If Delta=left|begin{array}{lll}1 & a & a^2 1 | vedclass

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(C) Given,$\Delta=\left|\begin{array}{lll}1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2\end{array}ight|$ and $\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1

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$\Delta_1=-\Delta$

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(C) Given,$\Delta=\left|\begin{array}{lll}1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2\end{array}ight|$ and $\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \ b c & c a & a b \ a & b & c\end{array}ight|$.First,we evaluate $\Delta$:$\Delta = (a-b)(b-c)(c-a)$.Now,for $\Delta_1$,we multiply and divide the determinant by $abc$ or perform row/column operations.Alternatively,multiply $R_1$ by $abc$,$R_2$ by $1$,and $R_3$ by $1$ is not efficient. Let us multiply $C_1$ by $a$,$C_2$ by $b$,and $C_3$ by $c$:$\Delta_1 = \frac{1}{abc} \left|\begin{array}{ccc}a & b & c \ abc & abc & abc \ a^2 & b^2 & c^2\end{array}ight| = \frac{abc}{abc} \left|\begin{array}{ccc}a & b & c \ 1 & 1 & 1 \ a^2 & b^2 & c^2\end{array}ight| = \left|\begin{array}{ccc}a & b & c \ 1 & 1 & 1 \ a^2 & b^2 & c^2\end{array}ight|$.Swapping $R_1$ and $R_2$ gives $-\left|\begin{array}{ccc}1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2\end{array}ight|$.Swapping $R_2$ and $R_3$ gives $(-1)^2 \left|\begin{array}{ccc}1 & 1 & 1 \ a^2 & b^2 & c^2 \ a & b & c\end{array}ight|$.Actually,the standard result for $\Delta_1$ is $-(a-b)(b-c)(c-a)$.Thus,$\Delta_1 = -\Delta$.

If $\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|$ and $\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \\ b c & c a & a b \\ a & b & c\end{array}\right|$,then

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