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]]`
The value of `abs(A^(50))` equals

Options

A. (a)`-1`
B. (b)0
C. (c)1
D. (d)25

Explanation

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

### Step 1: Define the Matrix A
The matrix \( A \) is given as:
\[ A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \]

### Step 2: Understand the Recurrence Relation
The recurrence relation provided is:
\[ A^n = A^{n-2} + A^2 - I \quad \text{for } n \geq 3 \]
where \( I \) is the identity matrix.

### Step 3: Calculate \( A^2 \)
First, we need to calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \]
Calculating this gives:
\[ A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \]
So, \( A^2 = A \).

### Step 4: Use the Recurrence Relation
Using the recurrence relation:
\[ A^n = A^{n-2} + A^2 - I \]
Since \( A^2 = A \), we can rewrite it as:
\[ A^n = A^{n-2} + A - I \]

### Step 5: Calculate Higher Powers
We can now compute \( A^3 \):
\[ A^3 = A^1 + A - I = A + A - I = 2A - I \]
Continuing this process, we can find:
\[ A^4 = A^2 + A^2 - I = 2A - I \]
The pattern suggests that:
\[ A^n = (n-1)A - (n-2)I \]

### Step 6: Calculate \( A^{50} \)
Using the derived formula:
\[ A^{50} = 49A - 48I \]

### Step 7: Substitute \( A \) and \( I \)
Substituting \( A \) and \( I \):
\[ A^{50} = 49 \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} - 48 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Calculating this gives:
\[ A^{50} = \begin{bmatrix} 49 & 0 & 0 \\ 49 & 0 & 49 \\ 0 & 49 & 0 \end{bmatrix} - \begin{bmatrix} 48 & 0 & 0 \\ 0 & 48 & 0 \\ 0 & 0 & 48 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 49 & -48 & 49 \\ 0 & 49 & -48 \end{bmatrix} \]

### Step 8: Calculate the Determinant
Now, we need to calculate the determinant of \( A^{50} \):
\[ \text{det}(A^{50}) = \text{det}\begin{bmatrix} 1 & 0 & 0 \\ 49 & -48 & 49 \\ 0 & 49 & -48 \end{bmatrix} \]
Using the determinant formula for a \( 3 \times 3 \) matrix:
\[ \text{det}(A^{50}) = 1 \cdot \text{det}\begin{bmatrix} -48 & 49 \\ 49 & -48 \end{bmatrix} = 1 \cdot (-48 \cdot -48 - 49 \cdot 49) = 2304 - 2401 = -97 \]

### Final Answer
Thus, the absolute value of the determinant is:
\[ \text{abs}(A^{50}) = 97 \]