Why is there another transformation matrix for the bases of the image and of the preimage of this mapping? There is the following Matrix:
\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix}
At a point it is needed to calculate the determinant of the matrix. In the official solution it is written:
$det\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix} = (c-b)(c-a)(b-a)$
And I don't see how they get this. If I calculate the determinant I am always getting this:
$det\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix} = (bc^2-b^2c)-(ac^2-a^2c)+(ab^2-a^2b)=c(b(c-b)-a(c-a)+ab(b-a)).$
But after that point I don't know how to proceed and get the form above. Can you help me?
 A: Substracting the first column from the second and third, you get
$\begin{vmatrix}
1 &  1&1 \\ 
 a&b  &c \\ 
a^{2} & b^{2} &c^{2} 
\end{vmatrix}=\begin{vmatrix}
1 &  0&0 \\ 
 a&b-a  &c-a \\ 
a^{2} & b^{2}-a^{2} &c^{2} -a^{2}
\end{vmatrix}=$
$=(b-a)\cdot (c-a)\begin{vmatrix}
1 &  0&0 \\ 
 a&1  &1 \\ 
a^{2} & b+a &c+a
\end{vmatrix}=(b-a)\cdot (c-a)\begin{vmatrix}
1 &1 \\ 
b+a &c+a
\end{vmatrix}=$
$=(b-a)\cdot (c-a)\cdot (c+a-b-a)=(b-a)\cdot (c-a)\cdot(c-b).$
A: Subtracting the first colum from the second and third columns we get
$$
\det\begin{pmatrix}1&1&1\\ a& b& c\\a^2& b^2& c^2\end{pmatrix}=\det\begin{pmatrix}1&0&0\\a&b-a&c-a\\ a^2&b^2-a^2&c^2-a^2\end{pmatrix}
$$
It follows that
\begin{eqnarray}
\det\begin{pmatrix}1&1&1\\ a& b& c\\a^2& b^2& c^2\end{pmatrix}&=&\det\begin{pmatrix}1&0&0\\a&b-a&c-a\\ a^2&b^2-a^2&c^2-a^2\end{pmatrix}=\det\begin{pmatrix}b-a&c-a\\ b^2-a^2&c^2-a^2\end{pmatrix}\\
&=&\det\begin{pmatrix}b-a&c-a\\ (b-a)(b+a)&(c-a)(c+a)\end{pmatrix}\\
&=&(b-a)(c-a)\det\begin{pmatrix}1&1\\ b+a&c+a\end{pmatrix}=(b-a)(c-a)(c+a-b-a)\\
&=&(b-a)(c-a)(c-b)
\end{eqnarray}
