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Question

For non zero, a, b, c if Δ=∣ ∣1+a1111+b1111+c∣ ∣=0, then the value of 1a+1b+1c=
  1. None of these
  2. abc
  3. 1abc
  4. a(a+b+c)


Solution

The correct option is A None of these
Δ=∣ ∣1+a1111+b1111+c∣ ∣
=abc∣ ∣ ∣ ∣1a+11b1c1a1b+11c1a1b1c+1∣ ∣ ∣ ∣, by C11aC1C21bC2C31cC3=abc∣ ∣ ∣ ∣ ∣(1+1a)1b1c(1+1a)1b+11c(1+1a)1b1c+1∣ ∣ ∣ ∣ ∣
by C1C1+C2+C3
=abc(1+1a)∣ ∣ ∣ ∣11b1c11b+11c11b1c+1∣ ∣ ∣ ∣
[By taking (1+1a) as common]
Δ=abc(1+1a)∣ ∣ ∣ ∣11b1c11b+11c11b1c+1∣ ∣ ∣ ∣,byR2R2R1R3R3R1=abc(1+1a+1b+1c).1, (by expanding along C1)
Therefore Δ=0 Either abc = 0 or 1+1a+1b+1c=0 
But a,b,c are non-zero and hence the product abc cannot be zero. So the only alternative is that 1a+1b+1c=1

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