The total number of matrices A = ( cases 0 & 2y & 1 2x & y & - 1 2x...

Given ${A}^{T}A = 3{I}_{3}$

\left($\begin{cases}0 & 2x & 2x \\ 2y & y & - y \\ 1 & - 1 & 1\end{cases}$\right)\left($\begin{cases}0 & 2y & 1 \\ 2x & y & - 1 \\ 2x & - y & 1\end{cases}$\right) = \left($\begin{cases}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{cases}$\right)</p><p>\left(\begin{cases}8{x}^{2} & 0 & 0 \\ 0 & 6{y}^{2} & 0 \\ 0 & 0 & 3\end{cases}\right) = \left(\begin{cases}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{cases}\right)

On comparing the two matrices, we get

$8{x}^{2} = 3$ and $6{y}^{2} = 3$

$⇒ {x}^{2} = \dfrac{3}{8}$ and ${y}^{2} = \dfrac{1}{2}$

$⇒ x = \pm \sqrt{\dfrac{3}{8}}$ and $y = \pm \sqrt{\dfrac{1}{2}}$

For each of $x \text{ and } y$ we have two possible values, hence there are total $2 \times 2 = 4$ combinations of values are possible.

Hence, total $4$ matrices are possible.