Question

For the matrix \[ \begin{bmatrix} 3 & 1 & 2 \\ 2 & -3 & -1 \\ 1 & 2 & 1 \end{bmatrix} \] find the ratio of the product of eigenvalues to the sum of eigenvalues (round off to nearest integer).

Hint:

Determinant = product of eigenvalues; trace = sum of eigenvalues.

Answer 1

Correct Answer: 8

For any matrix:
\[ \text{Product of eigenvalues} = \det(A) \] \[ \text{Sum of eigenvalues} = \text{trace}(A) \] Trace: \[ 3 + (-3) + 1 = 1 \] Determinant:
\[ \det(A) = 3(-3\cdot1 - (-1)\cdot2) - 1(2\cdot1 - (-1)\cdot1) + 2(2\cdot2 - (-3)\cdot1) \] \[ = 3(-3 + 2) - 1(2 + 1) + 2(4 + 3) \] \[ = 3(-1) - 3 + 14 = -3 - 3 + 14 = 8 \] Thus ratio: \[ \frac{\det(A)}{\text{trace}(A)} = \frac{8}{1} = 8. \]