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Properties of Determinants

Show that b...

Question

Show that $∣ ∣ ∣ ∣ \begin{matrix} b c & b + c & 1 c a & c + a & 1 a b & a + b & 1 \end{matrix} ∣ ∣ ∣ ∣ = (a - b) (b - c) (c - a)$ .

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Solution

$∣ ∣ ∣ ∣ \begin{matrix} b c & b + c & 1 c a & c + a & 1 a b & a + b & 1 \end{matrix} ∣ ∣ ∣ ∣$ .
Using $R_{1} \to R_{1} - R_{2}$ and $R_{2} \to R_{2} - R_{3}$

We get,
$∣ ∣ ∣ ∣ \begin{matrix} b c - c a & b - a & 0 c a - a b & c - b & 0 a b & a + b & 1 \end{matrix} ∣ ∣ ∣ ∣$ .
= $(b c - c a) (c - b) - (c a - a b) (b - a)$
= $(a - b) (b - c) (c - a)$

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Properties of Determinants

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