lambda)(2 - \lambda) - 3 \times 1 \] \[ = (2 - \lambda)^2 - 3 \] \[ = 4 - 4\lambda + \lambda^2 - 3 \] \[ = \lambda^2 - 4\lambda + 1 \] Set the determinant equal to zero: \[ \lambda^2 - 4\lambda + 1 = 0 \] Solve using the quadratic formula: \[ \lambda = \frac{4 \pm \sqrt{16 - 4}}{2} \] \[ \lambda = \frac{4 \pm \sqrt{12}}{2} \] \[ \lambda = \frac{4 \pm 3.464}{2} \] Thus, the eigenvalues are: \[ \lambda_1 = 3.73, \quad \lambda_2 = 0.27 \] Therefore, the second eigenvalue is 3.73.