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The total number of matrices $A = \left[ {\begin{array}{*{20}{c}}
0&{2x}&{2x}\\
{2y}&y&{ - y}\\
1&{ - 1}&1
\end{array}} \right];\,\left( {x,y \in R,\,x \ne y} \right)$ for which ${A^T}A = 3{I_3}$
$6$
$2$
$3$
$4$
Solution
${A^T}A = 3{I_3}$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
0&{2x}&{2x}\\
{2y}&y&{ – y}\\
1&{ – 1}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&{2y}&1\\
{2x}&y&{ – 1}\\
{2x}&{ – y}&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3&0&0\\
0&3&0\\
0&0&3
\end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
{8{x^2}}&0&0\\
0&{6{y^2}}&0\\
0&0&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3&0&0\\
0&3&0\\
0&0&3
\end{array}} \right]$
$ \Rightarrow 8{x^2} = 3 \Rightarrow x = \pm \sqrt {\frac{3}{8}} $
$ \Rightarrow 6{y^2} = 3 \Rightarrow y = \pm \sqrt {\frac{1}{2}} $
$4$ matrices are possible