Question

Prove that :

\[\begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix} = \begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{vmatrix}\]

 

Answer 1

\[\text{ Let LHS }= ∆ = \begin{vmatrix} 1 & a & bc\\1 & b & ca \\1 & c & ab \end{vmatrix}\] 
\[ = \frac{1}{abc}\begin{vmatrix} a & a^2 & abc\\b & b^2 & bca \\c & c^2 & abc \end{vmatrix} \left[\text{ Applying }R_1 \to a R_1 , R_2 \to b R_2\text{ and }R_3 \to c R_3\text{ and then dividing it by abc }\right] \] 
\[ = \frac{abc}{abc}\begin{vmatrix} a & a^2 & 1\\b & b^2 & 1\\c & c^2 & 1 \end{vmatrix} \left[\text{ Taking out abc common from }C_3 \right]\] 
\[ = \left( - 1 \right) \begin{vmatrix} 1 & a^2 & a\\1 & b^2 & b\\1 & c^2 & c \end{vmatrix} \left[\text{ Interchanging }C_3 \text{ and }C_1\text{ to get - ve value of original determinant }\right]\] 
\[ = \left( - 1 \right)\left( - 1 \right)\begin{vmatrix} 1 & a & a^2 \\1 & b^{} & b^2 \\1 & c & c^2 \end{vmatrix} \left[\text{ Applying }C_2 \leftrightarrow C_3 \right]\] 
\[ = \begin{vmatrix} 1 & a & a^2 \\1 & b^{} & b^2 \\1 & c & c \end{vmatrix}\] 
\[ = RHS\]