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.