Question

Show that $\left|\begin{array}{ccc}bc&b+c&1\\ca&c+a&1\\ab&a+b&1\end{array}\right|$ = (a - b)(b - c)(c - a).

Answer 1

Answer:

Step-by-step explanation:

$\left|\begin{array}{ccc}bc&b+c&1\\ca&c+a&1\\ab&a+b&1\end{array}\right|$

it can be easily solved by applying some elementary row operations

$R_{3} -> R_{3}-R_{1}\\\\R_{2} -> R_{2}-R_{1}\\\\\\\left|\begin{array}{ccc}bc&b+c&1\\ca-bc&c+a-b-c&1-1\\ab-bc&a+b-b-c&1-1\end{array}\right|\\\\\\\left|\begin{array}{ccc}bc&b+c&1\\c(a-b)&a-b&0\\b(a-c)&a-c&0\end{array}\right|$

now expand the determinant along C3

$=1[b(a-c)(a-b)-c(a-b)(a-c)]\\\\=(a-c)(a-b)(a-c)$

hence proved