Question

Let $A = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ satisfy $A^2 + \alpha(\text{adj}(\text{adj}(A))) + \beta(\text{adj}(A)(\text{adj}(\text{adj}(A)))) = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$ for some $\alpha, \beta \in \mathbb{R}$.
Then $(\alpha - \beta)^2$ is equal to __________.

Hint:

Use the property that $\text{adj}(\text{adj}(A)) = |A|^{n-2} A$ and $\text{adj}(A) \cdot \text{adj}(\text{adj}(A)) = |\text{adj}(A)| I$.

Answer 1

Correct Answer: 4

Alternatively, we can evaluate the entire matrix expression step-by-step. Let $A = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}$.
Step 1: Calculate $A^2$:
$$ A^2 = \begin{bmatrix} (-1)(-1)+1(1)+(-1)(0) & (-1)(1)+1(0)+(-1)(0) & (-1)(-1)+1(1)+(-1)(1) \\ 1(-1)+0(1)+1(0) & 1(1)+0(0)+1(0) & 1(-1)+0(1)+1(1) \\ 0(-1)+0(1)+1(0) & 0(1)+0(0)+1(0) & 0(-1)+0(1)+1(1) \end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$
Step 2: Find $\text{adj}(A)$. We first find the cofactor matrix $C$:
$C_{11} = 0, C_{12} = -1, C_{13} = 0, C_{21} = -1, C_{22} = -1, C_{23} = 0, C_{31} = 1, C_{32} = 0, C_{33} = -1$.
Thus, $\text{adj}(A) = C^T = \begin{bmatrix} 0 & -1 & 1 \\ -1 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$.
Step 3: Find $\text{adj}(\text{adj}(A))$ similarly from $\text{adj}(A)$:
The cofactor matrix for $\text{adj}(A)$ has elements:
$C'_{11} = 1, C'_{12} = -1, C'_{13} = 0, C'_{21} = -1, C'_{22} = 0, C'_{23} = 0, C'_{31} = 1, C'_{32} = -1, C'_{33} = -1$.
So $\text{adj}(\text{adj}(A)) = \begin{bmatrix} 1 & -1 & 1 \\ -1 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix} = -A$.
Step 4: Find the product of adjoints:
$$ \text{adj}(A) \cdot \text{adj}(\text{adj}(A)) = \begin{bmatrix} 0 & -1 & 1 \\ -1 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ -1 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I $$
Step 5: Form the final equation:
$$ \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \alpha \begin{bmatrix} 1 & -1 & 1 \\ -1 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix} $$
Equating corresponding entries:
Row 1, Col 2: $-1 - \alpha = -2 \Rightarrow \alpha = 1$
Row 2, Col 2: $1 + \beta = 0 \Rightarrow \beta = -1$
Then $(\alpha - \beta)^2 = (1 - (-1))^2 = 4$.