1. Key Concepts and Formulas

To effectively solve this problem, a solid understanding of determinant properties for an $n \times n$ matrix $X$ is essential. For the given matrix $A$, the order is $n=3$.

• Determinant of a Scalar Multiple: If $k$ is a scalar, then $\det(kX) = k^n \det(X)$.
• Determinant of the Adjoint Matrix: $\det(\text{adj}(X)) = (\det(X))^{n-1}$.
• Determinant of the Adjoint of Adjoint Matrix: Applying the previous formula twice, we get: $\det(\text{adj}(\text{adj}(X))) = (\det(\text{adj}(X)))^{n-1} = ((\det(X))^{n-1})^{n-1} = (\det(X))^{(n-1)^2}$.
• Effect of Elementary Row/Column Operations on Determinant: Operations of the form $R_i \rightarrow R_i + kR_j$ (or $C_i \rightarrow C_i + kC_j$) do not change the determinant's value. This property is crucial for simplifying matrices before calculating their determinants.

For a $3 \times 3$ matrix ($n=3$), the exponent $(n-1)^2$ becomes $(3-1)^2 = 2^2 = 4$.

2. Step-by-Step Solution

Our goal is to determine the value of $\det(\text{adj}(\text{adj}(3A)))$ and express it in the form $2^m \cdot 3^n$ to find $m+n$. We will achieve this by first calculating $\det(A)$ and then systematically applying the determinant properties.

Step 1: Calculate $\det(A)$

We are given the matrix $A = \begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix}$. To simplify the calculation of $\det(A)$, we use elementary row operations to create zeros in the first column, which will allow us to expand the determinant easily.

$\det(A) = \left|\begin{array}{ccc} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{array}\right|$

Apply the following row operations, which do not change the determinant's value:

1. $R_2 \rightarrow R_2 - 2R_1$: This operation eliminates the element in position $(2,1)$.

   • New $R_2$: $[4-2(2), (6+2p)-2(2+p), (8+3p+2q)-2(2+p+q)]$
   • New $R_2$: $[0, 6+2p-4-2p, 8+3p+2q-4-2p-2q]$
   • New $R_2$: $[0, 2, 4+p]$

2. $R_3 \rightarrow R_3 - 3R_1$: This operation eliminates the element in position $(3,1)$.

   • New $R_3$: $[6-3(2), (12+3p)-3(2+p), (20+6p+3q)-3(2+p+q)]$
   • New $R_3$: $[0, 12+3p-6-3p, 20+6p+3q-6-3p-3q]$
   • New $R_3$: $[0, 6, 14+3p]$

The determinant of the simplified matrix is: $\det(A) = \left|\begin{array}{ccc} 2 & 2+p & 2+p+q \\ 0 & 2 & 4+p \\ 0 & 6 & 14+3p \end{array}\right|$

Now, expand the determinant along the first column (which contains two zeros): $\det(A) = 2 \cdot \left|\begin{array}{cc} 2 & 4+p \\ 6 & 14+3p \end{array}\right| - 0 \cdot (\text{minor}) + 0 \cdot (\text{minor})$ $\det(A) = 2 \cdot [2(14+3p) - 6(4+p)]$ $\det(A) = 2 \cdot [28+6p - 24-6p]$ $\det(A) = 2 \cdot [4]$ $\det(A) = 8$ We can express this as a power of 2: $\det(A) = 2^3$.

Step 2: Apply Determinant Properties to $\det(\text{adj}(\text{adj}(3A)))$

We need to evaluate $\det(\text{adj}(\text{adj}(3A)))$. Let $X = 3A$. Since $A$ is a $3 \times 3$ matrix, $n=3$.

Using the formula for the determinant of the adjoint of adjoint matrix: $\det(\text{adj}(\text{adj}(X))) = (\det(X))^{(n-1)^2}$ Substitute $X=3A$ and $n=3$: $\det(\text{adj}(\text{adj}(3A))) = (\det(3A))^{(3-1)^2} = (\det(3A))^{2^2} = (\det(3A))^4$

Next, we evaluate $\det(3A)$ using the scalar multiple property $\det(kX) = k^n \det(X)$: Substitute $k=3$, $X=A$, and $n=3$: $\det(3A) = 3^3 \det(A)$ We found $\det(A) = 8 = 2^3$. Substitute this value: $\det(3A) = 3^3 \cdot 2^3 = (3 \cdot 2)^3 = 6^3$

Finally, substitute this result back into the expression for $\det(\text{adj}(\text{adj}(3A)))$: $\det(\text{adj}(\text{adj}(3A))) = (6^3)^4$ $\det(\text{adj}(\text{adj}(3A))) = 6^{3 \cdot 4} = 6^{12}$

To match the target form $2^m \cdot 3^n$, we express $6^{12}$ as: $6^{12} = (2 \cdot 3)^{12} = 2^{12} \cdot 3^{12}$

Step 3: Determine $m$ and $n$, then calculate $m+n$

We have $\det(\text{adj}(\text{adj}(3A))) = 2^{12} \cdot 3^{12}$. Comparing this with the given form $2^m \cdot 3^n$, we identify the values of $m$ and $n$: $m = 12$ $n = 12$

Both $m$ and $n$ are natural numbers ($\mathbb{N}$), as specified in the question. Now, calculate their sum: $m+n = 12 + 12 = 24$

3. Common Mistakes & Tips

• Incorrect Order 'n': Always correctly identify the order of the matrix, $n$, as it is fundamental to all determinant properties involving scalar multiples and adjoints.
• Misapplication of Scalar Multiple Property: A common error is to write $\det(kA) = k \det(A)$ instead of the correct $\det(kA) = k^n \det(A)$.
• Exponent for Adjoints: Ensure the correct exponent is used for $\det(\text{adj}(X)) = (\det(X))^{n-1}$ and especially for nested adjoints like $\det(\text{adj}(\text{adj}(X))) = (\det(X))^{(n-1)^2}$.
• Inefficient Determinant Calculation: For larger matrices, avoid direct expansion initially. Prioritize elementary row/column operations to create zeros, making the expansion much simpler and less prone to errors.

4. Summary

This problem tests the comprehensive understanding and application of determinant properties, particularly those involving scalar multiplication and adjoint matrices. The solution strategy involved first calculating the determinant of the base matrix $A$ efficiently using row operations. Then, the properties $\det(kX) = k^n \det(X)$ and $\det(\text{adj}(\text{adj}(X))) = (\det(X))^{(n-1)^2}$ were systematically applied to simplify the complex expression $\det(\text{adj}(\text{adj}(3A)))$. Finally, the result was expressed in the desired $2^m \cdot 3^n$ format to find $m+n$.

The final answer is $\boxed{\text{24}}$, which corresponds to option (A).