The total number of matrices `A = [{:(0, 2y, 1), (2x, y, -1), (2x, -y, 1):}] (x, y in R, x ne y)` for which `A^(T)A = 3I_(3)` is

Options

A. 2
B. 4
C. 3
D. 6

Answer

To solve the problem, we need to find the total number of matrices \( A \) such that:

\[ A = \begin{pmatrix} 0 & 2y & 1 \\ 2x & y & -1 \\ 2x & -y & 1 \end{pmatrix} \]

and it satisfies the condition \( A^T A = 3 I_3 \), where \( I_3 \) is the identity matrix of order 3.

### Step 1: Find the Transpose of Matrix A

The transpose of matrix \( A \) is given by:

\[ A^T = \begin{pmatrix} 0 & 2x & 2x \\ 2y & y & -y \\ 1 & -1 & 1 \end{pmatrix} \]

### Step 2: Compute \( A^T A \)

Now, we need to compute the product \( A^T A \):

\[ A^T A = \begin{pmatrix} 0 & 2x & 2x \\ 2y & y & -y \\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} 0 & 2y & 1 \\ 2x & y & -1 \\ 2x & -y & 1 \end{pmatrix} \]

Calculating each element of the resulting matrix:

1. **First Row, First Column:**
\[ 0 \cdot 0 + 2x \cdot 2x + 2x \cdot 2x = 4x^2 + 4x^2 = 8x^2 \]

2. **First Row, Second Column:**
\[ 0 \cdot 2y + 2x \cdot y + 2x \cdot (-y) = 2xy - 2xy = 0 \]

3. **First Row, Third Column:**
\[ 0 \cdot 1 + 2x \cdot (-1) + 2x \cdot 1 = -2x + 2x = 0 \]

4. **Second Row, First Column:**
\[ 2y \cdot 0 + y \cdot 2x + (-y) \cdot 2x = 2xy - 2xy = 0 \]

5. **Second Row, Second Column:**
\[ 2y \cdot 2y + y \cdot y + (-y) \cdot (-y) = 4y^2 + y^2 + y^2 = 6y^2 \]

6. **Second Row, Third Column:**
\[ 2y \cdot 1 + y \cdot (-1) + (-y) \cdot 1 = 2y - y - y = 0 \]

7. **Third Row, First Column:**
\[ 1 \cdot 0 + (-1) \cdot 2x + 1 \cdot 2x = -2x + 2x = 0 \]

8. **Third Row, Second Column:**
\[ 1 \cdot 2y + (-1) \cdot y + 1 \cdot (-y) = 2y - y - y = 0 \]

9. **Third Row, Third Column:**
\[ 1 \cdot 1 + (-1) \cdot (-1) + 1 \cdot 1 = 1 + 1 + 1 = 3 \]

So, we have:

\[ A^T A = \begin{pmatrix} 8x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \]

### Step 3: Set \( A^T A \) Equal to \( 3 I_3 \)

We know that:

\[ 3 I_3 = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} \]

Setting the matrices equal gives us the following equations:

1. \( 8x^2 = 3 \)
2. \( 6y^2 = 3 \)
3. The third equation is satisfied as it is \( 3 = 3 \).

### Step 4: Solve for \( x \) and \( y \)

From \( 8x^2 = 3 \):
\[ x^2 = \frac{3}{8} \implies x = \pm \sqrt{\frac{3}{8}} = \pm \frac{\sqrt{3}}{2\sqrt{2}} = \pm \frac{\sqrt{6}}{4} \]

From \( 6y^2 = 3 \):
\[ y^2 = \frac{3}{6} = \frac{1}{2} \implies y = \pm \frac{1}{\sqrt{2}} \]

### Step 5: Count the Valid Combinations

We have two possible values for \( x \) and two possible values for \( y \):
- \( x = \frac{\sqrt{6}}{4} \) or \( x = -\frac{\sqrt{6}}{4} \)
- \( y = \frac{1}{\sqrt{2}} \) or \( y = -\frac{1}{\sqrt{2}} \)

Since \( x \neq y \), we need to ensure that the combinations do not lead to \( x = y \).

The valid combinations are:
1. \( x = \frac{\sqrt{6}}{4}, y = \frac{1}{\sqrt{2}} \)
2. \( x = \frac{\sqrt{6}}{4}, y = -\frac{1}{\sqrt{2}} \)
3. \( x = -\frac{\sqrt{6}}{4}, y = \frac{1}{\sqrt{2}} \)
4. \( x = -\frac{\sqrt{6}}{4}, y = -\frac{1}{\sqrt{2}} \)

Thus, we have a total of **4 valid matrices**.

### Final Answer

The total number of matrices \( A \) is **4**.