Let `A = [[1,0,0],[1,0,1], [0,1,0]] " satisfies " A^(n) = A^(n-2) + A^(2 ) -I` for `nge 3` and
consider matrix `underset(3xx3)(U)` with its columns as `U_(1), U_(2), U_(3),` such that
`A^(50)U_(1)=[[1],[25],[25]],A^(50) U_(2)=[[0],[1],[0]]and A^(50) U_(3)[[0],[0],[1]]`
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 and the properties of the matrix \( A \).

### Step-by-step Solution:

1. **Understand the Recurrence Relation**:
We are given that:
\[ A^n = A^{n-2} + A^2 - I \quad \text{for } n \geq 3 \]
This means we can express \( A^{50} \) in terms of previous powers of \( A \).

2. **Express \( A^{50} \)**:
Using the recurrence relation, we can express \( A^{50} \) as:
\[ A^{50} = A^{48} + A^2 - I \]
Continuing this process, we can express \( A^{48} \) in terms of \( A^{46} \):
\[ A^{48} = A^{46} + A^2 - I \]
Repeating this, we eventually express \( A^{50} \) in terms of \( A^2 \) and the identity matrix \( I \).

3. **Calculate the Pattern**:
Continuing this pattern down to \( A^2 \):
\[ A^{50} = A^2 + A^2 + A^2 + ... - 24I \]
Since we have \( 25 \) terms of \( A^2 \) (from \( A^2 \) to \( A^{50} \)), we can simplify this to:
\[ A^{50} = 25A^2 - 24I \]

4. **Calculate \( A^2 \)**:
Now we need to find \( A^2 \):
\[ A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \]
To calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \]
Performing the multiplication:
\[ A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \]

5. **Substituting \( A^2 \) into \( A^{50} \)**:
Now substituting \( A^2 \) back into the equation for \( A^{50} \):
\[ A^{50} = 25 \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} - 24I \]
Where \( I \) is the identity matrix:
\[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Thus:
\[ A^{50} = \begin{bmatrix} 25 & 0 & 0 \\ 25 & 0 & 25 \\ 0 & 25 & 0 \end{bmatrix} - \begin{bmatrix} 24 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 25 & -24 & 25 \\ 0 & 25 & -24 \end{bmatrix} \]

6. **Finding the Trace of \( A^{50} \)**:
The trace of a matrix is the sum of its diagonal elements:
\[ \text{Trace}(A^{50}) = 1 + (-24) + (-24) = 1 - 24 - 24 = 1 - 48 = -47 \]

### Final Answer:
The trace of \( A^{50} \) is \( -47 \).