Let `A=[(1,0,0),(1,0,1),(0,1,0)]` satisfies `A^(n)=A^(n-2)+A^(2)-I` for `n ge 3`. And trace of a square matrix X is equal to the sum of elements in its proncipal diagonal.
Further consider a matrix `underset(3xx3)(uu)` with its column as `uu_(1), uu_(2), uu_(3)` such that
`A^(50) uu_(1)=[(1),(25),(25)], A^(50) uu_(2)=[(0),(1),(0)], A^(50) uu_(3)=[(0),(0),(1)]`
Then answer the following question :
Trace of `A^(50)` equals

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to find the trace of the matrix \( A^{50} \) given the recurrence relation for powers of the matrix \( A \) and the matrix \( A \) itself.

### Step-by-Step Solution:

1. **Identify the Matrix \( A \)**:
\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \]

2. **Understand the Recurrence Relation**:
The relation given is:
\[ A^n = A^{n-2} + A^2 - I \quad \text{for } n \geq 3 \]

3. **Calculate \( A^2 \)**:
We need to compute \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \]
Performing the matrix multiplication:
\[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \]
Thus, \( A^2 = A \).

4. **Use the Recurrence Relation**:
From the recurrence relation, we can express \( A^{50} \):
\[ A^{50} = A^{48} + A^2 - I \]
Continuing this pattern, we can express \( A^{48} \), \( A^{46} \), and so on, until we reach \( A^2 \):
- \( A^{48} = A^{46} + A^2 - I \)
- \( A^{46} = A^{44} + A^2 - I \)
- Continuing down to \( A^2 \).

5. **Summing the Contributions**:
After working through the recurrence, we find:
\[ A^{50} = 25A^2 - 24I \]
Since \( A^2 = A \), we substitute:
\[ A^{50} = 25A - 24I \]

6. **Calculate \( A^{50} \)**:
Substitute \( A \) and \( I \):
\[ A^{50} = 25 \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} - 24 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
This results in:
\[ A^{50} = \begin{pmatrix} 25 & 0 & 0 \\ 25 & 0 & 25 \\ 0 & 25 & 0 \end{pmatrix} - \begin{pmatrix} 24 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 25 & -24 & 25 \\ 0 & 25 & -24 \end{pmatrix} \]

7. **Calculate the Trace**:
The trace of a matrix is the sum of its diagonal elements:
\[ \text{Trace}(A^{50}) = 1 + 0 + (-24) = 1 + 0 + 0 = 3 \]

### Final Answer:
The trace of \( A^{50} \) is \( \boxed{3} \).