Let a , b , c be such that ( b + c ) ≠ 0 and a a + 1 a - 1 - b b + 1 b - 1 c c - 1 c + 1 + a + 1 b + 1…

Let a,b,c be such that (b+c)0 and

aa+1a-1-bb+1b-1cc-1c+1+a+1b+1c-1a-1b-1c+1-1n+2a-1n-1b-1nc=0

Then the value of n is
  1. Zero
  2. Any even integer
  3. Any odd integer
  4. Any integer

Solution


aa+1a-1-bb+1b-1cc-1c+1+a+1b+1c-1a-1b-1c+1-1n+2a-1n-1b-1nc=0

aa+1a-1-bb+1b-1cc-1c+1+-1na+1b+1c-1a-1b-1c+1a-bc=0

aa+1a-1-bb+1b-1cc-1c+1+-1na+1a-1ab+1b-1-bc-1c+1c=0

aa+1a-1-bb+1b-1cc-1c+1+-1naa+1a-1-bb+1b-1cc-1c+1=0

(1+-1n)aa+1a-1-bb+1b-1cc-1c+1=0

Hence, n is an odd integer.

Asked in: AP EAMCET 2021 (19 Aug Shift 2)

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