Key Concepts and Formulas Used:

This problem requires a solid understanding of several fundamental properties of determinants and matrices. For a square matrix $M$ of order $n \times n$:

1. Determinant of a product: If $P$ and $Q$ are two $n \times n$ matrices, then $|PQ| = |P||Q|$.
2. Determinant of the identity matrix: The determinant of the identity matrix $I$ is always 1, i.e., $|I| = 1$.
3. Determinant of a scalar multiple: If $k$ is a scalar, then $|kM| = k^n|M|$.
4. Determinant of the adjoint matrix: The determinant of the adjoint of a matrix $M$ is given by $|adj(M)| = |M|^{n-1}$.

Step-by-step Elaborate Solution:

We are given that A and B are two $3 \times 3$ matrices (so, the order $n=3$). We are also given $AB = I$ and $|A| = \frac{1}{8}$. Our goal is to find the value of $|adj\,(B\,adj(2A))|$.

Step 1: Determine the value of $|B|$.

Given $AB = I$. Taking the determinant on both sides: $|AB| = |I|$ Using the property of the determinant of a product, $|AB| = |A||B|$: $|A||B| = |I|$ Since $I$ is the identity matrix, $|I|=1$. $|A||B| = 1$ We are given $|A| = \frac{1}{8}$. Substitute this value: $\left(\frac{1}{8}\right)|B| = 1$ Multiplying both sides by 8, we get: $|B| = 8$

Step 2: Apply the adjoint determinant formula to the outermost adjoint.

We need to evaluate $|adj\,(B\,adj(2A))|$. Let $X = B\,adj(2A)$. Then the expression becomes $|adj(X)|$. Using the formula $|adj(M)| = |M|^{n-1}$, and since $n=3$ for our matrices: $|adj(X)| = |X|^{3-1} = |X|^2$ Substituting $X = B\,adj(2A)$ back: $|adj\,(B\,adj(2A))| = |B\,adj(2A)|^2$

Step 3: Simplify the determinant inside the square.

Now we need to evaluate $|B\,adj(2A)|$. Using the determinant of a product property, $|PQ| = |P||Q|$: $|B\,adj(2A)| = |B| \cdot |adj(2A)|$ So, the expression from Step 2 becomes: $|adj\,(B\,adj(2A))| = (|B| \cdot |adj(2A)|)^2 = |B|^2 \cdot |adj(2A)|^2$

Step 4: Evaluate $|adj(2A)|$.

We now need to find $|adj(2A)|$. Again, using the formula $|adj(M)| = |M|^{n-1}$ for $M=2A$ and $n=3$: $|adj(2A)| = |2A|^{3-1} = |2A|^2$ Substitute this back into the expression from Step 3: $|B|^2 \cdot (|2A|^2)^2 = |B|^2 \cdot |2A|^4$

Step 5: Evaluate $|2A|$.

We need to find $|2A|$. Using the property of the determinant of a scalar multiple, $|kA| = k^n|A|$: For $k=2$ and $n=3$: $|2A| = 2^3|A| = 8|A|$ We are given $|A| = \frac{1}{8}$. Substitute this value: $|2A| = 8 \left(\frac{1}{8}\right) = 1$

Step 6: Substitute all values and calculate the final result.

Now we have all the components:

• $|B| = 8$
• $|2A| = 1$

Substitute these into the expression from Step 4, which was $|B|^2 \cdot |2A|^4$: $|adj\,(B\,adj(2A))| = (8)^2 \cdot (1)^4$ $= 64 \cdot 1$ $= 64$

Thus, the value of $|adj\,(B\,adj(2A))|$ is 64.

Tips and Common Mistakes to Avoid:

• Order of Operations: Always work from the innermost part of the expression outwards, or systematically break down the problem into smaller, manageable steps.
• Exponent Confusion: Be extremely careful with the exponents in the formulas. For an $n \times n$ matrix:
  • $|kA| = k^n|A|$ (the scalar $k$ is raised to the power of $n$)
  • $|adj(A)| = |A|^{n-1}$ (the determinant $|A|$ is raised to the power of $n-1$)
  • A common mistake is to confuse $k^n$ with $k^{n-1}$ or to incorrectly apply the power to $k$ inside the adjoint. For example, $|adj(kA)| \neq k^{n-1}|adj(A)|$. Instead, $|adj(kA)| = |kA|^{n-1} = (k^n|A|)^{n-1} = k^{n(n-1)}|A|^{n-1}$.
• Matrix Order: Always pay attention to the order $n$ of the matrix, as it directly affects the exponents in the formulas. Here, $n=3$.
• Basic Properties: Don't forget fundamental properties like $|AB|=|A||B|$ and $|I|=1$. These are often the starting points for such problems.

Summary and Key Takeaway:

This problem is a good test of your ability to systematically apply multiple determinant properties. The key takeaway is to break down complex determinant expressions involving adjoints and scalar multiples into simpler, known formulas. By carefully applying the rules for $|AB|$, $|I|$, $|kA|$, and $|adj(A)|$ in the correct sequence, even a seemingly complicated expression can be simplified to a numerical value. Always double-check the exponents ($n$ vs. $n-1$) based on the matrix order.

The final answer is $\boxed{\text{64}}$.