Question

If a + b + c = 4 and ab + bc + ca = 0, then what is the value of the following determinant?

\(\left| {\begin{array}{*{20}{c}} {{a}}&{{b}}&{{c}}\\ {{b}}&{{c}}&{{a}}\\ {{c}}&{{a}}&{{b}} \end{array}} \right|\)

1. 32
2. -64
3. -128
4. 64

Answer 1

Correct Answer - Option 2 : -64

Concept:

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

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

Calculation:

Let Δ  = \(\left| {\begin{array}{*{20}{c}} {{a}}&{{b}}&{{c}}\\ {{b}}&{{c}}&{{a}}\\ {{c}}&{{a}}&{{b}} \end{array}} \right|\) 

= a (cb - a²) - b (b² - ca) + c (ba - c²)

= abc - a³ - b³ + abc + abc - c³

= -[a³ + b³ + c³ - 3abc]

= -(a + b + c)(a² + b² + c² - ab - bc - ca)      ....(i)

As we know,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ 4² = a2 + b2 + c2 + 2 × 0 

⇒ 16 = a2 + b2 + c2

From equ (i)

Δ = -[4 × (16 - 0)]

= -64