Here's a detailed and educational solution to the given problem.

1. Key Concepts and Formulas

For a square matrix $X$ of order $n$:

• Determinant of a scalar multiple: $\operatorname{det}(kX) = k^n \operatorname{det}(X)$. This formula is crucial when a scalar $k$ is multiplied by a matrix $X$ before taking the determinant.
• Determinant of the adjoint: $\operatorname{det}(\operatorname{adj}(X)) = (\operatorname{det}(X))^{n-1}$. This formula relates the determinant of the adjoint of a matrix to the determinant of the matrix itself.
• Given order of the matrix: In this problem, the matrix $A$ is of order $n=3$. We will use this value throughout the calculations.

2. Step-by-Step Solution

Our goal is to simplify the complex determinant expression $\operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A)))$ and then use the given equality to find the values of $m$ and $n$.

Given:

• Order of matrix $A$, $n=3$.
• $\operatorname{det}(A) = -2$.
• $\operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A))) = 2^{m+n} \cdot 3^{m n}$, with $m>n$.

Step 1: Simplify the determinant of the innermost expression, $\operatorname{det}(3A)$. We start by evaluating the determinant of $3A$.

• What we are doing: Applying the determinant of a scalar multiple formula.
• Why: This is the innermost part of the expression, and its determinant will be needed for subsequent steps involving its adjoint.
• Math: $\operatorname{det}(3A) = 3^n \operatorname{det}(A)$ Since $n=3$ and $\operatorname{det}(A)=-2$: $\operatorname{det}(3A) = 3^3 \cdot (-2) = 27 \cdot (-2) = -54$

Step 2: Simplify $\operatorname{det}(\operatorname{adj}(3A))$. Next, we find the determinant of the adjoint of $3A$.

• What we are doing: Applying the determinant of the adjoint formula.
• Why: This step builds upon the previous result and moves outwards in the given expression.
• Math: $\operatorname{det}(\operatorname{adj}(X)) = (\operatorname{det}(X))^{n-1}$ Here, $X = 3A$ and $n=3$. So, $n-1 = 3-1 = 2$. Using the result from Step 1, $\operatorname{det}(3A) = -54$: $\operatorname{det}(\operatorname{adj}(3A)) = (\operatorname{det}(3A))^2 = (-54)^2$ To simplify $(-54)^2$, we factorize 54: $54 = 2 \cdot 27 = 2 \cdot 3^3$. $\operatorname{det}(\operatorname{adj}(3A)) = (-(2 \cdot 3^3))^2 = (2 \cdot 3^3)^2 = 2^2 \cdot (3^3)^2 = 2^2 \cdot 3^6$

Step 3: Simplify $\operatorname{det}(-6 \operatorname{adj}(3A))$. Now we consider the next layer, which involves multiplying the matrix $\operatorname{adj}(3A)$ by a scalar $-6$.

• What we are doing: Applying the determinant of a scalar multiple formula again.
• Why: This step calculates the determinant of the matrix whose adjoint will be taken in the next step.
• Math: Let $B = \operatorname{adj}(3A)$. We need to find $\operatorname{det}(-6B)$. $\operatorname{det}(-6B) = (-6)^n \operatorname{det}(B)$ Since $n=3$ and from Step 2, $\operatorname{det}(\operatorname{adj}(3A)) = 2^2 \cdot 3^6$: $\operatorname{det}(-6 \operatorname{adj}(3A)) = (-6)^3 \cdot (2^2 \cdot 3^6)$ Factorize $6 = 2 \cdot 3$: $\operatorname{det}(-6 \operatorname{adj}(3A)) = (-(2 \cdot 3))^3 \cdot (2^2 \cdot 3^6)$ $= (-1)^3 \cdot 2^3 \cdot 3^3 \cdot 2^2 \cdot 3^6$ $= -1 \cdot 2^{3+2} \cdot 3^{3+6}$ $= -2^5 \cdot 3^9$

Step 4: Simplify $\operatorname{det}(\operatorname{adj}(-6 \operatorname{adj}(3A)))$. This step involves taking the adjoint of the matrix from Step 3.

• What we are doing: Applying the determinant of the adjoint formula.
• Why: This is the second-to-last step in simplifying the full expression.
• Math: Let $C = -6 \operatorname{adj}(3A)$. We need to find $\operatorname{det}(\operatorname{adj}(C))$. $\operatorname{det}(\operatorname{adj}(C)) = (\operatorname{det}(C))^{n-1}$ Here, $n=3$, so $n-1=2$. Using the result from Step 3, $\operatorname{det}(-6 \operatorname{adj}(3A)) = -2^5 \cdot 3^9$: $\operatorname{det}(\operatorname{adj}(-6 \operatorname{adj}(3A))) = (\operatorname{det}(-6 \operatorname{adj}(3A)))^2$ $= (-2^5 \cdot 3^9)^2$ $= (2^5 \cdot 3^9)^2$ $= 2^{5 \cdot 2} \cdot 3^{9 \cdot 2}$ $= 2^{10} \cdot 3^{18}$

Step 5: Simplify the entire expression, $\operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3A)))$. Finally, we apply the scalar multiple determinant formula one last time.

• What we are doing: Applying the determinant of a scalar multiple formula.
• Why: This gives us the full simplified value of the left-hand side of the given equation.
• Math: Let $D = \operatorname{adj}(-6 \operatorname{adj}(3A))$. We need to find $\operatorname{det}(3D)$. $\operatorname{det}(3D) = 3^n \operatorname{det}(D)$ Since $n=3$ and from Step 4, $\operatorname{det}(\operatorname{adj}(-6 \operatorname{adj}(3A))) = 2^{10} \cdot 3^{18}$: $\operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3A))) = 3^3 \cdot (2^{10} \cdot 3^{18})$ $= 2^{10} \cdot 3^{3+18}$ $= 2^{10} \cdot 3^{21}$

Step 6: Equate the simplified expression with the given form and find $m$ and $n$. We are given that $\operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A))) = 2^{m+n} \cdot 3^{m n}$.

• What we are doing: Comparing the powers of 2 and 3 to form a system of equations.
• Why: This allows us to determine the values of $m$ and $n$.
• Math: From Step 5, we have $2^{10} \cdot 3^{21}$. Therefore, by comparing the exponents: $m+n = 10 \quad \text{(Equation 1)}$ $mn = 21 \quad \text{(Equation 2)}$ We can solve this system by considering a quadratic equation whose roots are $m$ and $n$: $x^2 - (m+n)x + mn = 0$. $x^2 - 10x + 21 = 0$ Factoring the quadratic equation: $(x-3)(x-7) = 0$ The roots are $x=3$ and $x=7$. We are given the condition $m>n$. Therefore: $m = 7$ $n = 3$

Step 7: Calculate $4m+2n$. Finally, we substitute the values of $m$ and $n$ into the expression $4m+2n$.

• What we are doing: Substituting the determined values of $m$ and $n$.
• Why: This is the final value requested by the problem.
• Math: $4m+2n = 4(7) + 2(3)$ $= 28 + 6$ $= 34$

3. Common Mistakes & Tips

• Incorrect Order ($n$): Ensure you consistently use the correct order of the matrix ($n=3$ in this case) for all determinant formulas. A common mistake is using $n-1$ or $n-2$ where $n$ is required, or vice-versa.
• Sign Errors: Be very careful with negative signs, especially when raising negative numbers to powers. For instance, $(-k)^2 = k^2$ but $(-k)^3 = -k^3$.
• Exponent Rules: Remember to correctly apply exponent rules like $(a^x)^y = a^{xy}$ and $a^x \cdot a^y = a^{x+y}$ when combining terms.
• Working from Inside Out: For nested expressions like this, always start with the innermost operation and work your way outwards. This methodical approach minimizes errors.

4. Summary

This problem involved systematically applying properties of determinants and adjoints of matrices. We began by simplifying the innermost determinant expression, $\operatorname{det}(3A)$, and then progressively evaluated the determinants of adjoints and scalar multiples, working outwards. Each step utilized the matrix order $n=3$ and the given $\operatorname{det}(A)=-2$. This led to the simplified expression $2^{10} \cdot 3^{21}$. By equating this to $2^{m+n} \cdot 3^{mn}$, we formed a system of equations $m+n=10$ and $mn=21$. Solving these, and considering the condition $m>n$, we found $m=7$ and $n=3$. Finally, substituting these values into $4m+2n$ yielded the result 34.

5. Final Answer

The final answer is $\boxed{34}$.