$∣ ∣ ∣ ∣ \begin{matrix} b + c & a & a b & c + a & b c & c & a + b \end{matrix} ∣ ∣ ∣ ∣ =$

abc

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2abc

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3abc

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4abc

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Solution

The correct option is D 4abc
$∣ ∣ ∣ ∣ \begin{matrix} b + c & a & a b & c + a & b c & c & a + b \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} 0 & - 2 c & - 2 b b & c + a & b c & c & a + b \end{matrix} ∣ ∣ ∣ ∣ {b y R_{1} \to R_{1} - (R_{2} + R_{3})} = 2 c . b (a + b - c) - 2 b . c (b - c - a) = 4 a b c$ .

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$∣ ∣ ∣ ∣ \begin{matrix} b + c & a & a b & c + a & b c & c & a + b \end{matrix} ∣ ∣ ∣ ∣$ $= 4 a b c$

|\begin{matrix} b + c & a & a \\ b & c + a & b \\ c & c & a + b \end{matrix}| = 3 a b c

∣ ∣ ∣ ∣ \begin{matrix} b + c & a & a b & c + a & b c & c & a + b \end{matrix} ∣ ∣ ∣ ∣ = 4 a b c

∣ ∣ ∣ ∣ \begin{matrix} b + c & a & a b & c + a & b c & c & a + b \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} b + c & a & a b & c + a & b c & c & a + b \end{matrix} ∣ ∣ ∣ ∣

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