Question

Let \( M \) be a real \( 6 \times 6 \) matrix. Let 2 and -1 be two eigenvalues of \( M. \) If \( M^5 = aI + bM \), where \( a, b \in \mathbb{R}, \) then

• A. \( a = 10, b = 11 \)
• B. \( a = -11, b = 10 \)
• C. \( a = -10, b = 11 \)
• D. \( a = 10, b = -11 \)

Hint:

For polynomial relations of matrices, substitute eigenvalues directly to get scalar equations for \( a, b. \)

Answer 1

The Correct Option is A

Step 1: Use property of eigenvalues. 
If \( \lambda \) is an eigenvalue of \( M \), then \( \lambda^5 \) is an eigenvalue of \( M^5 \). Hence, for eigenvalue relation \( M^5 = aI + bM \), we get: \[ \lambda^5 = a + b\lambda. \]

Step 2: Substitute eigenvalues. 
For \( \lambda = 2 \): \( 2^5 = a + 2b \Rightarrow 32 = a + 2b. \) For \( \lambda = -1 \): \( (-1)^5 = a - b \Rightarrow -1 = a - b. \)

Step 3: Solve for \( a, b. \) 
Subtract equations: \[ (32 - (-1)) = (a + 2b) - (a - b) \Rightarrow 33 = 3b \Rightarrow b = 11. \] Substitute in \( a - b = -1 \Rightarrow a = 10. \)

Final Answer: \( a = 10, b = 11. \)