Question:

Step 1: Calculate the determinant of $A$. Given the matrix $A = \begin{bmatrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2p & 8 + 3p + 2q \\ 6 & 12 + 3p & 20 + 6p + 3q \end{bmatrix}$. Apply row operations to simplify the determinant: $R_2 \to R_2 - 2R_1$ $R_3 \to R_3 - 3R_1$ $$|A| = \begin{vmatrix} 2 & 2 + p & 2 + p + q \\ 4 - 2(2) & (6 + 2p) - 2(2 + p) & (8 + 3p + 2q) - 2(2 + p + q) \\ 6 - 3(2) & (12 + 3p) - 3(2 + p) & (20 + 6p + 3q) - 3(2 + p + q) \end{vmatrix}$$ $$|A| = \begin{vmatrix} 2 & 2 + p & 2 + p + q \\ 0 & 6 + 2p - 4 - 2p & 8 + 3p + 2q - 4 - 2p - 2q \\ 0 & 12 + 3p - 6 - 3p & 20 + 6p + 3q - 6 - 3p - 3q \end{vmatrix}$$ $$|A| = \begin{vmatrix} 2 & 2 + p & 2 + p + q \\ 0 & 2 & 4 + p \\ 0 & 6 & 14 + 3p \end{vmatrix}$$ Expand the determinant along the first column: $$|A| = 2 \begin{vmatrix} 2 & 4 + p \\ 6 & 14 + 3p \end{vmatrix}$$ $$|A| = 2 [2(14 + 3p) - 6(4 + p)]$$ $$|A| = 2 [28 + 6p - 24 - 6p]$$ $$|A| = 2 [4]$$ $$|A| = 8 = 2^3$$ Step 2: Calculate $\det(\text{adj}(\text{adj}(3A)))$. For an $n \times n$ matrix $X$, the following properties hold: 1. $\det(\text{adj}(X)) = (\det(X))^{n-1}$ 2. $\det(kX) = k^n \det(X)$ In this problem, $A$ is a $3 \times 3$ matrix, so $n=3$. First, find $\det(\text{adj}(3A))$: Using property 1, with $X = 3A$: $$\det(\text{adj}(3A)) = (\det(3A))^{3-1} = (\det(3A))^2$$ Next, find $\det(3A)$ using property 2: $$\det(3A) = 3^3 \det(A)$$ Substitute $\det(A) = 2^3$: $$\det(3A) = 3^3 \cdot 2^3$$ Now substitute this back into the expression for $\det(\text{adj}(3A))$: $$\det(\text{adj}(3A)) = (3^3 \cdot 2^3)^2 = 3^{3 \cdot 2} \cdot 2^{3 \cdot 2} = 3^6 \cdot 2^6$$ Finally, calculate $\det(\text{adj}(\text{adj}(3A)))$. Let $Y = \text{adj}(3A)$. Using property 1 again: $$\det(\text{adj}(Y)) = (\det(Y))^{n-1}$$ $$\det(\text{adj}(\text{adj}(3A))) = (\det(\text{adj}(3A)))^{3-1} = (\det(\text{adj}(3A)))^2$$ Substitute the value of $\det(\text{adj}(3A))$: $$\det(\text{adj}(\text{adj}(3A))) = (3^6 \cdot 2^6)^2$$ $$\det(\text{adj}(\text{adj}(3A))) = 3^{6 \cdot 2} \cdot 2^{6 \cdot 2}$$ $$\det(\text{adj}(\text{adj}(3A))) = 3^{12} \cdot 2^{12}$$ Step 3: Determine $m+n$. We are given that $\det(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n$. Comparing this with our result $2^{12} \cdot 3^{12}$: $m = 12$ $n = 12$ Therefore, $m+n = 12 + 12 = 24$.