JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 16)
The total number of matrices
$$A = \left( {\matrix{ 0 & {2y} & 1 \cr {2x} & y & { - 1} \cr {2x} & { - y} & 1 \cr } } \right)$$
(x, y $$ \in $$ R,x $$ \ne $$ y) for which ATA = 3I3 is :-
$$A = \left( {\matrix{ 0 & {2y} & 1 \cr {2x} & y & { - 1} \cr {2x} & { - y} & 1 \cr } } \right)$$
(x, y $$ \in $$ R,x $$ \ne $$ y) for which ATA = 3I3 is :-
3
4
2
6
Explanation
Given ATA = 3I3
$$ \Rightarrow $$ $$\left[ {\matrix{ 0 & {2x} & {2x} \cr {2y} & y & { - y} \cr 1 & { - 1} & 1 \cr } } \right]\left[ {\matrix{ 0 & {2y} & 1 \cr {2x} & y & { - 1} \cr {2x} & { - y} & 1 \cr } } \right]$$
= $$3\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$
$$ \Rightarrow $$ $$\left[ {\matrix{ {8{x^2}} & 0 & 0 \cr 0 & {6{y^2}} & 0 \cr 0 & 0 & 3 \cr } } \right]$$ = $$\left[ {\matrix{ 3 & 0 & 0 \cr 0 & 3 & 0 \cr 0 & 0 & 3 \cr } } \right]$$
$$ \therefore $$ 8x2 = 3, 6y2 = 3
$$ \Rightarrow $$ x = $$ \pm \sqrt {{3 \over 8}} $$, y = $$ \pm \sqrt {{1 \over 2}} $$
Total possible combination of x and y = 2 $$ \times $$ 2 = 4
$$ \Rightarrow $$ $$\left[ {\matrix{ 0 & {2x} & {2x} \cr {2y} & y & { - y} \cr 1 & { - 1} & 1 \cr } } \right]\left[ {\matrix{ 0 & {2y} & 1 \cr {2x} & y & { - 1} \cr {2x} & { - y} & 1 \cr } } \right]$$
= $$3\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$
$$ \Rightarrow $$ $$\left[ {\matrix{ {8{x^2}} & 0 & 0 \cr 0 & {6{y^2}} & 0 \cr 0 & 0 & 3 \cr } } \right]$$ = $$\left[ {\matrix{ 3 & 0 & 0 \cr 0 & 3 & 0 \cr 0 & 0 & 3 \cr } } \right]$$
$$ \therefore $$ 8x2 = 3, 6y2 = 3
$$ \Rightarrow $$ x = $$ \pm \sqrt {{3 \over 8}} $$, y = $$ \pm \sqrt {{1 \over 2}} $$
Total possible combination of x and y = 2 $$ \times $$ 2 = 4
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