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 :
The value of `|uu|` equals

Options

A. 0
B. 1
C. 2
D. `-1`

Explanation

To solve the problem step by step, let's go through the solution systematically.

### Step 1: Understand the given matrix A and the recurrence relation
We have the matrix:
\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \]
The recurrence relation given is:
\[ A^n = A^{n-2} + A^2 - I \quad \text{for } n \geq 3 \]
where \( I \) is the identity matrix.

### Step 2: Calculate \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \]

Calculating the multiplication:
- First row: \( (1*1 + 0*1 + 0*0, 1*0 + 0*0 + 0*1, 1*0 + 0*1 + 0*0) = (1, 0, 0) \)
- Second row: \( (1*1 + 0*1 + 1*0, 1*0 + 0*0 + 1*1, 1*0 + 0*1 + 1*0) = (1, 1, 1) \)
- Third row: \( (0*1 + 1*1 + 0*0, 0*0 + 1*0 + 0*1, 0*0 + 1*1 + 0*0) = (1, 0, 0) \)

Thus, we have:
\[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Use the recurrence relation to find \( A^{50} \)
Using the recurrence relation, we can find \( A^{50} \):
\[ A^{50} = A^{48} + A^2 - I \]
We need to find a pattern or a general formula for \( A^n \) using the recurrence relation.

Continuing this process, we can derive:
\[ A^{n} = kA^2 + (n-2)I \]
for some integer \( k \).

### Step 4: Find \( A^{50} \)
Using the derived formula:
\[ A^{50} = 25A^2 - 24I \]
Substituting \( A^2 \):
\[ A^{50} = 25 \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} - 24 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
Calculating this gives:
\[ A^{50} = \begin{pmatrix} 25 & 0 & 0 \\ 25 & 25 & 25 \\ 0 & 0 & 25 \end{pmatrix} - \begin{pmatrix} 24 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 25 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 5: Solve for \( U \)
We have:
\[ A^{50} U_1 = \begin{pmatrix} 1 \\ 25 \\ 25 \end{pmatrix}, \quad A^{50} U_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \quad A^{50} U_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \]
Let \( U_1 = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \).

Setting up the equations:
1. \( A^{50} U_1 = \begin{pmatrix} 1 \\ 25 \\ 25 \end{pmatrix} \) leads to \( U_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \)
2. \( U_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \)
3. \( U_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \)

### Step 6: Form the matrix \( U \)
Thus, the matrix \( U \) is:
\[ U = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 7: Calculate the determinant of \( U \)
The determinant of \( U \) is:
\[ |U| = 1 \]

### Final Answer
The value of \( |U| \) equals **1**.