By using the properties of determinants show that $(i) \quad \begin{vmatrix} 1&a&a^2 \\ 1&b&b^2 \\ 1&c&c^2 \end{vmatrix} = (a-b)(b-c)(c-a)$

Note: This is part 1 of a 2  part question, split as 2 separate questions here.

Toolbox:
• If each element of a row (or column) of a determinant is multiplied by a constant k ,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any one column of the determinant.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns. rows.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
By using properties of determinant ,show that

(i) $\begin{vmatrix}1 & a & a^2\\1 & b & b^2\\1 &c & c^2\end{vmatrix}=(a-b)(b-c)(c-a)$

Let $\bigtriangleup=\begin{vmatrix}1 & a & a^2\\1 & b & b^2\\1 &c & c^2\end{vmatrix}$

Let us apply $R_1\rightarrow R_1-R_3$ and $R_2\rightarrow R_2-R_3$

$\bigtriangleup=\begin{vmatrix}0 & (a-c) & (a^2-c^2)\\0 & (b-c) & (b^2-c^2)\\1 & c & c^2\end{vmatrix}$

$a^2-c^2=(a+c)(a-c)$

$b^2-c^2=(b+c)(b-c)$

Let us take (a-c) as a common factor from $R_1$ and (b-c) from $R_2$ we get

$(a-c)(b-c)\begin{vmatrix}0 & 1 & a+c\\0 & 1 & b+c\\1 &c &c^2\end{vmatrix}$

Now applying $R_1\rightarrow R_1-R_2$

$\bigtriangleup=(a-c)(b-c)\begin{vmatrix}0 & 0 & a-b\\0 & 1 & b+c\\1 & c & c^2\end{vmatrix}$

Taking (a-b) as a factor from $R_1$

$(a-c)(b-c)(a-b)\begin{vmatrix}0 & 0 & 1\\0 & 1 & b+c\\1 & c & c^2\end{vmatrix}$

Now expanding along $C_1$ we get

$(a-c)(b-c)(a-b)1\mid 0(b+c)-1\mid$

$\;\;\;=(-1)(a-b)(b-c)(a-c)$

$\;\;\;=(a-b)(b-c)(c-a)$

Hence $\begin{vmatrix}1 & a & a^2\\1 & b & b^2\\1 & c& c^2\end{vmatrix}=(a-b)(b-c)(c-a)$

Hence proved.

edited Feb 24, 2013