Let `A =[[2,1,0],[1,2,-1],[0,-1,2]]`. If `|adj(adj(adj2A)|=(16)^(n)`, then `n` is equal to

Options

A. `12`
B. `10`
C. `8`
D. `9`

Answer

To solve the problem, we need to find the value of \( n \) such that \( | \text{adj}(\text{adj}(\text{adj}(2A))) | = 16^n \).

### Step-by-Step Solution:

1. **Define the Matrix \( A \)**:
\[ A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix} \]

2. **Calculate the Determinant of \( A \)**:
We can calculate the determinant \( |A| \) using the formula for the determinant of a 3x3 matrix:
\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \).

For our matrix:
\[ |A| = 2 \cdot (2 \cdot 2 - (-1) \cdot (-1)) - 1 \cdot (1 \cdot 2 - (-1) \cdot 0) + 0 \cdot (1 \cdot (-1) - 2 \cdot 0) \]
\[ = 2 \cdot (4 - 1) - 1 \cdot (2 - 0) + 0 \]
\[ = 2 \cdot 3 - 2 = 6 - 2 = 4 \]
Thus, \( |A| = 4 \).

3. **Calculate the Determinant of \( 2A \)**:
The determinant of a scalar multiple of a matrix is given by:
\[ |kA| = k^n |A| \]
where \( n \) is the order of the matrix (which is 3 in this case). Therefore:
\[ |2A| = 2^3 |A| = 8 \cdot 4 = 32 \]

4. **Calculate the Determinant of \( \text{adj}(2A) \)**:
The determinant of the adjugate of a matrix is given by:
\[ |\text{adj}(A)| = |A|^{n-1} \]
Thus:
\[ |\text{adj}(2A)| = |2A|^{2} = 32^2 = 1024 \]

5. **Calculate the Determinant of \( \text{adj}(\text{adj}(2A)) \)**:
Again applying the property of the adjugate:
\[ |\text{adj}(\text{adj}(2A))| = |\text{adj}(A)|^{n-1} = |\text{adj}(2A)|^{2} = 1024^2 = 1048576 \]

6. **Calculate the Determinant of \( \text{adj}(\text{adj}(\text{adj}(2A))) \)**:
Finally, we calculate:
\[ |\text{adj}(\text{adj}(\text{adj}(2A)))| = |\text{adj}(\text{adj}(2A))|^{2} = 1048576^2 = 1099511627776 \]

7. **Express in terms of \( 16^n \)**:
We know that \( 16 = 2^4 \), so:
\[ 16^n = (2^4)^n = 2^{4n} \]
We need to find \( n \) such that:
\[ 2^{40} = 2^{4n} \]
This gives us:
\[ 40 = 4n \implies n = 10 \]

### Final Answer:
Thus, the value of \( n \) is \( \boxed{10} \).