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# By using properties of determinants show that determinant 1 a a^2 1 b b^2 q c c^2 =(a-b)(b-c)(c-a) Ex: 4.2 Q: 8 (i)

Q : 8        By using properties of determinants, show that:

(i)      $\dpi{100} \begin{vmatrix} 1 &a &a^2 \\ 1 &b &b^2 \\ 1 &c &c^2 \end{vmatrix}=(a-b)(b-c)(c-a)$

Views

We have the determinant $\dpi{100} \begin{vmatrix} 1 &a &a^2 \\ 1 &b &b^2 \\ 1 &c &c^2 \end{vmatrix}$

Applying the row transformations $R_{1} \rightarrow R_{1} -R_{2}$  and then $R_{2} \rightarrow R_{2} -R_{3}$  we have:

$\dpi{100} \triangle = \begin{vmatrix} 0 &a-b &a^2-b^2 \\ 0 &b-c &b^2-c^2 \\ 1 &c &c^2 \end{vmatrix}$

$\dpi{100} = \begin{vmatrix} 0 &a-b &(a-b)(a+b) \\ 0 &b-c &(b-c)(b+c) \\ 1 &c &c^2 \end{vmatrix} = (a-b)(b-c)\begin{vmatrix} 0 &1 &(a+b) \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix}$

Now, applying $R_{1} \rightarrow R_{1} -R_{2}$ we have:

$\dpi{100} = (a-b)(b-c)\begin{vmatrix} 0 &0 &(a-c) \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix}$    or    $\dpi{100} = (a-b)(b-c)(a-c)\begin{vmatrix} 0 &0 &1 \\ 0 &1 &(b+c) \\ 1 &c &c^2 \end{vmatrix} =(a-b)(b-c)(a-c)\begin{vmatrix} 0 &1 \\ 1 & c \end{vmatrix}$

$\dpi{100} = (a-b)(b-c)(c-a)$

Hence proved.

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