AP EAMCET20224 Jul 2022Morning ShiftMathematicsDeterminantsActual
For i = 1 , 2 , 3 and j = 1 , 2 , 3 . If a i 2 + b i 2 + c i 2 = 1 , a i a j + b i b j + c i c j = 0 , ∀ i ≠ j and A = a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 then det A A T =
Options
- A0
- B1
- C- 1
- D3
Correct answer
B. 1
Step-by-step solution
We know A A T = A A T = A 2 i.e. a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 2 = a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 = Σ a 1 2 Σ a 1 a 2 Σ a 1 a 3 Σ a 2 a 1 Σ a 2 2 Σ a 2 a 3 Σ a 3 a 1 Σ a 3 a 2 Σ a 3 2 = 1 0 0 0 1 0 0 0 1 = 1