Question
If the system of linear equation x+2ay+az = 0, x+3by+bz = 0, x+4cy+cz = 0 has a non zero solution, then a, b, c
Answer: Option C
:
C
∣∣
∣∣12aa13bb14cc∣∣
∣∣=0,[C2→C2−2C3]⇒∣∣
∣∣10a1bb12cc∣∣
∣∣=0,[R3→R3−R2,R2→R2−R1]⇒∣∣
∣∣10a0bb−a02c−bc−b∣∣
∣∣=0;b(c−b)−(b−a)(2c−b)=0
On simplification,2b=1a+1c
∴ a, b, c are in Harmonic progression.
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:
C
∣∣
∣∣12aa13bb14cc∣∣
∣∣=0,[C2→C2−2C3]⇒∣∣
∣∣10a1bb12cc∣∣
∣∣=0,[R3→R3−R2,R2→R2−R1]⇒∣∣
∣∣10a0bb−a02c−bc−b∣∣
∣∣=0;b(c−b)−(b−a)(2c−b)=0
On simplification,2b=1a+1c
∴ a, b, c are in Harmonic progression.
Was this answer helpful ?
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