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

To solve this problem, we start by utilizing the fundamental properties of the adjoint of a matrix.
First, let's find the determinant of matrix $A$:
$$ |A| = \begin{vmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{vmatrix} $$
Expanding along the third row (which has the most zeros):
$$ |A| = 0 - 0 + 1 \cdot \begin{vmatrix} -1 & 1 \\ 1 & 0 \end{vmatrix} = 1(0 - 1) = -1 $$
Next, we use the property $\text{adj}(\text{adj}(A)) = |A|^{n-2} A$. Here, $n=3$, so:
$$ \text{adj}(\text{adj}(A)) = |A|^{3-2} A = |A| A = (-1) A = -A $$
Now, for the second part of the expression, $\text{adj}(A)(\text{adj}(\text{adj}(A)))$. By definition, for any square matrix $B$, $B \cdot \text{adj}(B) = |B| I$. Letting $B = \text{adj}(A)$, we have:
$$ \text{adj}(A) \cdot \text{adj}(\text{adj}(A)) = |\text{adj}(A)| I $$
We know that $|\text{adj}(A)| = |A|^{n-1} = (-1)^{3-1} = 1$. Thus:
$$ \text{adj}(A) \cdot \text{adj}(\text{adj}(A)) = 1 \cdot I = I $$
Substituting these results back into the given equation:
$$ A^2 + \alpha(-A) + \beta(I) = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix} $$
$$ A^2 - \alpha A + \beta I = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix} $$
Calculating $A^2$:
$$ A^2 = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$
Now substitute $A^2$, $A$, and $I$ into the equation:
$$ \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix} $$
Comparing the element at position (1, 2):
$$ -1 - \alpha(1) + 0 = -2 \Rightarrow -1 - \alpha = -2 \Rightarrow \alpha = 1 $$
Comparing the element at position (2, 2):
$$ 1 - \alpha(0) + \beta = 0 \Rightarrow 1 + \beta = 0 \Rightarrow \beta = -1 $$
Finally, calculate $(\alpha - \beta)^2$:
$$ (1 - (-1))^2 = (2)^2 = 4 $$