JEE MAIN - Mathematics (2020 - 4th September Evening Slot - No. 5)
If the system of equations
x+y+z=2
2x+4y–z=6
3x+2y+$$\lambda $$z=$$\mu $$
has infinitely many solutions, then
x+y+z=2
2x+4y–z=6
3x+2y+$$\lambda $$z=$$\mu $$
has infinitely many solutions, then
2$$\lambda $$ - $$\mu $$ = 5
$$\lambda $$ - 2$$\mu $$ = -5
2$$\lambda $$ + $$\mu $$ = 14
$$\lambda $$ + 2$$\mu $$ = 14
Explanation
$$D = 0\,\left| {\matrix{
1 & 1 & 1 \cr
2 & 4 & { - 1} \cr
3 & 2 & \lambda \cr
} } \right| = 0$$
$$ \Rightarrow $$ $$(4\lambda + 2) - 1(2\lambda + 3) + 1(4 - 12) = 0$$
$$ \Rightarrow $$ $$4\lambda + 2$$ $$ - 2\lambda - 3$$$$ - $$8$$ = 0$$
$$ \Rightarrow $$ $$2\lambda = 9 \Rightarrow \lambda = {9 \over 2}$$
$${D_x} = \left| {\matrix{ 2 & 1 & 1 \cr 6 & 4 & { - 1} \cr \mu & 2 & { - 9/2} \cr } } \right| = 0$$
$$ \Rightarrow \mu = 5$$
By checking all the options we find (C) is correct
$$2\lambda + \mu = 14$$
$$ \Rightarrow $$ $$(4\lambda + 2) - 1(2\lambda + 3) + 1(4 - 12) = 0$$
$$ \Rightarrow $$ $$4\lambda + 2$$ $$ - 2\lambda - 3$$$$ - $$8$$ = 0$$
$$ \Rightarrow $$ $$2\lambda = 9 \Rightarrow \lambda = {9 \over 2}$$
$${D_x} = \left| {\matrix{ 2 & 1 & 1 \cr 6 & 4 & { - 1} \cr \mu & 2 & { - 9/2} \cr } } \right| = 0$$
$$ \Rightarrow \mu = 5$$
By checking all the options we find (C) is correct
$$2\lambda + \mu = 14$$
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