JEE Main20202 Sep 2020Evening ShiftMathematicsMatricesActual
Let a , b , c ∈ R be all non-zero and satisfies a 3 + b 3 + c 3 = 2 . If the matrix A = a b c b c a c a b satisfies A T A = I , then a value of a b c can be
Options
- A- 1 3
- B1 3
- C3
- D2 3
Correct answer
B. 1 3
Step-by-step solution
∵ A T A = I ⇒ a b c b c a c a b a b c b c a c a b = 1 0 0 0 1 0 0 0 1 ⇒ a 2 + b 2 + c 2 a b + b c + a c a b + b c + a c a b + b c + a c a 2 + b 2 + c 2 a b + b c + a c a b + b c + a c a b + b c + a c a 2 + b 2 + c 2 = 1 0 0 0 1 0 0 0 1 On comparing each element both sides, we get a 2 + b 2 + c 2 = 1   &   a b + b c + c a = 0   . . . . . . . . . i We know that a 3 + b 3 + c 3 - 3 a b c = a + b + c a 2 + b 2 + c 2 - a b - b c - a c . ⇒ 2 - 3 a b c = a + b + c 1 - 0 (from equ