Let $$a \in \mathbb{R}$$ and A be a matrix of order $$3 \times 3$$ such that $$\det(A) = -4$$ and $$A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$$, where I is the identity matrix of order $$3 \times 3$$. If $$\det((a+1)\operatorname{adj}((a-1)A))$$ is $$2^m 3^n$$, $$m, n \in \{0,1,2,\ldots,20\}$$, then $$m + n$$ is equal to:

We are given

$$A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}, \qquad \det(A) = -4.$$

Subtracting the identity matrix $$I$$, we find

$$A = \begin{bmatrix} 1-1 & a-0 & 1-0 \\ 2-0 & 1-1 & 0-0 \\ a-0 & 1-0 & 2-1 \end{bmatrix} = \begin{bmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{bmatrix}.$$

Step 1: Determine $$a$$ using $$\det(A) = -4$$.

Compute the determinant:

$$\det(A) = 0 \;-\; a\begin{vmatrix} 2 & 0 \\ a & 1 \end{vmatrix} \;+\; 1\begin{vmatrix} 2 & 0 \\ a & 1 \end{vmatrix}$$

$$\det(A) = -a(2\cdot1 - 0\cdot a) + (2\cdot1 - 0\cdot a) = -2a + 2 = 2(1-a).$$

Given $$\det(A) = -4$$, we have

$$2(1-a) = -4 \;\Longrightarrow\; 1-a = -2 \;\Longrightarrow\; a = 3.$$

Step 2: Set up the required determinant.

We need

$$D = \det\!\bigl((a+1)\,\operatorname{adj}((a-1)A)\bigr).$$

Property used: For an $$n \times n$$ matrix $$B$$,

$$\det\!\bigl(\operatorname{adj}(B)\bigr)=\det(B)^{\,n-1}.$$

Here $$n = 3,$$ so $$\det(\operatorname{adj}(B)) = \det(B)^{2}.$$

Also, for a scalar $$k,$$

$$\det(kB) = k^{\,n}\det(B).$$

Step 3: Apply the properties.

Let $$B = (a-1)A.$$ Then

$$D = (a+1)^{3}\,\det\!\bigl(\operatorname{adj}(B)\bigr) = (a+1)^{3}\,\det(B)^{2}.$$

Compute $$\det(B):$$

$$\det(B) = \det\bigl((a-1)A\bigr) = (a-1)^{3}\det(A).$$

Step 4: Insert $$a = 3$$ and $$\det(A) = -4.$

$$a+1 = 4,\quad a-1 = 2.$$

$$\det(B) = 2^{3}(-4) = 8(-4) = -32.$$

Hence

$$D = 4^{3}\,(-32)^{2} = 64 $$\times$$ 1024 = 65536.$$

Step 5: Express $$D$$ as $$2^{m}3^{n}.$$

$$65536 = 2^{16},$$ so $$m = 16,\; n = 0.$$

Step 6: Compute $$m+n.$$

$$m+n = 16 + 0 = 16.$$

Therefore, the required value is $$16$$, which corresponds to Option D.