If k ∈ R and det A = a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 = K then det B = a 1 b 1 c 1 a 2 + 2 a 1 b 2…

If kR and detA=a1b1c1a2b2c2a3b3c3=K then detB=a1b1c1a2+2a1b2+2b1c2+2c1a3b3c3 is equal to
  1. 0
  2. 2 K
  3. K
  4. K2

Solution

detA=a1b1c1a2b2c2a3b3c3=K

 detB=a1b1c1a2+2a1b2+2b1c2+2c1a3b3c3 

R2R2-2R1

detB=a1b1c1a2+2a1-2a1b2+2b1-2b1c2+2c1-2c1a3b3c3

detB=a1b1c1a2  b2  c2a3b3c3

detB=K

Asked in: AP EAMCET 2021 (20 Aug Shift 2)

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