Question

The characteristic polynomial of a \(3 \times 3\) matrix \( A \) is \[ |A - \lambda I| = \lambda^3 - 9\lambda^2 + 23\lambda - 15. \] Let \( X = \text{trace}(A) \) and \( Y = \det(A) \), then:

• A. \( X = Y = 3 \)
• B. \( \dfrac{X}{Y} = \dfrac{9}{15} \)
• C. \( \dfrac{X}{Y} = \dfrac{15}{9} \)
• D. \( X = 15, Y = 8 \)

Hint:

In a characteristic polynomial of a \(3 \times 3\) matrix, the coefficient of \(\lambda^2\) gives the negative of the trace, and the constant term gives the determinant.

Answer 1

The Correct Option is B

For a \(3 \times 3\) matrix \(A\), the characteristic polynomial is generally given by: \[ |A - \lambda I| = \lambda^3 - (\text{trace } A)\lambda^2 + (\text{sum of principal minors})\lambda - \det(A). \] Comparing the given polynomial: \[ \lambda^3 - 9\lambda^2 + 23\lambda - 15 \] with the general form, we get:
- Trace \(X = \text{trace}(A) = 9\)
- Determinant \(Y = \det(A) = 15\)
So, \[ \dfrac{X}{Y} = \dfrac{9}{15} \]