Question

The total number of matrices
$A=\left[\begin{array}{ccc}0& 2y& 1\\ 2x& y& -1\\ 2x& -y& 1\end{array}\right],(x,y\in R,x\ne y)$

for which $A^{T}A=3I_3$ is :

• A. $2$
• B. $4$
• C. $6$
• D. $3$

Answer 1

The correct option is B $4$

$A=\left[\begin{array}{ccc}0& 2y& 1\\ 2x& y& -1\\ 2x& -y& 1\end{array}\right]$

$\therefore A^T=\left[\begin{array}{ccc}0& 2x& 2x\\ 2y& y& -y\\ 1& -1& 1\end{array}\right]$

Now, $A^{T}A=3I_3$
$\Rightarrow \left[\begin{array}{ccc}0& 2x& 2x\\ 2y& y& -y\\ 1& -1& 1\end{array}\right]\left[\begin{array}{ccc}0& 2y& 1\\ 2x& y& -1\\ 2x& -y& 1\end{array}\right]=\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}8x^2& 0& 0\\ 0& 6y^2& 0\\ 0& 0& 3\end{array}\right]=\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]$

$\therefore 8x^2=3\Rightarrow x=\pm \frac{\sqrt{3}}{2\sqrt{2}}$
and $6y^2=3\Rightarrow y=\pm \frac{1}{\sqrt{2}}$

Therefore, $4$ such matrices are possible.