Question

The eigenvalues of the matrix \[ A = \begin{bmatrix} 3 & -1 & 1 \\ -1 & 5 & -1 \\ 1 & -1 & 3 \end{bmatrix} \] are \(\lambda_1, \lambda_2, \lambda_3\). The value of \(\lambda_1 \lambda_2 \lambda_3 (\lambda_1 + \lambda_2 + \lambda_3)\) is:

• A. 11
• B. 45
• C. 396
• D. 495

Hint:

For eigenvalue-based expressions, use trace for the sum and determinant for the product of eigenvalues to simplify calculations without solving the characteristic equation.

Answer 1

The Correct Option is D

Step 1: Recall eigenvalue properties.
For any square matrix \(A\): - Sum of eigenvalues = \(\text{Tr}(A)\) (trace). - Product of eigenvalues = \(\det(A)\). 

Step 2: Compute trace.
\[ \text{Tr}(A) = 3 + 5 + 3 = 11 \] So, \[ \lambda_1 + \lambda_2 + \lambda_3 = 11 \] 

Step 3: Compute determinant.
\[ \det(A) = 3\begin{vmatrix} 5 & -1 \\ -1 & 3 \end{vmatrix} - (-1)\begin{vmatrix} -1 & -1 \\ 1 & 3 \end{vmatrix} + 1\begin{vmatrix} -1 & 5 \\ 1 & -1 \end{vmatrix} \] 

First minor: \((5)(3) - (-1)(-1) = 15 - 1 = 14\). Second minor: \((-1)(3) - (-1)(1) = -3 + 1 = -2\). Third minor: \((-1)(-1) - (5)(1) = 1 - 5 = -4\). \[ \det(A) = 3(14) + 1(-2) + 1(-4) = 42 - 2 - 4 = 36 \] So, \[ \lambda_1 \lambda_2 \lambda_3 = 36 \] 

Step 4: Final computation.
\[ \lambda_1 \lambda_2 \lambda_3 (\lambda_1 + \lambda_2 + \lambda_3) = 36 \times 11 = 396 \] Wait — check carefully. Options contain 495, not 396. Let's confirm. 

Step 5: Recalculate determinant.
Expanding again: \[ \det(A) = 3((5)(3) - (-1)(-1)) - (-1)((-1)(3) - (-1)(1)) + (1)((-1)(-1) - (5)(1)) \] \[ = 3(15 - 1) - (-1)(-3 + 1) + (1)(1 - 5) \] \[ = 3(14) - (-1)(-2) + (-4) \] \[ = 42 - (2) - 4 = 36 \] So indeed, \(\det(A) = 36\). Thus, \[ \lambda_1 \lambda_2 \lambda_3 (\lambda_1 + \lambda_2 + \lambda_3) = 36 \times 11 = 396 \] 

Final Answer: \[ \boxed{396} \]