Key Concept: Properties of Adjoint and Determinant

The core concept utilized in this problem is the relationship between the determinant of a matrix and the determinant of its adjoint. For an $n \times n$ square matrix $A$, the determinant of its adjoint, denoted as $\operatorname{adj}(A)$, is given by the formula: $|\operatorname{adj}(A)| = |A|^{n-1}$ Here, $n$ represents the order (dimension) of the matrix $A$.

Step-by-Step Solution

1. Determine the Order of Matrix A and Apply the Adjoint Property

• Understanding the setup: We are given matrix $B$ which is the adjoint of matrix $A$, i.e., $B = \operatorname{adj}(A)$. We are also given that $|A|=2$.
• Finding the order of A: Matrix $B$ is a $3 \times 3$ matrix. Since $B$ is the adjoint of $A$, matrix $A$ must also be a $3 \times 3$ matrix. Therefore, the order $n=3$.
• Applying the formula: Using the property $|\operatorname{adj}(A)| = |A|^{n-1}$, we can substitute $n=3$ and $|A|=2$: $|\operatorname{adj}(A)| = |A|^{3-1} = |A|^2 = 2^2 = 4$
• Relating to B: Since $B = \operatorname{adj}(A)$, we have: $|B| = 4$ This is a crucial piece of information that will help us find the value of $\alpha$.

2. Calculate the Determinant of Matrix B in terms of $\alpha$

• Matrix B: $B=\left[\begin{array}{lll} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{array}\right]$
• Expanding the determinant: We calculate $|B|$ by expanding along the first row (or any row/column): $|B| = 1 \cdot \det\left[\begin{array}{ll} 2 & 3 \\ \alpha & 4 \end{array}\right] - 3 \cdot \det\left[\begin{array}{ll} 1 & 3 \\ \alpha & 4 \end{array}\right] + \alpha \cdot \det\left[\begin{array}{ll} 1 & 2 \\ \alpha & \alpha \end{array}\right]$ $|B| = 1( (2)(4) - (3)(\alpha) ) - 3( (1)(4) - (3)(\alpha) ) + \alpha( (1)(\alpha) - (2)(\alpha) )$ $|B| = 1(8 - 3\alpha) - 3(4 - 3\alpha) + \alpha(\alpha - 2\alpha)$ $|B| = 8 - 3\alpha - 12 + 9\alpha + \alpha(-\alpha)$ $|B| = 8 - 3\alpha - 12 + 9\alpha - \alpha^2$ $|B| = -\alpha^2 + 6\alpha - 4$

3. Solve for $\alpha$ using the Determinant Equation

• Equating $|B|$: From Step 1, we know $|B|=4$. From Step 2, we found $|B| = -\alpha^2 + 6\alpha - 4$. Equating these two expressions: $-\alpha^2 + 6\alpha - 4 = 4$
• Rearranging to a quadratic equation: $-\alpha^2 + 6\alpha - 8 = 0$ Multiplying by $-1$ to make the leading coefficient positive: $\alpha^2 - 6\alpha + 8 = 0$
• Factoring the quadratic: We look for two numbers that multiply to 8 and add to -6. These are -2 and -4. $(\alpha - 2)(\alpha - 4) = 0$
• Possible values for $\alpha$: $\alpha = 2 \quad \text{or} \quad \alpha = 4$
• Applying the given condition: The problem states that $\alpha > 2$. Therefore, we must choose $\alpha = 4$.

Common Mistake Alert: Always remember to use all given conditions, like $\alpha > 2$, to narrow down the possible solutions. Failing to do so might lead to incorrect results.

4. Substitute $\alpha=4$ into the Expression and Perform Matrix Multiplication

• The expression to evaluate: We need to find the value of: $\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c}\alpha \\ -2 \alpha \\ \alpha\end{array}\right]$

• Substitute $\alpha=4$ into the row and column matrices: $\left[\begin{array}{ccc}4 & -2(4) & 4\end{array}\right] B\left[\begin{array}{c}4 \\ -2(4) \\ 4\end{array}\right] = \left[\begin{array}{ccc}4 & -8 & 4\end{array}\right] B\left[\begin{array}{c}4 \\ -8 \\ 4\end{array}\right]$

• Substitute $\alpha=4$ into matrix B: $B=\left[\begin{array}{lll} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{array}\right]$

• The full expression with numerical values: $\left[\begin{array}{ccc}4 & -8 & 4\end{array}\right] \left[\begin{array}{lll} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{array}\right] \left[\begin{array}{c}4 \\ -8 \\ 4\end{array}\right]$

• Perform the first matrix multiplication: Multiply the row matrix by $B$. Let $R_1 = \left[\begin{array}{ccc}4 & -8 & 4\end{array}\right]$. $R_1 B = \left[\begin{array}{ccc}4 & -8 & 4\end{array}\right] \left[\begin{array}{lll} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{array}\right]$ The resulting matrix will be a $1 \times 3$ matrix:

  • First element: $(4)(1) + (-8)(1) + (4)(4) = 4 - 8 + 16 = 12$
  • Second element: $(4)(3) + (-8)(2) + (4)(4) = 12 - 16 + 16 = 12$
  • Third element: $(4)(4) + (-8)(3) + (4)(4) = 16 - 24 + 16 = 8$ So, $R_1 B = \left[\begin{array}{ccc}12 & 12 & 8\end{array}\right]$.

• Perform the second matrix multiplication: Now multiply the result from the previous step by the column matrix $C_1 = \left[\begin{array}{c}4 \\ -8 \\ 4\end{array}\right]$. $\left[\begin{array}{ccc}12 & 12 & 8\end{array}\right] \left[\begin{array}{c}4 \\ -8 \\ 4\end{array}\right]$ The result will be a $1 \times 1$ matrix (a scalar value): $(12)(4) + (12)(-8) + (8)(4)$ $= 48 - 96 + 32$ $= 80 - 96$ $= -16$

Tip for Matrix Multiplication: Be meticulous with your calculations, especially with signs. A common error is miscalculating products or sums.

Conclusion

The value of the given expression is $-16$.

Final Answer: The final answer is $\boxed{\text{-16}}$.

Key Takeaway: This problem effectively tests your understanding of the properties of the adjoint matrix and determinants, specifically $|\operatorname{adj}(A)| = |A|^{n-1}$, along with your ability to perform matrix determinant calculations and matrix multiplication accurately. Always pay attention to given conditions like inequalities for variables.