JEE MAIN - Mathematics (2020 - 5th September Evening Slot - No. 13)
If the system of linear equations
x + y + 3z = 0
x + 3y + k2z = 0
3x + y + 3z = 0
has a non-zero solution (x, y, z) for some k $$ \in $$ R, then x + $$\left( {{y \over z}} \right)$$ is equal to :
x + y + 3z = 0
x + 3y + k2z = 0
3x + y + 3z = 0
has a non-zero solution (x, y, z) for some k $$ \in $$ R, then x + $$\left( {{y \over z}} \right)$$ is equal to :
9
3
-9
-3
Explanation
x + y + 3z = 0 .....(i)
x + 3y + k2z = 0 .........(ii)
3x + y + 3z = 0 ......(iii)
$$\left| {\matrix{ 1 & 1 & 3 \cr 1 & 3 & {{k^2}} \cr 3 & 1 & 3 \cr } } \right|$$ = 0
$$ \Rightarrow $$ 9 + 3 + 3k2 – 27 – k2 – 3 = 0
$$ \Rightarrow $$ k2 = 9
Perform (i) – (iii),
–2x = 0 $$ \Rightarrow $$ x = 0
Now from (i), y + 3z = 0
$$ \Rightarrow $$ $${y \over z} = - 3$$
$$ \therefore $$ x + $$\left( {{y \over z}} \right)$$ = -3
x + 3y + k2z = 0 .........(ii)
3x + y + 3z = 0 ......(iii)
$$\left| {\matrix{ 1 & 1 & 3 \cr 1 & 3 & {{k^2}} \cr 3 & 1 & 3 \cr } } \right|$$ = 0
$$ \Rightarrow $$ 9 + 3 + 3k2 – 27 – k2 – 3 = 0
$$ \Rightarrow $$ k2 = 9
Perform (i) – (iii),
–2x = 0 $$ \Rightarrow $$ x = 0
Now from (i), y + 3z = 0
$$ \Rightarrow $$ $${y \over z} = - 3$$
$$ \therefore $$ x + $$\left( {{y \over z}} \right)$$ = -3
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