The number of `2xx2` matrices `A`, that are there with the elements as real numbers satisfying `A+A^(T)=I` and `A A^(T)=I` is

Options

A. zero
B. one
C. two
D. infinite

Explanation

To solve the problem, we need to find the number of \(2 \times 2\) matrices \(A\) with real number elements that satisfy the equations \(A + A^T = I\) and \(AA^T = I\).

Let's denote the matrix \(A\) as:

\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

### Step 1: Use the first equation \(A + A^T = I\)

The transpose of \(A\) is:

\[ A^T = \begin{pmatrix} a & c \\ b & d \end{pmatrix} \]

Now we can add \(A\) and \(A^T\):

\[ A + A^T = \begin{pmatrix} a & b \\ c & d \end{pmatrix} + \begin{pmatrix} a & c \\ b & d \end{pmatrix} = \begin{pmatrix} 2a & b + c \\ b + c & 2d \end{pmatrix} \]

Setting this equal to the identity matrix \(I\):

\[ \begin{pmatrix} 2a & b + c \\ b + c & 2d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

From this, we can derive the following equations:

1. \(2a = 1\) → \(a = \frac{1}{2}\)
2. \(b + c = 0\) → \(c = -b\)
3. \(2d = 1\) → \(d = \frac{1}{2}\)

### Step 2: Substitute values into the matrix

Now substituting \(a\) and \(d\) into the matrix \(A\):

\[ A = \begin{pmatrix} \frac{1}{2} & b \\ -b & \frac{1}{2} \end{pmatrix} \]

### Step 3: Use the second equation \(AA^T = I\)

Now we need to check the second condition \(AA^T = I\):

Calculating \(AA^T\):

\[ A A^T = \begin{pmatrix} \frac{1}{2} & b \\ -b & \frac{1}{2} \end{pmatrix} \begin{pmatrix} \frac{1}{2} & -b \\ b & \frac{1}{2} \end{pmatrix} \]

Calculating the product:

\[ = \begin{pmatrix} \frac{1}{4} + b^2 & -\frac{1}{2}b + \frac{1}{2}b \\ -\frac{1}{2}b + \frac{1}{2}b & b^2 + \frac{1}{4} \end{pmatrix} = \begin{pmatrix} \frac{1}{4} + b^2 & 0 \\ 0 & b^2 + \frac{1}{4} \end{pmatrix} \]

Setting this equal to the identity matrix \(I\):

\[ \begin{pmatrix} \frac{1}{4} + b^2 & 0 \\ 0 & b^2 + \frac{1}{4} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

From this, we derive:

1. \(\frac{1}{4} + b^2 = 1\) → \(b^2 = 1 - \frac{1}{4} = \frac{3}{4}\)
2. \(b^2 + \frac{1}{4} = 1\) → \(b^2 = 1 - \frac{1}{4} = \frac{3}{4}\) (this is consistent)

### Step 4: Solve for \(b\)

From \(b^2 = \frac{3}{4}\):

\[ b = \pm \frac{\sqrt{3}}{2} \]

### Step 5: Determine the matrices

Now substituting back for \(b\):

1. If \(b = \frac{\sqrt{3}}{2}\):

\[ A_1 = \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \]

2. If \(b = -\frac{\sqrt{3}}{2}\):

\[ A_2 = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \]

### Conclusion

Thus, the number of \(2 \times 2\) matrices \(A\) that satisfy both conditions is **2**.