JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 17)
The values of $$\lambda $$ and $$\mu $$ for which the system of linear equations
x + y + z = 2
x + 2y + 3z = 5
x + 3y + $$\lambda $$z = $$\mu $$
has infinitely many solutions are, respectively:
x + y + z = 2
x + 2y + 3z = 5
x + 3y + $$\lambda $$z = $$\mu $$
has infinitely many solutions are, respectively:
6 and 8
5 and 8
5 and 7
4 and 9
Explanation
For infinite many solutions
D = D1 = D2 = D3 = 0
Now D = $$\left| {\matrix{ 1 & 1 & 1 \cr 1 & 2 & 3 \cr 1 & 3 & \lambda \cr } } \right|$$ = 0
$$ \Rightarrow $$ 1. (2$$\lambda $$ – 9) –1.($$\lambda $$ – 3) + 1.(3 – 2) = 0
$$ \Rightarrow $$ $$\lambda $$ = 5
Now D1 = $$\left| {\matrix{ 2 & 1 & 1 \cr 5 & 2 & 3 \cr \mu & 3 & 5 \cr } } \right|$$ = 0
$$ \Rightarrow $$ 2(10 – 9) –1(25 – 3$$\mu $$) + 1(15 – 2$$\mu $$) = 0
$$ \Rightarrow $$ $$\mu $$ = 8
D = D1 = D2 = D3 = 0
Now D = $$\left| {\matrix{ 1 & 1 & 1 \cr 1 & 2 & 3 \cr 1 & 3 & \lambda \cr } } \right|$$ = 0
$$ \Rightarrow $$ 1. (2$$\lambda $$ – 9) –1.($$\lambda $$ – 3) + 1.(3 – 2) = 0
$$ \Rightarrow $$ $$\lambda $$ = 5
Now D1 = $$\left| {\matrix{ 2 & 1 & 1 \cr 5 & 2 & 3 \cr \mu & 3 & 5 \cr } } \right|$$ = 0
$$ \Rightarrow $$ 2(10 – 9) –1(25 – 3$$\mu $$) + 1(15 – 2$$\mu $$) = 0
$$ \Rightarrow $$ $$\mu $$ = 8
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