**23.** Let \( A \) be a square matrix of order 3 such that \( \det(A) = -2 \) and \[ \det\left( 3 \, \text{adj}(-6 \, \text{adj}(3A)) \right) = 2^{m+n} \cdot 3^{mn}, \quad m > n. \] Then \( 4m + 2n \) is equal to _____.

Key ideas you must know

1. Determinant scaling by a scalar
   • For an $n\times n$ matrix $M$, if you multiply the whole matrix by a scalar $k$, the determinant multiplies by $k^n$.
   • Formula: $\det(kM)=k^n\det(M)$.
2. Adjoint and determinant
   • For an $n\times n$ matrix $M$, the adjoint (also called adjugate), noted $\operatorname{adj}(M)$, satisfies
   $M\,\operatorname{adj}(M)=\det(M)\,I$
   • Very useful property:
   $\det\big(\operatorname{adj}(M)\big)=\big(\det(M)\big)^{\,n-1}.$
3. Chaining operations
   When you perform several steps (scalar multiplication, then adjoint, then scalar again, etc.), you can track the determinant step by step—you never need to write the whole matrix.

Logical chain to attack the problem

1. Start with $A$ whose determinant is given ($-2$).
2. Step-1: Create $3A$. Use the scalar rule to get its determinant.
3. Step-2: Take the adjoint of that. Use the adjoint rule to get its new determinant.
4. Step-3: Multiply by $-6$. Again apply the scalar rule.
5. Step-4: Take the adjoint once more. Apply the adjoint rule again.
6. Step-5: Multiply by $3$. One last scalar rule.
7. When you finish, you will have the determinant in the form $2^{\text{(something)}}3^{\text{(something)}}$. Compare powers with $2^{m+n}3^{mn}$ to find $m$ and $n$.
8. Finally compute $4m+2n$.

What the question says

Imagine you have a magic 3×3 number‐box (that’s the matrix A) and its secret code number (its determinant) is −2.
We do some cooking with this box:

1. Triple it (multiply by 3).
2. Find its twin box called the adjoint (it’s like a special rearrangement of the numbers).
3. Multiply that twin by −6.
4. Again find the twin of this new box.
5. Finally triple the result once more.

The question asks: “What is the secret code number (determinant) of the final box?”
That number must look exactly like $2^{m+n} \times 3^{mn}$, and you have to figure out the two little numbers m and n.
When you know them, plug them into $4m + 2n$ and write the answer.

So it’s a fancy game of multiplying and taking twins, and all we really track is how the determinant changes each time.

Step-by-step calculation

1. Original matrix $A$ : $\det(A)=-2$.

2. Multiply by 3
   $B = 3A$
   $\det(B)=3^3\,\det(A)=27\times(-2)=-54$

3. Take adjoint
   $C = \operatorname{adj}(B)=\operatorname{adj}(3A)$
   For order $n=3$:
   $\det(C)=\big(\det(B)\big)^{n-1}=(-54)^{2}=2916$

4. Multiply by $-6$
   $D = -6C$
   $\det(D)=(-6)^3\,\det(C)=-216\times2916$
   Factorising:
   $-216=-(2^3\,3^3), \quad 2916=2^2\,3^6$
   $\det(D) = -2^{3+2}\,3^{3+6}= -2^{5}\,3^{9}$

5. Take adjoint again
   $E = \operatorname{adj}(D)=\operatorname{adj}(-6\,\operatorname{adj}(3A))$
   $\det(E)=\big(\det(D)\big)^{n-1}=\big(-2^{5}\,3^{9}\big)^{2}=2^{10}\,3^{18}$

6. Multiply by 3
   $F = 3E$
   $\det(F)=3^3\,\det(E)=3^{3}\,2^{10}\,3^{18}=2^{10}\,3^{21}$

7. Match with $2^{m+n}\,3^{mn}$
   [ m+n = 10, \quad mn = 21 ]
   The positive integer pair meeting these is $(m,n)=(7,3)$ with $m>n$.

8. Compute $4m+2n$
   $4m+2n = 4\times7 + 2\times3 = 28+6 = 34$

Final answer: 34