To solve for the value of $ m+n $, we first establish the matrix $ A $ and determine the value of $ a $. The condition given is:

$ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} $

Since $ I $ is the identity matrix of size $ 3 \times 3 $, we have:

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

We know $\operatorname{det}(A) = -4$. Calculating the determinant:

$ \det(A) = 0(0 \times 1 - 0 \times 1) - a(2 \times 1 - 0 \times a) + 1(2 \times 1 - 0 \times 1) = -2a + 2 $

Setting $-2a + 2 = -4$, we solve for $ a $:

$ -2a + 2 = -4 $

$ -2a = -6 $

$ a = 3 $

With $ a = 3 $, the problem is to find $\operatorname{det}((a+1) \operatorname{adj}((a-1) A))$. With $ a = 3 $:

Calculate $(a+1) \operatorname{adj}((a-1) A)$:

$ = 4 \operatorname{adj}(3A) $

Calculate the determinant:

$ \det(4 \operatorname{adj}(3A)) = 4^3 \times \det(\operatorname{adj}(3A)) $

Using the property $\operatorname{det}(\operatorname{adj}(A)) = \det(A)^{n-1}$ for a $ n \times n $ matrix:

$ \operatorname{adj}(3A) = (3A)^{2-1} = (3 \times \det(A))^2 $

$ = (3^3 \times (-4))^2 $

Therefore, calculate $ \det(3A)^2 $:

$ \det((3A)^2) = (3^3 \times (-4))^2 = 3^6 \times 4^2 $

Substitute back:

$ 4^3 \times \det((3A)^2) = 4^3 \times 3^6 \times 4^2 = 2^6 \times 3^6 \times 2^4 $

Simplifying gives:

$ = 2^{10} \times 3^6 $

Finally, we have powers $ m = 10 $ and $ n = 6 $. Therefore:

$ m+n = 10 + 6 = 16 $