JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 3)
Let S be the set of all real values of k for which the systemof linear equations
x + y + z = 2
2x + y $$-$$ z = 3
3x + 2y + kz = 4
has a unique solution. Then S is :
x + y + z = 2
2x + y $$-$$ z = 3
3x + 2y + kz = 4
has a unique solution. Then S is :
an empty set
equal to {0}
equal to R
equal to R $$-$$ {0}
Explanation
As system of linear equations have unique solutions so, determinant of coefficient $$ \ne $$ 0
$$ \therefore $$ $$\left| {\matrix{ 1 & 1 & 1 \cr 2 & 1 & { - 1} \cr 3 & 2 & k \cr } } \right|$$ $$ \ne $$ 0
$$ \Rightarrow $$ k + 2 - (2k + 3) + 1 $$ \ne $$ 0
$$ \Rightarrow $$ k $$ \ne $$ 0
$$ \therefore $$ k $$ \in $$ R - {0}
$$ \therefore $$ $$\left| {\matrix{ 1 & 1 & 1 \cr 2 & 1 & { - 1} \cr 3 & 2 & k \cr } } \right|$$ $$ \ne $$ 0
$$ \Rightarrow $$ k + 2 - (2k + 3) + 1 $$ \ne $$ 0
$$ \Rightarrow $$ k $$ \ne $$ 0
$$ \therefore $$ k $$ \in $$ R - {0}
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