3 and 4 .Determinants and Matrices
hard

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 ${A^T}A = 3{I_3}$

A

$6$

B

$2$

C

$3$

D

$4$

(JEE MAIN-2019)

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 

Standard 12
Mathematics

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