JEE MAIN - Mathematics (2020 - 2nd September Morning Slot - No. 1)
Let S be the set of all $$\lambda $$ $$ \in $$ R for which the system
of linear equations
2x – y + 2z = 2
x – 2y + $$\lambda $$z = –4
x + $$\lambda $$y + z = 4
has no solution. Then the set S :
2x – y + 2z = 2
x – 2y + $$\lambda $$z = –4
x + $$\lambda $$y + z = 4
has no solution. Then the set S :
contains more than two elements.
contains exactly two elements.
is a singleton.
is an empty set.
Explanation
For no solution :
$$\Delta $$ = 0 and $$\Delta $$1/$$\Delta $$2/$$\Delta $$3 $$ \ne $$ 0
$$\Delta $$ = $$\left| {\matrix{ 2 & { - 1} & 2 \cr 1 & { - 2} & \lambda \cr 1 & \lambda & 1 \cr } } \right|$$ = 0
$$ \Rightarrow $$ 2(–2 – $$\lambda $$2) + 1 (1 – $$\lambda $$) + 2($$\lambda $$ + 2) = 0
$$ \Rightarrow $$ –2$$\lambda $$2 + $$\lambda $$ + 1 = 0
$$ \Rightarrow $$ $$\lambda $$ = 1, $$ - {1 \over 2}$$
When $$\lambda $$ = 1
2x – y + 2z = 2 ...(1)
x – 2y + z = –4 ...(2)
x + y + z = 4 ...(3)
Adding (2) and (3), we get
2x – y + 2z = 0 (contradiction) hence no solution.
$$ \therefore $$ $$\lambda $$ = 1 belongs to set S.
When $$\lambda $$ = $$ - {1 \over 2}$$
2x – y + 2z = 2 ...(1)
x – 2y $$ - {1 \over 2}$$z = –4 ...(2)
x $$ - {1 \over 2}$$y + z = 4 ...(3)
(1) and (3) contradict each other, hence no solution.
$$ \therefore $$ $$\lambda $$ = $$ - {1 \over 2}$$ belongs to set S.
$$\Delta $$ = 0 and $$\Delta $$1/$$\Delta $$2/$$\Delta $$3 $$ \ne $$ 0
$$\Delta $$ = $$\left| {\matrix{ 2 & { - 1} & 2 \cr 1 & { - 2} & \lambda \cr 1 & \lambda & 1 \cr } } \right|$$ = 0
$$ \Rightarrow $$ 2(–2 – $$\lambda $$2) + 1 (1 – $$\lambda $$) + 2($$\lambda $$ + 2) = 0
$$ \Rightarrow $$ –2$$\lambda $$2 + $$\lambda $$ + 1 = 0
$$ \Rightarrow $$ $$\lambda $$ = 1, $$ - {1 \over 2}$$
When $$\lambda $$ = 1
2x – y + 2z = 2 ...(1)
x – 2y + z = –4 ...(2)
x + y + z = 4 ...(3)
Adding (2) and (3), we get
2x – y + 2z = 0 (contradiction) hence no solution.
$$ \therefore $$ $$\lambda $$ = 1 belongs to set S.
When $$\lambda $$ = $$ - {1 \over 2}$$
2x – y + 2z = 2 ...(1)
x – 2y $$ - {1 \over 2}$$z = –4 ...(2)
x $$ - {1 \over 2}$$y + z = 4 ...(3)
(1) and (3) contradict each other, hence no solution.
$$ \therefore $$ $$\lambda $$ = $$ - {1 \over 2}$$ belongs to set S.
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