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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

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Best answer
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

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