Question

Let a, b, c ∈R be all non-zero and satisfy a³ + b³ + c³ = 2. If the matrix

satisfies A^TA = I, then a value of abc can be :

(1)  \(\frac{2}{3}\)

(2)  \(\frac{1}{3}\)

(3)  3

(4)  \(\frac{1}{3}\)

Answer 1

Answer is (4)  \(\frac{1}{3}\)

A^TA = I 

⇒ a² + b² + c² = 1

and ab + bc + ca = 0

Now, (a + b + c) 2 = 1

⇒a + b + c = ± 1

So, a³ + b³ + c³ – 3abc

= (a + b + c)(a² + b² + c² – ab – bc – ca)

= ± 1 (1 – 0) = ± 1

⇒ 3 abc = 2 ± 1 = 3, 1

⇒ abc = 1,\(\frac{1}{3}\)