Let `S` be the set of `2xx2` matrices given by `S={A=[[a,b],[c,d]],"where" a,b,c,d,in I}`, such that `A^(T)=A^(-1)` Then

Options

A. number of matrices in set s is equal to 6
B. number of matrices in set S such that `|A-I_(2)| ne 0` is equal to 3
C. symmetric matrices are more than skew -symmetric matrices in set S
D. all matrices in set S are singular

Explanation

To solve the problem, we need to find the set of \(2 \times 2\) matrices \(A\) such that \(A^T = A^{-1}\).

### Step-by-Step Solution:

1. **Understanding the Condition**:
We start with the condition \(A^T = A^{-1}\). This implies that the matrix \(A\) is orthogonal, meaning that the transpose of \(A\) is equal to its inverse.

2. **Matrix Representation**:
Let \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), where \(a, b, c, d \in \mathbb{I}\) (the set of integers).

3. **Transpose of A**:
The transpose of \(A\) is given by:
\[ A^T = \begin{bmatrix} a & c \\ b & d \end{bmatrix} \]

4. **Inverse of A**:
The inverse of \(A\) is calculated as follows:
\[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]
For \(A\) to be invertible, we need \(ad - bc \neq 0\).

5. **Setting the Condition**:
From the condition \(A^T = A^{-1}\), we equate the two matrices:
\[ \begin{bmatrix} a & c \\ b & d \end{bmatrix} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

6. **Equating Elements**:
This gives us the following equations:
- \(a = \frac{d}{ad - bc}\)
- \(c = \frac{-b}{ad - bc}\)
- \(b = \frac{-c}{ad - bc}\)
- \(d = \frac{a}{ad - bc}\)

7. **Analyzing the Equations**:
From the equations, we can derive that:
- \(ad - bc = 1\) (since \(A\) is orthogonal, its determinant must be \(1\) or \(-1\)).
- This leads us to \(ad - bc = 1\) or \(-1\).

8. **Possible Values**:
Since \(a, b, c, d \in \mathbb{I}\), we can analyze the possible combinations of \(a, b, c, d\) that satisfy the determinant condition.

9. **Finding Matrices**:
The possible matrices that satisfy the conditions are:
- \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
- \(\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\)
- \(\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}\)

10. **Conclusion**:
Thus, there are a total of 8 matrices in the set \(S\) that satisfy the condition \(A^T = A^{-1}\).