Let $$A$$ be a non-singular matrix of order 3. If $$\det(3 \text{ adj}(2 \text{ adj}((\det A)A))) = 3^{-13} \cdot 2^{-10}$$ and $$\det(3 \text{ adj}(2A)) = 2^m \cdot 3^n$$, then $$|3m + 2n|$$ is equal to ________

Let $$A$$ be a non-singular $$3 \times 3$$ matrix and denote $$\det A$$ by $$d$$.

Case 1: Compute $$\det\Bigl(3\,\operatorname{adj}\!\bigl(2\,\operatorname{adj}((\det A)A)\bigr)\Bigr)$$.

Step 1: Inner-most matrix.
$$(\det A)A = dA$$

Step 2: Adjugate of a scalar multiple (order $$n=3$$).
Formula: $$\operatorname{adj}(kB)=k^{\,n-1}\operatorname{adj}(B)\,\,\,-(1)$$
Applying (1): $$\operatorname{adj}(dA)=d^{2}\operatorname{adj}(A)$$

Step 3: Multiply by $$2$$, then take adjugate again.
First form the matrix $$2\,\operatorname{adj}(dA)=2d^{2}\operatorname{adj}(A)$$.
Using (1) once more with scalar $$2d^{2}$$:
$$\operatorname{adj}\!\bigl(2d^{2}\operatorname{adj}(A)\bigr)=(2d^{2})^{2}\operatorname{adj}(\operatorname{adj}(A))=4d^{4}\operatorname{adj}(\operatorname{adj}(A))$$

Step 4: Evaluate $$\operatorname{adj}(\operatorname{adj}(A))$$.
For an invertible $$n \times n$$ matrix, $$\operatorname{adj}(\operatorname{adj}(A))=(\det A)^{\,n-2}A\,\,\,-(2)$$.
With $$n=3$$, (2) gives $$\operatorname{adj}(\operatorname{adj}(A))=dA$$.

Hence the matrix inside the determinant becomes
$$3 \times 4d^{4} \times dA = 12d^{5}A$$

Step 5: Determinant of a scalar multiple.
For $$n=3$$, $$\det(kB)=k^{3}\det(B)$$.
Therefore,
$$\det(12d^{5}A)=(12d^{5})^{3}\det(A)=12^{3}d^{15}\,d=12^{3}d^{16}$$

The question states that this determinant equals $$3^{-13}\,2^{-10}$$, so
$$12^{3}d^{16}=3^{-13}2^{-10}\,\,\,-(3)$$

Factorising $$12^{3}$$(since $$12=2^{2}\,3$$):
$$12^{3}=2^{6}3^{3}$$.
Substituting in (3): $$2^{6}3^{3}d^{16}=2^{-10}3^{-13}$$
Equating powers of $$2$$ and $$3$$ separately,
$$d^{16}=2^{-16}3^{-16}$$

Taking the $$\tfrac{1}{8}$$-th power on both sides,
$$d^{2}=2^{-2}3^{-2}\,\,\,-(4)$$

Case 2: Compute $$\det\bigl(3\,\operatorname{adj}(2A)\bigr)$$.

Step 1: Evaluate $$\operatorname{adj}(2A)$$ using (1).
$$\operatorname{adj}(2A)=2^{2}\operatorname{adj}(A)=4\operatorname{adj}(A)$$

Step 2: Multiply by $$3$$.
$$3\,\operatorname{adj}(2A)=12\,\operatorname{adj}(A)$$

Step 3: Determinant.
$$\det\!\bigl(12\,\operatorname{adj}(A)\bigr)=12^{3}\det(\operatorname{adj}(A))$$
For a $$3 \times 3$$ matrix, $$\det(\operatorname{adj}(A))=(\det A)^{2}=d^{2}$$.
Hence
$$\det\!\bigl(3\,\operatorname{adj}(2A)\bigr)=12^{3}d^{2}$$

Substitute $$12^{3}=2^{6}3^{3}$$ and $$d^{2}$$ from (4):
$$12^{3}d^{2}=2^{6}3^{3}\times 2^{-2}3^{-2}=2^{4}3^{1}$$

Thus $$m=4$$ and $$n=1$$ (since $$2^{m}3^{n}=2^{4}3^{1}$$).

Finally,
$$|3m+2n|=|3(4)+2(1)|=|12+2|=14$$

Therefore, the required value is $$14$$.