Question

If the characteristic polynomial of the matrix \( A_{3\times3} \) is given by \( f(\lambda) = \lambda^3 - 10\lambda^2 + 27\lambda - 18 \), then trace of A and determinant of A respectively are

• A. 10 and 18
• B. 10 and -18
• C. -10 and 18
• D. -10 and -18

Hint:

The coefficient of \( \lambda^2 \) gives the negative trace, and the constant term gives \( -\det(A) \).

Answer 1

The Correct Option is A

The general form of the characteristic polynomial of a \( 3 \times 3 \) matrix is
\[ f(\lambda) = \lambda^3 - (\text{tr}A)\lambda^2 + (\text{sum of principal minors})\lambda - \det(A) \]
Comparing with \( f(\lambda) = \lambda^3 - 10\lambda^2 + 27\lambda - 18 \), we get:
\[ \text{tr}A = 10, \quad \det(A) = 18 \]