For a $3 \times 3$ matrix $M$, let $\text{trace}(M)$ denote the sum of all the diagonal elements of $M$. Let $A$ be a $3 \times 3$ matrix such that $|A|=\frac{1}{2}$ and $\text{trace}(A)=3$. If $B=\text{adj}(\text{adj}(2 A))$, then the value of $|B|+\text{trace}(B)$ equals:

Step 1: Calculate the determinant of $2A$

For an $n \times n$ matrix $A$ and a scalar $k$, the determinant of $kA$ is given by $k^n |A|$. In this problem, $A$ is a $3 \times 3$ matrix, so $n=3$. We are given that $|A| = \frac{1}{2}$.

Step 2: Calculate the determinant of $\text{adj}(2A)$

The determinant of the adjoint of a matrix $M$ is equal to the determinant of $M$ raised to the power of $(n-1)$, where $n$ is the order of the matrix. Here, $M = 2A$ and $n=3$. We found $|2A|=4$.

Step 3: Calculate the determinant of $B$

We need to find the determinant of $B = \text{adj}(\text{adj}(2A))$. Using the property $|\text{adj}(M)| = |M|^{n-1}$ twice, we get $|B| = |\text{adj}(2A)|^{n-1}$. Since $|\text{adj}(2A)| = |2A|^{n-1}$, we have $|B| = (|2A|^{n-1})^{n-1} = |2A|^{(n-1)^2}$. For $n=3$, this simplifies to $|2A|^{(3-1)^2} = |2A|^{2^2} = |2A|^4$. We know $|2A|=4$.

Step 4: Calculate the trace of $B$

For an $n \times n$ matrix $M$, the adjoint of the adjoint of $M$ is given by $|M|^{n-2} M$. Here, $M = 2A$ and $n=3$. So, $B = \text{adj}(\text{adj}(2A)) = |2A|^{3-2} (2A) = |2A| (2A)$. We know $|2A|=4$.

Step 5: Calculate $|B|+\text{trace}(B)$

We have calculated $|B|=256$ and $\text{trace}(B)=24$. Now we sum these two values.

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