Question

Prove that :

$\left|\begin{array}{ccc}1& a& bc\\ 1& b& ca\\ 1& c& ab\end{array}\right|=\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{LHS}=∆=\left|1abc \\ 1bca \\ 1cab\right| \\ =\frac{1}{\mathrm{abc}}\left|a{a}^{2}abc \\ b{b}^{2}bca \\ c{c}^{2}abc\right|\left[\mathrm{Applying}{\mathrm{R}}_1\to \mathrm{a}{\mathrm{R}}_1,{\mathrm{R}}_2\to \mathrm{b}{\mathrm{R}}_2\mathrm{and}{\mathrm{R}}_3\to \mathrm{c}{\mathrm{R}}_3\mathrm{and}\mathrm{then}\mathrm{dividing}\mathrm{it}\mathrm{by}abc\right] \\ =\frac{abc}{abc}\left|a{a}^{2}1 \\ b{b}^{2}1 \\ c{c}^{2}1\right|\left[\mathrm{Taking}\mathrm{out}abc\mathrm{common}\mathrm{from}{\mathrm{C}}_3\right] \\ =\left(-1\right)\left|1a^{2}a \\ 1b^{2}b \\ 1c^{2}c\right|\left[\mathrm{Interchanging}{\mathrm{C}}_3\mathrm{and}{\mathrm{C}}_1\mathrm{to}\mathrm{get}-\mathrm{ve}\mathrm{value}\mathrm{of}\mathrm{original}\mathrm{determinant}\right] \\ =\left(-1\right)\left(-1\right)\left|1a{a}^2 \\ 1b^{}{b}^2 \\ 1c{c}^2\right|\left[\mathrm{Applying}{\mathrm{C}}_2\leftrightarrow {\mathrm{C}}_3\right] \\ =\left|1a{a}^2 \\ 1b^{}{b}^2 \\ 1cc\right| \\ =\mathrm{RHS} \end{gathered}$$