Question

Let \( M \) be a \(3 \times 3\) real matrix. Let \[ \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \quad \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \begin{pmatrix} 0 \\ -1 \\ \alpha \end{pmatrix} \] be eigenvectors of \(M\) corresponding to three distinct eigenvalues. Then, which of the following is NOT a possible value of \(\alpha\)?

• A. 0
• B. 1
• C. -2
• D. 2

Hint:

Eigenvectors corresponding to distinct eigenvalues must be linearly independent, so their determinant should not vanish.

Answer 1

The Correct Option is C

Step 1: Property of eigenvectors.
Eigenvectors corresponding to distinct eigenvalues must be linearly independent.
Step 2: Check for linear independence.
Form the matrix with the given vectors as columns: \[ A = \begin{pmatrix} 1 & 1 & 0 \\ 2 & 1 & -1 \\ 3 & 1 & \alpha \end{pmatrix} \] For independence, \(\det(A) \neq 0\).
Step 3: Compute determinant.
\[ \det(A) = 1(\alpha \cdot 1 - (-1)\cdot 1) - 1(2\alpha - (-1)\cdot 3) + 0(...) = (\alpha + 1) - (2\alpha + 3) = -\alpha - 2 \] Set \(\det(A) = 0 \Rightarrow \alpha = -2\).
Step 4: Conclusion.
If \(\alpha = -2\), the determinant becomes 0, meaning the vectors are linearly dependent. Thus, \(\alpha = -2\) is NOT allowed. Final Answer: \[ \boxed{-2} \]