mple \( 1.9 \)Evaluate \( \left|\begin{array}{lll}1 & a & a^{2} \\ 1 &

1.	mple \( 1.9 \)Evaluate \( \left\|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right\|=(a-b)(b-c)(c-a) \
Answer» \(\begin{vmatrix}1& a & a^2 \\[0.3em]1 & b &b^2 \\[0.3em]1 & c & c^2\end{vmatrix}\) = \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & b-a &b^2-c^2 \\[0.3em]0 & c-a & c^2-a^2\end{vmatrix}\) (By applying R₂→R₂-R₁ and R₃→R₃-R₁) = \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & b-a &(b-a)(b+a) \\[0.3em]0 & c-a & (c-a)(c+a)\end{vmatrix}\) = (b-a) (c-a) \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & 1 &b+a \\[0.3em]0 & 1 &c+a\end{vmatrix}\) (By applying R₂→\(\frac{R_2}{b-a}\) and R₃→\(\frac{R_3}{c-a}\)) = (b-a) (c-a) \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & 1 &b+a \\[0.3em]0 & 0 &c-b\end{vmatrix}\) (By applying R₃→R₃-R₂) = (b-a) (c-a) (c-b) \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & 1 &b+a \\[0.3em]0 & 0 &1\end{vmatrix}\) (By applying R₃→\(\frac{R_3}{c-b}\)) = - (a-b) (c-a) x -(b-c) x 1 (By expanding determinant) = (a-b) (b-c) (c-a).

mple \( 1.9 \)Evaluate \( \left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a-b)(b-c)(c-a) \

Answer»

\(\begin{vmatrix}1& a & a^2 \\[0.3em]1 & b &b^2 \\[0.3em]1 & c & c^2\end{vmatrix}\)

= \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & b-a &b^2-c^2 \\[0.3em]0 & c-a & c^2-a^2\end{vmatrix}\)

(By applying R₂→R₂-R₁ and R₃→R₃-R₁)

= \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & b-a &(b-a)(b+a) \\[0.3em]0 & c-a & (c-a)(c+a)\end{vmatrix}\)

= (b-a) (c-a) \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & 1 &b+a \\[0.3em]0 & 1 &c+a\end{vmatrix}\)

(By applying R₂→\(\frac{R_2}{b-a}\) and R₃→\(\frac{R_3}{c-a}\))

= (b-a) (c-a) \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & 1 &b+a \\[0.3em]0 & 0 &c-b\end{vmatrix}\)

(By applying R₃→R₃-R₂)

= (b-a) (c-a) (c-b) \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & 1 &b+a \\[0.3em]0 & 0 &1\end{vmatrix}\)

(By applying R₃→\(\frac{R_3}{c-b}\))

= - (a-b) (c-a) x -(b-c) x 1

(By expanding determinant)

= (a-b) (b-c) (c-a).