Question

Using the property of determinants and without expanding, prove that: $\begin{vmatrix}1&bc&a(b+c)\\\1&ca&b(c+a)\\\1&ab&c(a+b)\end{vmatrix}$=0

Answer 1

$\triangle=\begin{vmatrix}1&bc&a(b+c)\\\1&ca&b(c+a)\\\1&ab&c(a+b)\end{vmatrix}$

By applying C3 $\to$ C3 + C2, we have:
$\triangle=\begin{vmatrix}1&bc&ab+bc+ca\\\1&ca&ab+bc+ca\\\1&ab&ab+bc+ca\end{vmatrix}$
Here, two columns C1 and C3 are proportional.
∆ = 0.