Question

If \( \lambda_1, \lambda_2, \lambda_3 \) are the eigenvalues of the matrix \( A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & -3 & 1 \\ 0 & 0 & 4 \end{pmatrix} \), then the values of \[ (\lambda_1 + \lambda_2 + \lambda_3) \ \text{and} \ (\lambda_1 \cdot \lambda_2 \cdot \lambda_3) \] are respectively

• A. 1 and 3
• B. -3 and 4
• C. 4 and 12
• D. 2 and -12

Hint:

The eigenvalues of a triangular matrix are the entries on its diagonal. Use this to quickly compute sums and products of eigenvalues.

Answer 1

The Correct Option is D

The eigenvalues of a triangular matrix (upper or lower) are its diagonal elements. For matrix \[ A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & -3 & 1 \\ 0 & 0 & 4 \end{pmatrix} \] the eigenvalues are: \[ \lambda_1 = 1, \ \lambda_2 = -3, \ \lambda_3 = 4 \] Now, \[ \lambda_1 + \lambda_2 + \lambda_3 = 1 + (-3) + 4 = 2 \] and \[ \lambda_1 \cdot \lambda_2 \cdot \lambda_3 = 1 \times (-3) \times 4 = -12 \]