$∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ =$

abc(a+b+c)

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3 a^{2} b^{2} c^{2}

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None of these

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Solution

The correct option is C \N
$Δ = (b - a) (b - a) . ∣ ∣ ∣ ∣ \begin{matrix} b & b - c & c a & a - b & b c & c - a & a \end{matrix} ∣ ∣ ∣ ∣ = (a - b)^{2} ∣ ∣ ∣ ∣ \begin{matrix} b & b & c a & a & b c & c & a \end{matrix} ∣ ∣ ∣ ∣ = 0, [b y C_{2} \to C_{2} + C_{3}]$ .

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Similar questions

∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ =

∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

|\begin{matrix} b^{2} - a b & b - c & b c - a c \\ a b - a^{2} & a - b & b^{2} - a b \\ b c - c a & c - a & a b - a^{2} \end{matrix}|

a b c (b - c) (c - a) (a - b)

(b - c) (c - a) (a - b)

(a + b + c) (b - c) (c - a) (a - b)

∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

c o s (i l o g \frac{a - i b}{a + i b})

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