JEE MAIN - Mathematics (2018 - 15th April Evening Slot - No. 3)
If the system of linear equations
x + ay + z = 3
x + 2y + 2z = 6
x + 5y + 3z = b
has no solution, then :
x + ay + z = 3
x + 2y + 2z = 6
x + 5y + 3z = b
has no solution, then :
a = $$-$$ 1, b = 9
a = $$-$$ 1, b $$ \ne $$ 9
a $$ \ne $$ $$-$$ 1, b = 9
a = 1, b $$ \ne $$ 9
Explanation
As the given system of equations has no solution then
$$\Delta $$ = 0 and at least one of $$\Delta $$1, $$\Delta $$2 and $$\Delta $$2 should not be zero.
$$ \therefore $$ $$\Delta $$ = $$\left| {\matrix{ 1 & a & 1 \cr 1 & 2 & 2 \cr 1 & 5 & 3 \cr } } \right| = 0$$
$$ \Rightarrow $$ - $$a$$ - 1 = 0
$$ \Rightarrow $$ a = - 1
$$\Delta $$2 = $$\left| {\matrix{ 1 & 3 & 1 \cr 1 & 6 & 2 \cr 1 & b & 3 \cr } } \right| \ne 0$$
$$ \Rightarrow $$ b $$ \ne $$ 0
$$\Delta $$ = 0 and at least one of $$\Delta $$1, $$\Delta $$2 and $$\Delta $$2 should not be zero.
$$ \therefore $$ $$\Delta $$ = $$\left| {\matrix{ 1 & a & 1 \cr 1 & 2 & 2 \cr 1 & 5 & 3 \cr } } \right| = 0$$
$$ \Rightarrow $$ - $$a$$ - 1 = 0
$$ \Rightarrow $$ a = - 1
$$\Delta $$2 = $$\left| {\matrix{ 1 & 3 & 1 \cr 1 & 6 & 2 \cr 1 & b & 3 \cr } } \right| \ne 0$$
$$ \Rightarrow $$ b $$ \ne $$ 0
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