Let matrix `A=[[x,y,-z],[1,2,3],[1,1,2]]` where `x,y,zepsilonN`. If `det.(adj(adj.A))=2^8.(3^4)` then the number of such matrices A is :

Options

A. 220
B. 45
C. 55
D. 110

Answer

To solve the problem, we need to find the number of matrices \( A = \begin{bmatrix} x & y & -z \\ 1 & 2 & 3 \\ 1 & 1 & 2 \end{bmatrix} \) such that \( \det(\text{adj}(\text{adj}(A))) = 2^8 \cdot 3^4 \), where \( x, y, z \) are natural numbers.

### Step 1: Determine the order of the matrix
The matrix \( A \) is a \( 3 \times 3 \) matrix, so the order \( n = 3 \).

### Step 2: Use the property of determinants
We know that:
\[ \det(\text{adj}(A)) = \det(A)^{n-1} \]
For \( n = 3 \):
\[ \det(\text{adj}(A)) = \det(A)^2 \]
Now, applying this to the adjoint of the adjoint:
\[ \det(\text{adj}(\text{adj}(A))) = \det(\text{adj}(A))^{n-1} = \det(A)^{(n-1)(n-1)} = \det(A)^4 \]

### Step 3: Set up the equation
From the problem statement, we have:
\[ \det(A)^4 = 2^8 \cdot 3^4 \]
Taking the fourth root of both sides gives:
\[ \det(A) = \sqrt[4]{2^8 \cdot 3^4} = 2^{8/4} \cdot 3^{4/4} = 2^2 \cdot 3^1 = 4 \cdot 3 = 12 \]

### Step 4: Calculate the determinant of matrix \( A \)
The determinant of \( A \) can be calculated using the formula for the determinant of a \( 3 \times 3 \) matrix:
\[ \det(A) = x \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} - y \begin{vmatrix} 1 & 3 \\ 1 & 2 \end{vmatrix} - z \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} \]
Calculating the minors:
\[ \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = (2 \cdot 2 - 3 \cdot 1) = 4 - 3 = 1 \]
\[ \begin{vmatrix} 1 & 3 \\ 1 & 2 \end{vmatrix} = (1 \cdot 2 - 3 \cdot 1) = 2 - 3 = -1 \]
\[ \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = (1 \cdot 1 - 2 \cdot 1) = 1 - 2 = -1 \]
Thus, substituting back:
\[ \det(A) = x(1) - y(-1) - z(-1) = x + y + z \]
Setting this equal to 12:
\[ x + y + z = 12 \]

### Step 5: Count the number of solutions
We need to find the number of solutions in natural numbers \( x, y, z \) such that \( x + y + z = 12 \).

Using the stars and bars combinatorial method, the number of solutions in natural numbers can be calculated using the formula:
\[ \text{Number of solutions} = \binom{m - 1}{n - 1} \]
where \( m = 12 \) and \( n = 3 \):
\[ \text{Number of solutions} = \binom{12 - 1}{3 - 1} = \binom{11}{2} \]
Calculating \( \binom{11}{2} \):
\[ \binom{11}{2} = \frac{11 \cdot 10}{2 \cdot 1} = 55 \]

### Final Answer
Thus, the number of such matrices \( A \) is \( \boxed{55} \).