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

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

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3ab

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3a+5b

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Solution

The correct option is D \N
$∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b b & b & b 2 b & 2 b & 2 b \end{matrix} ∣ ∣ ∣ ∣ = 0 b y {\begin{matrix} R_{2} \to R_{2} - R_{1} R_{3} \to R_{3} - R_{2} \end{matrix}}$
Trick : Putting a=1=b. The determinant will be $∣ ∣ ∣ ∣ \begin{matrix} 2 & 3 & 4 3 & 4 & 5 5 & 6 & 7 \end{matrix} ∣ ∣ ∣ ∣ = 0$ .

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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} ∣ ∣ ∣ ∣ =

|\begin{matrix} a & a + b & a + 2 b \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{matrix}| = 9 (a + b) b^{2}

∣ ∣ ∣ ∣ \begin{matrix} a & a + b & a + 2 b a + 2 b & a & a + b a + b & a + 2 b & a \end{matrix} ∣ ∣ ∣ ∣ = 9 b^{2} (a + b)

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