JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 19)
If the system of equations
kx + y + 2z = 1
3x $$-$$ y $$-$$ 2z = 2
$$-$$2x $$-$$2y $$-$$4z = 3
has infinitely many solutions, then k is equal to __________.
kx + y + 2z = 1
3x $$-$$ y $$-$$ 2z = 2
$$-$$2x $$-$$2y $$-$$4z = 3
has infinitely many solutions, then k is equal to __________.
Answer
21
Explanation
D = 0
$$ \Rightarrow \left| {\matrix{ k & 1 & 2 \cr 3 & { - 1} & { - 2} \cr { - 2} & { - 2} & { - 4} \cr } } \right| = 0$$
$$ \Rightarrow $$ k (4 $$-$$ 4) $$-$$ 1 ($$-$$ 12 $$-$$ 4) + 2 ($$-$$ 6 $$-$$ 2)
$$ \Rightarrow $$ 16 $$-$$ 16 = 0
Also, $${D_1} = {D_2} = {D_3} = 0$$
$$ \Rightarrow {D_2} = \left| {\matrix{ k & 1 & 2 \cr 3 & 2 & { - 2} \cr { - 2} & 3 & { - 4} \cr } } \right| = 0$$
$$ \Rightarrow $$ k($$-$$8 + 6) $$-$$ 1($$-$$ 12 $$-$$ 4) + 2(9 + 4) = 0
$$ \Rightarrow $$ $$-$$ 2k + 16 + 26 = 0
$$ \Rightarrow $$ 2k = 42
$$ \Rightarrow $$ k = 21
$$ \Rightarrow \left| {\matrix{ k & 1 & 2 \cr 3 & { - 1} & { - 2} \cr { - 2} & { - 2} & { - 4} \cr } } \right| = 0$$
$$ \Rightarrow $$ k (4 $$-$$ 4) $$-$$ 1 ($$-$$ 12 $$-$$ 4) + 2 ($$-$$ 6 $$-$$ 2)
$$ \Rightarrow $$ 16 $$-$$ 16 = 0
Also, $${D_1} = {D_2} = {D_3} = 0$$
$$ \Rightarrow {D_2} = \left| {\matrix{ k & 1 & 2 \cr 3 & 2 & { - 2} \cr { - 2} & 3 & { - 4} \cr } } \right| = 0$$
$$ \Rightarrow $$ k($$-$$8 + 6) $$-$$ 1($$-$$ 12 $$-$$ 4) + 2(9 + 4) = 0
$$ \Rightarrow $$ $$-$$ 2k + 16 + 26 = 0
$$ \Rightarrow $$ 2k = 42
$$ \Rightarrow $$ k = 21
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