Let `A=[(1,1,1),(1,-1,0),(0,1,-1)], A_(1)` be a matrix formed by the cofactors of the elements of the matrix A and `A_(2)` be a matrix formed by the cofactors of the elements of matrix `A_(1)`. Similarly, If `A_(10)` be a matrrix formed by the cofactors of the elements of matrix `A_(9)`, then the value of `|A_(10)|` is

Options

A. `3^(10)`
B. `3^(20)`
C. 9
D. `3^(1024)`

Answer

To solve the problem, we need to find the determinant of the matrix \( A_{10} \) formed by the cofactors of the elements of the matrix \( A_{9} \). We will use the property of determinants related to cofactors.

### Step-by-Step Solution:

1. **Calculate the Determinant of Matrix \( A \)**:
Given the matrix:
\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 0 & 1 & -1 \end{pmatrix} \]
We can calculate the determinant using the formula:
\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
Expanding along the first row:
\[ |A| = 1 \cdot \begin{vmatrix} -1 & 0 \\ 1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} \]
Calculating the minors:
\[ |A| = 1 \cdot (-1 \cdot -1 - 0 \cdot 1) - 1 \cdot (1 \cdot -1 - 0 \cdot 0) + 1 \cdot (1 \cdot 1 - 0 \cdot -1) \]
\[ = 1 \cdot 1 - 1 \cdot (-1) + 1 \cdot 1 = 1 + 1 + 1 = 3 \]

2. **Apply the Cofactor Property**:
The property states that if \( B \) is formed by the cofactors of \( A \), then:
\[ |B| = |A|^2 \]
Therefore, we can write:
\[ |A_1| = |A|^2 = 3^2 = 9 \]

3. **Continue Applying the Property**:
Similarly, for \( A_2 \):
\[ |A_2| = |A_1|^2 = 9^2 = 81 \]
For \( A_3 \):
\[ |A_3| = |A_2|^2 = 81^2 = 6561 \]

4. **Generalize the Pattern**:
Continuing this process, we can see a pattern emerging:
\[ |A_n| = |A|^{2^n} \]
Specifically, we have:
\[ |A_1| = 3^{2^1}, \quad |A_2| = 3^{2^2}, \quad |A_3| = 3^{2^3}, \quad \ldots \]
Thus, for \( A_{10} \):
\[ |A_{10}| = |A|^{2^{10}} = 3^{2^{10}} = 3^{1024} \]

5. **Final Answer**:
Therefore, the value of \( |A_{10}| \) is:
\[ |A_{10}| = 3^{1024} \]