Let matrix `A=[(x,y,-z),(1,2,3),(1,1,2)]`, where `x, y, z in N`. If `|adj(adj (adj(adjA)))|=4^(8).5^(16)`, then the number of such matrices A is equal to (where, `|M|` represents determinant of a matrix M)

Answer

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 \( x, y, z \in \mathbb{N} \) (natural numbers), given that

\[ |adj(adj(adj(adj(A))))| = 4^8 \cdot 5^{16}. \]

### Step 1: Understanding the Determinant of the Adjoint

We know that for a \( 3 \times 3 \) matrix \( A \):

\[ |adj(A)| = |A|^{n-1} \]

where \( n \) is the order of the matrix. Here, \( n = 3 \), so:

\[ |adj(A)| = |A|^{2}. \]

### Step 2: Finding the Determinant of the Fourth Adjoint

Applying this repeatedly, we have:

\[ |adj(adj(A))| = |adj(A)|^{2} = |A|^{4}, \]
\[ |adj(adj(adj(A)))| = |adj(adj(A))|^{2} = |A|^{8}, \]
\[ |adj(adj(adj(adj(A))))| = |adj(adj(adj(A)))|^{2} = |A|^{16}. \]

### Step 3: Setting Up the Equation

From the problem, we have:

\[ |adj(adj(adj(adj(A))))| = |A|^{16} = 4^8 \cdot 5^{16}. \]

### Step 4: Simplifying the Right Side

We can express \( 4^8 \) as \( (2^2)^8 = 2^{16} \), so:

\[ |A|^{16} = 2^{16} \cdot 5^{16}. \]

Taking the 16th root on both sides gives:

\[ |A| = 2 \cdot 5 = 10. \]

### Step 5: Finding the Determinant of Matrix \( A \)

Next, we compute the determinant of \( A \):

\[ |A| = \begin{vmatrix} x & y & -z \\ 1 & 2 & 3 \\ 1 & 1 & 2 \end{vmatrix}. \]

Expanding the determinant along the first row:

\[ |A| = x \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} - y \begin{vmatrix} 1 & 3 \\ 1 & 2 \end{vmatrix} - z \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix}. \]

Calculating the minors:

1. \( \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = (2)(2) - (3)(1) = 4 - 3 = 1 \)
2. \( \begin{vmatrix} 1 & 3 \\ 1 & 2 \end{vmatrix} = (1)(2) - (3)(1) = 2 - 3 = -1 \)
3. \( \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = (1)(1) - (2)(1) = 1 - 2 = -1 \)

Thus, we have:

\[ |A| = x(1) - y(-1) - z(-1) = x + y + z. \]

### Step 6: Setting the Determinant Equal to 10

Now we set the determinant equal to 10:

\[ x + y + z = 10. \]

### Step 7: Counting the Natural Number Solutions

We need to find the number of solutions in natural numbers (positive integers) for the equation \( x + y + z = 10 \).

Using the stars and bars combinatorial method, the number of solutions in natural numbers is given by:

\[ \text{Number of solutions} = \binom{n-1}{k-1} = \binom{10-1}{3-1} = \binom{9}{2}. \]

Calculating \( \binom{9}{2} \):

\[ \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36. \]

### Conclusion

Thus, the number of such matrices \( A \) is \( \boxed{36} \).