$∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣ =$

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

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3 a b

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3 a + 5 b

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0

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Solution

The correct option is D $0$
$∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣$
$R_{3} \to R_{3} - R_{2}$ and $R_{2} \to R_{2} - R_{1}$
$= ∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b b & b & b 2 b & 2 b & 2 b \end{matrix} ∣ ∣ ∣ ∣$
$R_{3} \to \frac{R_{3}}{2}$
$= 2 ∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b b & b & b b & b & b \end{matrix} ∣ ∣ ∣ ∣$
Since $R_{2}$ and $R_{3}$ are identical,we have
$∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b b & b & b b & b & b \end{matrix} ∣ ∣ ∣ ∣ = 0$
$∴ ∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣ = 0$

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Similar questions

∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣ =

∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣ =

(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0

(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0

a, b, c

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