1. Key Concepts and Formulae

This problem requires a solid understanding of fundamental matrix operations and properties.

• Transpose of a Matrix ($A^T$): The transpose of a matrix $A$ is obtained by interchanging its rows and columns. If $A = [a_{ij}]$, then $A^T = [a_{ji}]$.
• Matrix Multiplication: For two matrices $A$ (of size $m \times n$) and $B$ (of size $n \times p$), their product $C = AB$ is an $m \times p$ matrix. The element $C_{ij}$ (in the $i$-th row and $j$-th column) is the dot product of the $i$-th row of $A$ and the $j$-th column of $B$.
• Identity Matrix ($I_n$): An $n \times n$ square matrix with ones on its main diagonal and zeros elsewhere. For a $3 \times 3$ matrix, $I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$.
• Scalar Multiplication of a Matrix: To multiply a matrix by a scalar (a number), multiply every element of the matrix by that scalar. For example, $3I_3 = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}$.
• Equality of Matrices: Two matrices are equal if and only if they have the same dimensions and their corresponding elements are equal.

2. Problem Setup and Given Information

We are given a $3 \times 3$ matrix $A$: $A = \begin{pmatrix} 0 & {2y} & 1 \\ {2x} & y & { - 1} \\ {2x} & { - y} & 1 \end{pmatrix}$ We are also given the condition $A^T A = 3I_3$, where $x, y \in \mathbb{R}$ and $x \ne y$. Our goal is to find the total number of such matrices $A$. Each unique pair of $(x, y)$ satisfying the conditions will define a unique matrix $A$.

3. Step-by-Step Solution with Explanations

Step 1: Determine the Transpose of Matrix A ($A^T$)

To calculate $A^T A$, we first need to find the transpose of $A$. We do this by swapping the rows and columns of $A$. The first row of $A$ becomes the first column of $A^T$. The second row of $A$ becomes the second column of $A^T$. The third row of $A$ becomes the third column of $A^T$.

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

Step 2: Calculate the Matrix Product $A^T A$

Next, we multiply $A^T$ by $A$. This is done to utilize the given condition $A^T A = 3I_3$. $A^T A = \begin{pmatrix} 0 & {2x} & {2x} \\ {2y} & y & { - y} \\ 1 & { - 1} & 1 \end{pmatrix} \begin{pmatrix} 0 & {2y} & 1 \\ {2x} & y & { - 1} \\ {2x} & { - y} & 1 \end{pmatrix}$

Let's compute each element of the resulting $3 \times 3$ matrix by taking the dot product of the rows of $A^T$ with the columns of $A$:

• Element (1,1): $(0)(0) + (2x)(2x) + (2x)(2x) = 0 + 4x^2 + 4x^2 = 8x^2$
• Element (1,2): $(0)(2y) + (2x)(y) + (2x)(-y) = 0 + 2xy - 2xy = 0$
• Element (1,3): $(0)(1) + (2x)(-1) + (2x)(1) = 0 - 2x + 2x = 0$
• Element (2,1): $(2y)(0) + (y)(2x) + (-y)(2x) = 0 + 2xy - 2xy = 0$
• Element (2,2): $(2y)(2y) + (y)(y) + (-y)(-y) = 4y^2 + y^2 + y^2 = 6y^2$
• Element (2,3): $(2y)(1) + (y)(-1) + (-y)(1) = 2y - y - y = 0$
• Element (3,1): $(1)(0) + (-1)(2x) + (1)(2x) = 0 - 2x + 2x = 0$
• Element (3,2): $(1)(2y) + (-1)(y) + (1)(-y) = 2y - y - y = 0$
• Element (3,3): $(1)(1) + (-1)(-1) + (1)(1) = 1 + 1 + 1 = 3$

So, the product matrix $A^T A$ is: $A^T A = \begin{pmatrix} {8x^2} & 0 & 0 \\ 0 & {6y^2} & 0 \\ 0 & 0 & 3 \end{pmatrix}$

• Tip: Notice that the resulting matrix is a diagonal matrix. This often happens when dealing with $A^T A$ or $A A^T$ for matrices with certain structures. The off-diagonal zeros confirm that rows of $A^T$ (which are columns of $A$) are orthogonal.

Step 3: Equate $A^T A$ with $3I_3$

We are given $A^T A = 3I_3$. We write out $3I_3$: $3I_3 = 3 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ Now, we equate the two matrices to find the values of $x$ and $y$: $\begin{pmatrix} {8x^2} & 0 & 0 \\ 0 & {6y^2} & 0 \\ 0 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ For these matrices to be equal, their corresponding elements must be equal. This gives us the following system of equations:

1. $8x^2 = 3$
2. $6y^2 = 3$
3. $3 = 3$ (This equation is consistent and does not provide new information for $x$ or $y$). All off-diagonal elements are $0=0$, which is also consistent.

Step 4: Solve for the values of $x$ and $y$

We solve the equations from Step 3 for $x$ and $y$:

1. From $8x^2 = 3$: $x^2 = \frac{3}{8}$ Taking the square root of both sides gives two possible values for $x$: $x = \pm \sqrt{\frac{3}{8}} = \pm \frac{\sqrt{3}}{\sqrt{8}} = \pm \frac{\sqrt{3}}{2\sqrt{2}}$ Rationalizing the denominator: $x = \pm \frac{\sqrt{3}\sqrt{2}}{2\sqrt{2}\sqrt{2}} = \pm \frac{\sqrt{6}}{4}$ So, $x$ can be $\frac{\sqrt{6}}{4}$ or $-\frac{\sqrt{6}}{4}$.

2. From $6y^2 = 3$: $y^2 = \frac{3}{6} = \frac{1}{2}$ Taking the square root of both sides gives two possible values for $y$: $y = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}$ Rationalizing the denominator: $y = \pm \frac{\sqrt{2}}{2}$ So, $y$ can be $\frac{\sqrt{2}}{2}$ or $-\frac{\sqrt{2}}{2}$.

Step 5: Apply the condition $x \ne y$ and count the valid matrices

The problem states that $x, y \in \mathbb{R}$ and $x \ne y$. We need to form all possible pairs $(x, y)$ from the values we found and then check if any pair violates the $x \ne y$ condition.

The possible values for $x$ are $x_1 = \frac{\sqrt{6}}{4}$ and $x_2 = -\frac{\sqrt{6}}{4}$. The possible values for $y$ are $y_1 = \frac{\sqrt{2}}{2}$ and $y_2 = -\frac{\sqrt{2}}{2}$.

Let's list all combinations and check $x \ne y$:

• Combination 1: $(x_1, y_1) = \left(\frac{\sqrt{6}}{4}, \frac{\sqrt{2}}{2}\right)$ To check if $x_1 = y_1$: $\frac{\sqrt{6}}{4}$ vs $\frac{\sqrt{2}}{2} = \frac{2\sqrt{2}}{4}$. Since $\sqrt{6} \ne 2\sqrt{2}$ (because $6 \ne 8$), $x_1 \ne y_1$. This pair is valid.
• Combination 2: $(x_1, y_2) = \left(\frac{\sqrt{6}}{4}, -\frac{\sqrt{2}}{2}\right)$ Here, $x_1$ is positive and $y_2$ is negative, so $x_1 \ne y_2$. This pair is valid.
• Combination 3: $(x_2, y_1) = \left(-\frac{\sqrt{6}}{4}, \frac{\sqrt{2}}{2}\right)$ Here, $x_2$ is negative and $y_1$ is positive, so $x_2 \ne y_1$. This pair is valid.
• Combination 4: $(x_2, y_2) = \left(-\frac{\sqrt{6}}{4}, -\frac{\sqrt{2}}{2}\right)$ To check if $x_2 = y_2$: $-\frac{\sqrt{6}}{4}$ vs $-\frac{\sqrt{2}}{2}$. Since $\frac{\sqrt{6}}{4} \ne \frac{\sqrt{2}}{2}$ (as shown in Combination 1), their negative counterparts are also not equal. So $x_2 \ne y_2$. This pair is valid.

All four possible pairs of $(x, y)$ satisfy the condition $x \ne y$. Each unique pair $(x, y)$ defines a unique matrix $A$. Therefore, there are 4 such matrices $A$.

• Common Mistake: Forgetting to check additional conditions given in the problem, such as $x \ne y$. While in this specific case, the condition doesn't eliminate any solutions, it's crucial to always verify it.

4. Summary and Key Takeaways

We systematically found the transpose of matrix $A$, performed matrix multiplication to compute $A^T A$, and then equated it to $3I_3$. This led to a system of equations for $x^2$ and $y^2$. Solving these equations yielded two possible real values for $x$ and two possible real values for $y$. By combining these values, we found four distinct pairs $(x, y)$. Crucially, we verified that for every such pair, the condition $x \ne y$ was satisfied. Each distinct pair $(x,y)$ corresponds to a unique matrix $A$. Therefore, there are 4 such matrices.

The final answer is $\boxed{4}$.