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Question
Evaluate :
\[\begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix}\]
Solution
\[∆ = \begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix} \]
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,
\[\begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix} = (a - b)(b - c)(c - a)\]
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