JEE Main 2020 · 2 September · evening · Single correct · +4 / −1

Let a, b, c ∈ R be all non-zero and satisfy
a³ + b³ + c³ = 2. If the matrix
A = ( a, b, c; b, c, a; c, a, b )
satisfies A T A = I, then a value of abc can be :

  1. (a) 3
  2. (b) 1/3
  3. (c) -1/3
  4. (d) 2/3
Answer and solution

Answer (b) 1/3

Solution

Maths sir · 4 steps

Turn AᵀA = I into two equations, then use the factorised form of a³+b³+c³-3abc.

  1. Read off conditions. AᵀA = I means every column has length 1 and any two columns are perpendicular. Column 1 with itself gives a²+b²+c² = 1. Column 1 with column 2 gives ab+bc+ca = 0.
  2. Square the sum. (a+b+c)² = a²+b²+c² + 2(ab+bc+ca) = 1 + 0 = 1, so a+b+c = ± 1.

    a³+b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ca) = (± 1)(1 - 0) = ± 1

  3. Use the given sum. Put a³+b³+c³ = 2 to get 2 - 3abc = ± 1. With +1 you get abc = 1/3. With -1 you get abc = 1, which is not an option.
  4. So. A possible value of abc is 1/3, option (b).

The idea to keep. Orthogonal means unit columns and perpendicular columns. A sum of squares plus a sum of products points straight to (a+b+c)² and the a³+b³+c³-3abc factorisation.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Matrices

Asked about 110 times in JEE Main since 2020. This question is from 2020.