Question

Evaluate :

\[\begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix}\]

Answer 1

\[∆ = \begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix} \]

When a = b, the first two rows become identical . Hence, a - b is a factor .

Similarly, when b = c and c = a, the second and third and third and first rows become identical . Hence, b - c and c - a are also factors . 

The degree of product of the diagonal elements is 3 . Hence, there are no other factors . 

\[ \begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix} = \lambda(a - b)(b - c)(c - a) \left[\text{ Where }\lambda\text{ is a constant }\right]\]

\[\begin{vmatrix}1 & 0 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 0\end{vmatrix} = 2\lambda \left[\text{ Putting }a = 0, b = 1 \text{ and }c = 2\text{ to find }\lambda \right]\]

\[ \Rightarrow 2 = 2\lambda\]

\[ \Rightarrow \lambda = 1\]

Hence,

\[\begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix} = (a - b)(b - c)(c - a)\]