Key Concepts and Formulas

1. Determinant of a Product: For any two square matrices $X$ and $Y$ of the same order $n$, $\det(XY) = \det(X) \det(Y)$. This property allows us to separate the determinant of a product into the product of individual determinants.
2. Determinant of an Adjoint: For any square matrix $M$ of order $n$, $\det(\operatorname{adj}(M)) = (\det(M))^{n-1}$. This formula directly links the determinant of the adjoint to the determinant of the original matrix.
3. Determinant of a Scalar Multiple: For any scalar $k$ and square matrix $M$ of order $n$, $\det(kM) = k^n \det(M)$. This is crucial for factoring out scalar multiples from a determinant.
4. Determinant of a Transpose: For any square matrix $M$, $\det(M^T) = \det(M)$. The determinant remains unchanged when a matrix is transposed.

In this problem, the matrix $A$ is of order $3 \times 3$, so $n=3$. Consequently, $n-1 = 2$.

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Step-by-Step Solution

Step 1: Deconstruct the Given Determinant Equation The problem provides the equation: $\det\left(\operatorname{adj}\left(2 A-A^T\right) \cdot \operatorname{adj}\left(A-2 A^T\right)\right)=2^8$ Our goal is to simplify this expression using the determinant properties. Let $M_1 = 2A - A^T$ and $M_2 = A - 2A^T$. The equation becomes $\det(\operatorname{adj}(M_1) \cdot \operatorname{adj}(M_2)) = 2^8$.

Applying the determinant of a product property: $\det(\operatorname{adj}(M_1)) \cdot \det(\operatorname{adj}(M_2)) = 2^8$

Now, applying the determinant of an adjoint property, $\det(\operatorname{adj}(M)) = (\det(M))^{n-1}$ with $n=3 \implies n-1=2$: $(\det(M_1))^2 \cdot (\det(M_2))^2 = 2^8$ This can be compactly written as: $(\det(M_1) \cdot \det(M_2))^2 = 2^8$

Taking the square root of both sides, we must consider both positive and negative roots: $\det(M_1) \cdot \det(M_2) = \pm \sqrt{2^8} = \pm 2^4 = \pm 16$

Step 2: Identify the Relationship Between $M_1$ and $M_2$ Let's examine the expressions for $M_1$ and $M_2$: $M_1 = 2A - A^T$ $M_2 = A - 2A^T$

To find a relationship, let's consider the transpose of $M_1$: $M_1^T = (2A - A^T)^T$ Using the property $(X-Y)^T = X^T - Y^T$ and $(kX)^T = kX^T$: $M_1^T = (2A)^T - (A^T)^T = 2A^T - A$ Now, compare $M_1^T$ with $M_2$: $M_1^T = 2A^T - A = -(A - 2A^T) = -M_2$ So, we have the crucial relationship: $\mathbf{M_1^T = -M_2}$.

Next, we relate their determinants using the properties: $\det(M_1) = \det(M_1^T)$ (using the determinant of a transpose property) Since $M_1^T = -M_2$, we have $\det(M_1^T) = \det(-M_2)$. Applying the determinant of a scalar multiple property, $\det(kM) = k^n \det(M)$ with $k=-1$ and $n=3$: $\det(-M_2) = (-1)^3 \det(M_2) = -\det(M_2)$ Combining these, we get: $\det(M_1) = -\det(M_2)$

Step 3: Solve for $\det(M_2)$ We have two key equations from Step 1 and Step 2:

1. $\det(M_1) \cdot \det(M_2) = \pm 16$
2. $\det(M_1) = -\det(M_2)$

Substitute the second equation into the first one: $(-\det(M_2)) \cdot (\det(M_2)) = \pm 16$ $-(\det(M_2))^2 = \pm 16$ Since $(\det(M_2))^2$ is a square, it must be non-negative. Therefore, $-(\det(M_2))^2$ must be non-positive. This implies that the right side of the equation must be negative. So, we choose $-16$: $-(\det(M_2))^2 = -16$ $(\det(M_2))^2 = 16$ Taking the square root of both sides: $\det(M_2) = \pm 4$ Thus, $\det(A - 2A^T) = \pm 4$.

Step 4: Calculate the Matrix $M_2$ and its Determinant First, let's write down matrix $A$ and its transpose $A^T$: $A = \left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]$ $A^T = \left[\begin{array}{lll}1 & 1 & 0 \\ 2 & 0 & 1 \\ \alpha & 1 & 2\end{array}\right]$ Now, calculate $2A^T$: $2A^T = \left[\begin{array}{ccc}2 & 2 & 0 \\ 4 & 0 & 2 \\ 2\alpha & 2 & 4\end{array}\right]$ Next, calculate $M_2 = A - 2A^T$: $M_2 = \left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right] - \left[\begin{array}{ccc}2 & 2 & 0 \\ 4 & 0 & 2 \\ 2\alpha & 2 & 4\end{array}\right] = \left[\begin{array}{ccc}1-2 & 2-2 & \alpha-0 \\ 1-4 & 0-0 & 1-2 \\ 0-2\alpha & 1-2 & 2-4\end{array}\right]$ $M_2 = \left[\begin{array}{ccc}-1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2\alpha & -1 & -2\end{array}\right]$ Now, we calculate the determinant of $M_2$. Expanding along the second column (which has two zeros) simplifies the calculation: $\det(M_2) = a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32}$ $\det(M_2) = (0)C_{12} + (0)C_{22} + (-1)C_{32}$ We only need to calculate the cofactor $C_{32}$. The element $a_{32} = -1$. The minor $M_{32}$ is the determinant of the submatrix obtained by removing row 3 and column 2: $M_{32} = \det\left(\begin{array}{cc}-1 & \alpha \\ -3 & -1\end{array}\right) = (-1)(-1) - (\alpha)(-3) = 1 + 3\alpha$ The cofactor $C_{32} = (-1)^{3+2} M_{32} = (-1)(1 + 3\alpha) = -1 - 3\alpha$. Therefore, $\det(M_2) = (-1) \cdot C_{32} = (-1)(-1 - 3\alpha) = 1 + 3\alpha$.

Step 5: Determine the Value of $\alpha$ From Step 3, we found $\det(M_2) = \pm 4$. From Step 4, we calculated $\det(M_2) = 1 + 3\alpha$. Equating these two results: $1 + 3\alpha = \pm 4$ This gives two possible cases:

Case 1: $1 + 3\alpha = 4$ $3\alpha = 4 - 1 \implies 3\alpha = 3 \implies \alpha = 1$

Case 2: $1 + 3\alpha = -4$ $3\alpha = -4 - 1 \implies 3\alpha = -5 \implies \alpha = -\frac{5}{3}$

The problem statement specifies that $\alpha \in (0, \infty)$, meaning $\alpha$ must be a positive real number. Thus, we select $\alpha = 1$ and discard $\alpha = -5/3$.

Step 6: Calculate $\det(A)$ and $(\det(A))^2$ Substitute the value $\alpha = 1$ back into the original matrix $A$: $A = \left[\begin{array}{lll}1 & 2 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]$ Now, calculate $\det(A)$ by expanding along the first row: $\det(A) = 1 \cdot \det\left(\begin{array}{cc}0 & 1 \\ 1 & 2\end{array}\right) - 2 \cdot \det\left(\begin{array}{cc}1 & 1 \\ 0 & 2\end{array}\right) + 1 \cdot \det\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)$ $\det(A) = 1 \cdot ((0)(2) - (1)(1)) - 2 \cdot ((1)(2) - (1)(0)) + 1 \cdot ((1)(1) - (0)(0))$ $\det(A) = 1 \cdot (-1) - 2 \cdot (2) + 1 \cdot (1)$ $\det(A) = -1 - 4 + 1$ $\det(A) = -4$ Finally, the problem asks for the value of $(\det(A))^2$: $(\det(A))^2 = (-4)^2 = 16$

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Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when calculating cofactors, determinants, and especially when taking square roots (remember $\pm$).
• Matrix Order (n): Always ensure you use the correct matrix order $n$ when applying formulas like $\det(kM) = k^n \det(M)$ and $\det(\operatorname{adj}(M)) = (\det(M))^{n-1}$. Here, $n=3$.
• Domain Restrictions: Never forget to check and apply any given constraints on variables, such as $\alpha \in (0, \infty)$, as they help in selecting the correct solution.

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Summary This problem requires a systematic application of determinant properties. We first simplified the given expression by using the determinant of a product and the determinant of an adjoint. Next, we identified a crucial relationship between $M_1 = 2A - A^T$ and $M_2 = A - 2A^T$ using transposes and scalar multiplication. This allowed us to solve for $\det(M_2)$. After calculating $M_2$ and its determinant in terms of $\alpha$, we equated the expressions to find possible values of $\alpha$. The given domain restriction for $\alpha$ helped us select the unique correct value. Finally, substituting $\alpha$ back into matrix $A$ allowed us to calculate $\det(A)$ and ultimately $(\det(A))^2$.

The final answer is $\boxed{\text{16}}$, which corresponds to option (A).