If `A` is square matrix of order `2xx2` such that `A^(2)=I,B=[{:(1,sqrt(2)),(0,1):}]` and `C=ABA` then

Options

A. `C^(2009)=A[{:(1,2009sqrt(2)),(0,1):}]`
B. `C^(2009)=A[{:(1,2009+sqrt(2)),(0,1):}]`
C. `|C|^(2009)=1`
D. `C^(2009)=A[{:(1,2009sqrt(2)),(0,1):}]`

Explanation

To solve the problem, we need to analyze the given matrices and perform the necessary calculations step by step.

### Given:
1. \( A \) is a square matrix of order \( 2 \times 2 \) such that \( A^2 = I \) (where \( I \) is the identity matrix).
2. \( B = \begin{pmatrix} 1 & \sqrt{2} \\ 0 & 1 \end{pmatrix} \)
3. \( C = ABA \)

### Step 1: Understanding the properties of matrix \( A \)
Since \( A^2 = I \), this implies that \( A \) is an involutory matrix. This means that \( A \) is its own inverse, i.e., \( A^{-1} = A \).

### Step 2: Calculate \( C = ABA \)
We need to find \( C \) by performing the multiplication \( ABA \).

1. **First, calculate \( BA \)**:
- Let \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \).
- Then,
\[ BA = \begin{pmatrix} 1 & \sqrt{2} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 \cdot a + \sqrt{2} \cdot c & 1 \cdot b + \sqrt{2} \cdot d \\ 0 \cdot a + 1 \cdot c & 0 \cdot b + 1 \cdot d \end{pmatrix} = \begin{pmatrix} a + \sqrt{2}c & b + \sqrt{2}d \\ c & d \end{pmatrix} \]

2. **Now calculate \( C = A(BA) \)**:
- We have \( BA = \begin{pmatrix} a + \sqrt{2}c & b + \sqrt{2}d \\ c & d \end{pmatrix} \).
- Now, multiply \( A \) with \( BA \):
\[ C = A \begin{pmatrix} a + \sqrt{2}c & b + \sqrt{2}d \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a + \sqrt{2}c & b + \sqrt{2}d \\ c & d \end{pmatrix} \]
- Performing this multiplication gives:
\[ C = \begin{pmatrix} a(a + \sqrt{2}c) + b(c) & a(b + \sqrt{2}d) + b(d) \\ c(a + \sqrt{2}c) + d(c) & c(b + \sqrt{2}d) + d(d) \end{pmatrix} \]

### Step 3: Find the modulus of \( C \)
To find the modulus (determinant) of \( C \), we can use the property that if \( A^2 = I \), then \( \text{det}(A) = \pm 1 \).

1. **Calculate \( \text{det}(C) \)**:
- Since \( C = ABA \) and using the property of determinants:
\[ \text{det}(C) = \text{det}(A) \cdot \text{det}(B) \cdot \text{det}(A) = (\text{det}(A))^2 \cdot \text{det}(B) \]
- Given \( \text{det}(A) = \pm 1 \) and \( \text{det}(B) = 1 \) (since \( B \) is an upper triangular matrix with diagonal entries 1), we have:
\[ \text{det}(C) = 1 \cdot 1 = 1 \]

### Step 4: Calculate \( C^{2009} \)
Since the determinant of \( C \) is 1, we can conclude that:
\[ C^{2009} = C \]
This means that the modulus of \( C^{2009} \) is also 1.

### Conclusion
Based on the calculations, we can conclude:
- \( C^{2009} = A \cdot B^{2009} \cdot A \)
- The modulus of \( C^{2009} \) is 1.

### Final Answer
The correct options are:
- Option 3: \( |C|^{2009} = 1 \)
- Option 4: \( C^{2009} = A \cdot B^{2009} \)