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NDA Math Determinants

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Question : 14 of 68

Marks: +1, -0

If

a+b+c=4

a b+b c+c a=0

, then what is the value of the following determinant?

\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}

32
$-64$
$-128$
64

Solution: 👈: Video Solution

Concept:

a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-c a)

(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)

Calculation:

Let

\Delta=\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}

=a(c b-a^{2})-b(b^{2}-c a)+c(b a-c^{2})

\;\; =a b c-a^{3}-b^{3}+a b c+a b c-c^{3}

\;\; =-[a^{3}+b^{3}+c^{3}-3 a b c]

\;\; =-(a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-c a)

As we know,

(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)

\Rightarrow 4^{2}=a^{2}+b^{2}+c^{2}+2 \times 0

\Rightarrow 16=a^{2}+b^{2}+c^{2}

From equ (i)

\Delta=-[4 \times (16-0)]

\;\; =-64

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