Question

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^TA = I, then a value of abc can be ______.

Options

• `-1/3`

• `1/3`

• 3

• `2/3`

Answer 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^TA = I, then a value of abc can be `underlinebb(1/3)`.

Explanation:

Given: A^TA = I

`\implies [(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)]`

`\implies [(suma^2, sumab, sumab),(sumab, suma^2, sumab),(sumab, sumab, suma^2)] = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

So `suma^2` = 1 and `sumab` = 0

Now, a³ + b³ + c³ – 3abc

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

= (a + b + c)(1 – 0)

= `sqrt((a + b + c)^2`

= `sqrt(suma^2 + 2sumab)`

= ±1

`\implies` 2 – 3abc = 1

`\implies` abc = `1/3`

or 2 – 3abc = –1

`\implies` abc = 1.