JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 7)
If the system of linear equations
2x + 2ay + az = 0
2x + 3by + bz = 0
2x + 4cy + cz = 0,
where a, b, c $$ \in $$ R are non-zero distinct; has a non-zero solution, then:
2x + 2ay + az = 0
2x + 3by + bz = 0
2x + 4cy + cz = 0,
where a, b, c $$ \in $$ R are non-zero distinct; has a non-zero solution, then:
$${1 \over a},{1 \over b},{1 \over c}$$ are in A.P.
a + b + c = 0
a, b, c are in G.P.
a,b,c are in A.P.
Explanation
For non-zero solution
$$\left| {\matrix{ 2 & {2a} & a \cr 2 & {3b} & b \cr 2 & {4c} & c \cr } } \right| = 0$$
$$ \Rightarrow $$ $$\left| {\matrix{ 1 & {2a} & a \cr 0 & {3b - 2a} & {b - a} \cr 0 & {4c - 2a} & {c - a} \cr } } \right| = 0$$
$$ \Rightarrow $$ (3b – 2a) (c –a) – (b – a) (4c – 2a) = 0
$$ \Rightarrow $$ 2ac = bc + ab
$$ \Rightarrow $$ $${2 \over b} = {1 \over a} + {1 \over c}$$
$$ \therefore $$ $${1 \over a},{1 \over b},{1 \over c}$$ are in A.P.
$$\left| {\matrix{ 2 & {2a} & a \cr 2 & {3b} & b \cr 2 & {4c} & c \cr } } \right| = 0$$
$$ \Rightarrow $$ $$\left| {\matrix{ 1 & {2a} & a \cr 0 & {3b - 2a} & {b - a} \cr 0 & {4c - 2a} & {c - a} \cr } } \right| = 0$$
$$ \Rightarrow $$ (3b – 2a) (c –a) – (b – a) (4c – 2a) = 0
$$ \Rightarrow $$ 2ac = bc + ab
$$ \Rightarrow $$ $${2 \over b} = {1 \over a} + {1 \over c}$$
$$ \therefore $$ $${1 \over a},{1 \over b},{1 \over c}$$ are in A.P.
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