Question

Suppose that 2 is an eigenvalue of the matrix \[ A = \begin{bmatrix} 0 & 3 & -\alpha\\ 0 & 1 & 0 \\1 & -1 & 3 \end{bmatrix}. \] Then the value of $\alpha$ is equal to (Answer in integer):

Hint:

When checking eigenvalue conditions, simplify determinant expansion carefully. Here, using row/column with zeros helps speed computation.

Answer 1

Correct Answer: 1

Step 1: Eigenvalue condition.
We need $\det(A-2I)=0$.

Step 2: Compute determinant.
\[ A-2I=\begin{bmatrix} -2 & 3 & -\alpha
0 & -1 & 0
1 & -1 & 1 \end{bmatrix}. \] Expand along row 2: \[ \det(A-2I)=(-1)\cdot \det\begin{bmatrix} -2 & -\alpha
1 & 1 \end{bmatrix}. \]

Step 3: Evaluate minor.
\[ \det\begin{bmatrix}-2 & -\alpha
1 & 1\end{bmatrix}=(-2)(1)-(-\alpha)(1)=-2+\alpha. \] So, \[ \det(A-2I)=(-1)(-2+\alpha)=2-\alpha. \]

Step 4: Solve for $\alpha$.
Eigenvalue condition: $2-\alpha=0 \;\Rightarrow\; \alpha=2$. Wait — careful! Let’s recompute with full expansion: \[ \det(A-2I)=\begin{vmatrix}-2 & 3 & -\alpha
0 & -1 & 0
1 & -1 & 1\end{vmatrix}. \] Expand across row 2: $(-1)(-1)^{2+2}\det\begin{bmatrix}-2 & -\alpha
1 & 1\end{bmatrix}$ indeed. So it is $(-1)((-2)(1)-(-\alpha)(1))=(-1)(-2+\alpha)=2-\alpha$. Set $2-\alpha=0 \Rightarrow \alpha=2$. Final Answer: $\boxed{2}$