Key Concepts and Formulas

This problem requires the application of fundamental properties of determinants, traces, and adjoints of matrices, especially for scalar multiplication and nested adjoint operations. For an $n \times n$ matrix $X$:

1. Determinant of a scalar multiple: If $k$ is a scalar, then $\det(kX) = k^n \det(X)$.
2. Trace of a scalar multiple: If $k$ is a scalar, then $\operatorname{trace}(kX) = k \operatorname{trace}(X)$.
3. Adjoint of an adjoint matrix: For an invertible $n \times n$ matrix $X$, $\operatorname{adj}(\operatorname{adj}(X)) = (\det(X))^{n-2} X$. For a $3 \times 3$ matrix ($n=3$), this identity simplifies to $\operatorname{adj}(\operatorname{adj}(X)) = (\det(X))^{3-2} X = \det(X) \cdot X$.

Step-by-Step Solution

We are given that $A$ is a $3 \times 3$ matrix (so $n=3$), $\det(A) = \frac{1}{2}$, and $\operatorname{trace}(A) = 3$. We need to find the value of $|B| + \operatorname{trace}(B)$, where $B = \operatorname{adj}(\operatorname{adj}(2A))$.

Step 1: Simplify the expression for matrix $B$. Our primary goal is to simplify the complex expression for $B$ into a more manageable form, ideally as a scalar multiple of $A$. We are given $B = \operatorname{adj}(\operatorname{adj}(2A))$. Let $X = 2A$. Since $A$ is a $3 \times 3$ matrix, $X$ is also a $3 \times 3$ matrix. Using the key identity for the adjoint of an adjoint of a $3 \times 3$ matrix, $\operatorname{adj}(\operatorname{adj}(X)) = \det(X) \cdot X$: $B = \operatorname{adj}(\operatorname{adj}(2A)) = \det(2A) \cdot (2A)$ Explanation: This is a crucial simplification step. The identity for nested adjoints directly transforms the expression involving two adjoint operations into a simpler form involving only the determinant of the inner matrix and the inner matrix itself.

Step 2: Calculate the determinant of $2A$. Before we can fully simplify $B$, we need to find the value of $\det(2A)$. We know $\det(A) = \frac{1}{2}$ and $A$ is a $3 \times 3$ matrix ($n=3$). Using the property $\det(kX) = k^n \det(X)$ with $k=2$ and $n=3$: $\det(2A) = 2^3 \det(A)$ $\det(2A) = 8 \cdot \frac{1}{2}$ $\det(2A) = 4$ Explanation: The determinant of a scalar multiple of a matrix scales by the scalar raised to the power of the matrix's dimension. Here, the scalar is 2 and the dimension is 3, so $2^3 = 8$.

Step 3: Express $B$ in terms of $A$. Now, substitute the calculated value of $\det(2A)$ back into the simplified expression for $B$ from Step 1: $B = \det(2A) \cdot (2A)$ $B = 4 \cdot (2A)$ $B = 8A$ Explanation: We have successfully simplified $B$ to a scalar multiple of $A$. This makes calculating $\det(B)$ and $\operatorname{trace}(B)$ straightforward, as we already know $\det(A)$ and $\operatorname{trace}(A)$.

Step 4: Calculate the determinant of $B$. We need to find $\det(B)$. Since $B=8A$, and $A$ is a $3 \times 3$ matrix: Using the property $\det(kX) = k^n \det(X)$ with $k=8$ and $n=3$: $\det(B) = \det(8A) = 8^3 \det(A)$ We are given $\det(A) = \frac{1}{2}$. $\det(B) = 512 \cdot \frac{1}{2}$ $\det(B) = 256$ Explanation: Similar to Step 2, the determinant of $8A$ involves $8^3$ because $A$ is a $3 \times 3$ matrix.

Step 5: Calculate the trace of $B$. Next, we need to find $\operatorname{trace}(B)$. Since $B=8A$: Using the property $\operatorname{trace}(kX) = k \operatorname{trace}(X)$ with $k=8$: $\operatorname{trace}(B) = \operatorname{trace}(8A) = 8 \cdot \operatorname{trace}(A)$ We are given $\operatorname{trace}(A) = 3$. $\operatorname{trace}(B) = 8 \cdot 3$ $\operatorname{trace}(B) = 24$ Explanation: The trace operator is linear. This means that the trace of a scalar multiple of a matrix is simply the scalar multiplied by the trace of the matrix.

Step 6: Compute the final value of $|B| + \operatorname{trace}(B)$. Finally, we sum the calculated values from Step 4 and Step 5: $|B| + \operatorname{trace}(B) = 256 + 24$ $|B| + \operatorname{trace}(B) = 280$ Explanation: This is the final required computation as per the question.

Common Mistakes & Tips

• Dimension dependence: Always pay close attention to the dimension $n$ of the matrix, especially when dealing with determinants of scalar multiples ($\det(kX) = k^n \det(X)$) and adjoint properties. For a $3 \times 3$ matrix, $n=3$.
• Adjoint identity: The formula $\operatorname{adj}(\operatorname{adj}(X)) = (\det(X))^{n-2} X$ is critical. Ensure you use the correct power for $n-2$. For $n=3$, it simplifies to $\det(X) \cdot X$.
• Trace vs. Determinant properties: Remember that while $\operatorname{trace}(kX) = k \operatorname{trace}(X)$, the determinant property is $\det(kX) = k^n \det(X)$. Do not confuse these.

Summary

This problem effectively tests your understanding and application of fundamental matrix properties. The key to solving it was to first simplify the expression for matrix $B$ using the identity for nested adjoints, which transformed $B = \operatorname{adj}(\operatorname{adj}(2A))$ into $B = \det(2A) \cdot (2A)$. Subsequent steps involved calculating $\det(2A)$ using the scalar multiple property of determinants, which led to $B=8A$. Finally, $\det(B)$ and $\operatorname{trace}(B)$ were calculated using the respective scalar multiple properties, and their sum yielded the final answer.

The final answer is $\boxed{\text{280}}$, which corresponds to option (D).