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Properties of Determinants

Using propert...

Question

Using properties of determinants, prove that:
$∣ ∣ ∣ ∣ \begin{matrix} 1 + a & 1 & 1 1 & 1 + b & 1 1 & 1 & 1 + c \end{matrix} ∣ ∣ ∣ ∣ = a b c + b c + c a + a b$

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Solution

$△ = ∣ ∣ ∣ ∣ \begin{matrix} 1 + a & 1 & 1 1 & 1 + b & 1 1 & 1 & 1 + c \end{matrix} ∣ ∣ ∣ ∣$
$R_{1} \to R_{1} - R_{2} R_{3} \to R_{3} - R_{2}$
$= ∣ ∣ ∣ ∣ \begin{matrix} a & - b & 0 1 & 1 + b & 1 0 & - b & c \end{matrix} ∣ ∣ ∣ ∣$
$= a (c + b c + b) + b c$
$= a b + b c + a c + a b c$

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