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)\,

Main Page:Linear Algebra

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