Question

Find the eigenvalues of the matrix:
Matrix \[ A = \begin{bmatrix} 4 & 1 & 2 \\ 1 & 3 & 0 \\ 2 & 0 & 5 \end{bmatrix} \]

• A. 1, 5, 6
• B. 3, 4, 7
• C. 2, 3, 8
• D. 1, 2, 9

Hint:

To find eigenvalues, solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( \lambda \) represents the eigenvalue. The solutions to this equation give the eigenvalues.

Answer 1

The Correct Option is C

Given matrix \( A \) is: \[ A = \begin{bmatrix} 4 & 1 & 2 \\ 1 & 3 & 0 \\ 2 & 0 & 5 \end{bmatrix} \] The eigenvalues of matrix \( A \) are the solutions to the characteristic equation: \[ \det(A - \lambda I) = 0 \] where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix. First, subtract \( \lambda \) from the diagonal elements of \( A \): \[ A - \lambda I = \begin{bmatrix} 4-\lambda & 1 & 2 \\ 1 & 3-\lambda & 0 \\ 2 & 0 & 5-\lambda \end{bmatrix} \] Now, calculate the determinant of the matrix: \[ \det(A - \lambda I) = \begin{vmatrix} 4-\lambda & 1 & 2 \\ 1 & 3-\lambda & 0 \\ 2 & 0 & 5-\lambda \end{vmatrix} \] Expanding the determinant, we get: \[ (4-\lambda) \begin{vmatrix} 3-\lambda & 0 \\ 0 & 5-\lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 0 \\ 2 & 5-\lambda \end{vmatrix} + 2 \begin{vmatrix} 1 & 3-\lambda \\ 2 & 0 \end{vmatrix} \] Simplifying this expression leads to the characteristic equation: \[ (\lambda - 2)(\lambda - 3)(\lambda - 8) = 0 \] Thus, the eigenvalues are \( \lambda = 2, 3, 8 \).