JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 21)
Let $$\lambda \in $$ R . The system of linear equations
2x1 - 4x2 + $$\lambda $$x3 = 1
x1 - 6x2 + x3 = 2
$$\lambda $$x1 - 10x2 + 4x3 = 3
is inconsistent for:
2x1 - 4x2 + $$\lambda $$x3 = 1
x1 - 6x2 + x3 = 2
$$\lambda $$x1 - 10x2 + 4x3 = 3
is inconsistent for:
exactly one positive value of $$\lambda $$
exactly one negative value of $$\lambda $$
exactly two values of $$\lambda $$
every value of $$\lambda $$
Explanation
D = $$\left| {\matrix{
2 & { - 4} & \lambda \cr
1 & { - 6} & 1 \cr
\lambda & { - 10} & 4 \cr
} } \right|$$ = 0
$$ \Rightarrow $$ $$\lambda $$ = 3, $$ - {2 \over 3}$$
D1 = $$\left| {\matrix{ 1 & { - 4} & \lambda \cr 2 & { - 6} & 1 \cr 3 & { - 10} & 4 \cr } } \right|$$
= 14 + 4(5) + $$\lambda $$(–2)
= –2$$\lambda $$ + 6
D2 = $$\left| {\matrix{ 2 & 1 & \lambda \cr 1 & 2 & 1 \cr \lambda & 3 & 4 \cr } } \right|$$
= –2($$\lambda $$ – 3)($$\lambda $$ + 1)
D3 = $$\left| {\matrix{ 2 & { - 4} & 1 \cr 1 & { - 6} & 2 \cr \lambda & { - 10} & 3 \cr } } \right|$$
= – 2$$\lambda $$ + 6
When , $$\lambda $$ = 3 then
D = D1 = D2 = D3 = 0
$$ \Rightarrow $$ Infinite many solution
When $$\lambda $$ = $$ - {2 \over 3}$$ then D1, D2, D3 none of them is zero so equations are inconsistant.
$$ \therefore $$ $$\lambda $$ = $$ - {2 \over 3}$$
$$ \Rightarrow $$ $$\lambda $$ = 3, $$ - {2 \over 3}$$
D1 = $$\left| {\matrix{ 1 & { - 4} & \lambda \cr 2 & { - 6} & 1 \cr 3 & { - 10} & 4 \cr } } \right|$$
= 14 + 4(5) + $$\lambda $$(–2)
= –2$$\lambda $$ + 6
D2 = $$\left| {\matrix{ 2 & 1 & \lambda \cr 1 & 2 & 1 \cr \lambda & 3 & 4 \cr } } \right|$$
= –2($$\lambda $$ – 3)($$\lambda $$ + 1)
D3 = $$\left| {\matrix{ 2 & { - 4} & 1 \cr 1 & { - 6} & 2 \cr \lambda & { - 10} & 3 \cr } } \right|$$
= – 2$$\lambda $$ + 6
When , $$\lambda $$ = 3 then
D = D1 = D2 = D3 = 0
$$ \Rightarrow $$ Infinite many solution
When $$\lambda $$ = $$ - {2 \over 3}$$ then D1, D2, D3 none of them is zero so equations are inconsistant.
$$ \therefore $$ $$\lambda $$ = $$ - {2 \over 3}$$
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