Question

What are the eigenvalues of the matrix \[ \begin{bmatrix} 2 & 1 & 1
1 & 4 & 1
1 & 1 & 2 \end{bmatrix} \] ?

• A. \( 1, 2, 5 \)
• B. \( 1, 3, 4 \)
• C. \( -5, 1, 2 \)
• D. \( -5, -1, 2 \)

Hint:

To find eigenvalues of a matrix, always solve the characteristic equation \( \det(A - \lambda I) = 0 \).

Answer 1

The Correct Option is A

Step 1: Given matrix: \[ A = \begin{bmatrix} 2 & 1 & 1
1 & 4 & 1
1 & 1 & 2 \end{bmatrix} \] The characteristic equation is obtained by solving: \[ \det(A - \lambda I) = 0 \] Step 2: Expanding the determinant: \[ \begin{vmatrix} 2 - \lambda & 1 & 1
1 & 4 - \lambda & 1
1 & 1 & 2 - \lambda \end{vmatrix} = 0 \] Expanding along the first row: \[ (2 - \lambda) \begin{vmatrix} 4 - \lambda & 1
1 & 2 - \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 1
1 & 2 - \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 4 - \lambda
1 & 1 \end{vmatrix} \] Step 3: Solving the determinant: \[ (2 - \lambda) [(4 - \lambda)(2 - \lambda) - 1] - 1(2 - \lambda - 1) + 1(1 - 4 + \lambda) \] Solving the resulting cubic equation: \[ (\lambda - 1)(\lambda - 2)(\lambda - 5) = 0 \] Conclusion: The eigenvalues are \( 1, 2, 5 \), which corresponds to option (A).