The total number of matrices A=( … ),(x,y∈ R,x≠ y) for which A^TA=3I_3 is:

Given ${A}^{T}A=3{I}_{3}$ATA=3I3​

$$(\begin{matrix}0 & 2x & 2x \\ 2y & y & -y \\ 1 & -1 & 1\end{matrix})(\begin{matrix}0 & 2y & 1 \\ 2x & y & -1 \\ 2x & -y & 1\end{matrix})=(\begin{matrix}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{matrix})$$(02y1​2xy−1​2x−y1​)(02x2x​2yy−y​1−11​)=(300​030​003​)

$$(\begin{matrix}8{x}^{2} & 0 & 0 \\ 0 & 6{y}^{2} & 0 \\ 0 & 0 & 3\end{matrix})=(\begin{matrix}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{matrix})$$(8x200​06y20​003​)=(300​030​003​)

On comparing the two matrices, we get

$8{x}^{2}=3$8x2=3 and $6{y}^{2}=3$6y2=3

$\Rightarrow {x}^{2}=\frac{3}{8}$⇒x2=83​ and ${y}^{2}=\frac{1}{2}$y2=21​

$\Rightarrow x=\pm \sqrt{\frac{3}{8}}$⇒x=±83​​ and $y=\pm \sqrt{\frac{1}{2}}$y=±21​​

For each of x&y we have two possible values, hence there are total $2\times 2=4$2×2=4 combinations of values are possible.

Hence, total $4$4 matrices are possible.