Question

Let A be a 3 \(\times\) 3 matrix such that the characteristic polynomial of A is f(x) \(=\) x\(^3\) \(-\) 4x\(^2\) \(-\) 11x \(+\) 30. If one of the eigenvalues of A is \(-\)3, then sum of the other two eigenvalues is

• A. \(-1\)
• B. \(-10\)
• C. 4
• D. 7

Hint:

For any characteristic equation \( x^3 - c_1 x^2 + c_2 x - c_3 = 0 \):
- Sum of roots (eigenvalues) \( = c_1 \) (coefficient of \( x^2 \) with sign changed).
- Product of roots (determinant) \( = c_3 \) (constant term with sign changed).
This property saves valuable time by avoiding the need to fully factor the polynomial.

Answer 1

The Correct Option is D

Step 1: Divide out the known root.
Since \( x = -3 \) is a root of \( f(x) = x^3 - 4x^2 - 11x + 30 \), dividing the cubic by \( (x + 3) \) using synthetic division with coefficients \( 1, -4, -11, 30 \) gives quotient coefficients \( 1, -7, 10 \), with zero remainder.

Step 2: Read off the reduced quadratic.
This means \( f(x) = (x + 3)(x^2 - 7x + 10) \), so the other two eigenvalues are the roots of:
\[ x^2 - 7x + 10 = 0 \]

Step 3: Use the sum-of-roots shortcut for a quadratic.
For \( x^2 - 7x + 10 = 0 \), the sum of the two roots is \( -(-7)/1 = 7 \), without needing to factor or solve for the individual roots.

Step 4: Final Answer.
\[ \boxed{\lambda_2 + \lambda_3 = 7} \]
Therefore, option (D) is correct.