LA2.1.7

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Show that {\begin{vmatrix}bc&b+c&1\\ca&c+a&1\\ab&a+b&1\end{vmatrix}}=(a-b)(b-c)(c-a)\,

Let A={\begin{vmatrix}bc&b+c&1\\ca&c+a&1\\ab&a+b&1\end{vmatrix}}\,

Applying R_{2}\rightarrow R_{2}-R_{1},R_{3}\rightarrow R_{3}-R_{2}\,

A={\begin{vmatrix}bc&b+c&1\\ca-bc&a-b&0\\ab-ca&a-c&0\end{vmatrix}}\,

Expanding by the elements of C3,we get A=1{\begin{vmatrix}c(a-b)&a-b\\a(b-c)&b-c\end{vmatrix}}=(a-b)(b-c)(c-a){\begin{vmatrix}c&1\\a&1\end{vmatrix}}\,

Therefore A=(a-b)(b-c)(c-a)\,

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