To solve the problem, we need to find the number of matrices \( A \) of the form
\[
A = \begin{pmatrix}
x & y & -z \\
1 & 2 & 3 \\
1 & 1 & 2
\end{pmatrix}
\]
where \( x, y, z \) are natural numbers, and we have the condition that
\[
\text{det}(\text{adj}(\text{adj}(A))) = 2^8 \cdot 3^4.
\]
### Step 1: Understanding the determinant of the adjoint
The formula for the determinant of the adjoint of a matrix \( A \) is given by:
\[
\text{det}(\text{adj}(A)) = (\text{det}(A))^{n-1}
\]
where \( n \) is the order of the matrix. For a \( 3 \times 3 \) matrix, \( n = 3 \), so we have:
\[
\text{det}(\text{adj}(A)) = (\text{det}(A))^{2}.
\]
### Step 2: Finding the determinant of the adjoint of the adjoint
Now, applying the formula again for the adjoint of the adjoint:
\[
\text{det}(\text{adj}(\text{adj}(A))) = (\text{det}(\text{adj}(A)))^{2} = ((\text{det}(A))^{2})^{2} = (\text{det}(A))^{4}.
\]
### Step 3: Setting up the equation
From the problem, we know that:
\[
\text{det}(\text{adj}(\text{adj}(A))) = 2^8 \cdot 3^4.
\]
Thus, we can set up the equation:
\[
(\text{det}(A))^{4} = 2^8 \cdot 3^4.
\]
Taking the fourth root of both sides gives us:
\[
\text{det}(A) = 2^{8/4} \cdot 3^{4/4} = 2^2 \cdot 3^1 = 4 \cdot 3 = 12.
\]
### Step 4: Calculating the determinant of matrix A
Next, we need to calculate the determinant of matrix \( A \):
\[
\text{det}(A) = \begin{vmatrix}
x & y & -z \\
1 & 2 & 3 \\
1 & 1 & 2
\end{vmatrix}.
\]
Using the determinant formula for a \( 3 \times 3 \) matrix, we expand:
\[
\text{det}(A) = x \begin{vmatrix}
2 & 3 \\
1 & 2
\end{vmatrix} - y \begin{vmatrix}
1 & 3 \\
1 & 2
\end{vmatrix} - z \begin{vmatrix}
1 & 2 \\
1 & 1
\end{vmatrix}.
\]
Calculating the minors:
1. \( \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = (2 \cdot 2 - 3 \cdot 1) = 4 - 3 = 1 \)
2. \( \begin{vmatrix} 1 & 3 \\ 1 & 2 \end{vmatrix} = (1 \cdot 2 - 3 \cdot 1) = 2 - 3 = -1 \)
3. \( \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = (1 \cdot 1 - 2 \cdot 1) = 1 - 2 = -1 \)
So, substituting back, we have:
\[
\text{det}(A) = x(1) - y(-1) - z(-1) = x + y + z.
\]
### Step 5: Setting the determinant equal to 12
Now we set up the equation:
\[
x + y + z = 12.
\]
### Step 6: Finding the number of solutions
We need to find the number of natural number solutions to the equation \( x + y + z = 12 \). This can be transformed into a combinatorial problem by letting \( x' = x - 1 \), \( y' = y - 1 \), \( z' = z - 1 \) (since \( x, y, z \) are natural numbers, \( x', y', z' \) are non-negative integers):
\[
x' + 1 + y' + 1 + z' + 1 = 12 \implies x' + y' + z' = 9.
\]
The number of non-negative integer solutions to this equation is given by the stars and bars theorem:
\[
\text{Number of solutions} = \binom{n+k-1}{k-1} = \binom{9 + 3 - 1}{3 - 1} = \binom{11}{2}.
\]
Calculating this:
\[
\binom{11}{2} = \frac{11 \times 10}{2} = 55.
\]
### Final Answer
Thus, the number of such matrices \( A \) is \( \boxed{55} \).
