Question

Without expanding the determinant, prove that
\(\begin{vmatrix} a &a^2  &bc \\   b&b^2 &ca \\   c&c^2  &ab  \end{vmatrix}=\begin{vmatrix} 1 &a^2  &a^3 \\   1&b^2  &b^3 \\   1&c^2  &c^3  \end{vmatrix}\)

Answer 1

L.H.S= \(\begin{vmatrix}     a& a^2& b_c\\    b & b^2 & ca \\    c& c^2& ab  \\    \end{vmatrix}  \)
\(=\frac{1}{abc}\begin{vmatrix}     a^2& a^3& abc\\    b^2 & b^3 & abc \\    c^2& c^3& abc  \\    \end{vmatrix}  [R_1\rightarrow{aR_1,R_2}\rightarrow{bR_2,R_3}\rightarrow{cR_3}]\)
\(=\frac{abc}{abc}\begin{vmatrix}     a^2& a^3& 1\\    b^2 & b^3 & 1 \\    c^2& c^3& 1 \\    \end{vmatrix}  =\begin{vmatrix}     a^2& a^3& 1\\    b^2 & b^3 & 1 \\    c^2& c^3& 1 \\    \end{vmatrix} \)     [Taking out factor ABC from C,]
 = \(\begin{vmatrix}     1& a^2& a^3\\    1 & b^2& b^3 \\   1& c^2& c^3 \\    \end{vmatrix} \)    [Applying \(C_1\rightarrow{C_1}\) \(C_2\rightarrow \)
=R.H.S
Hence, the given result is proved.