Why is there another transformation matrix for the bases of the image and of the preimage of this mapping?

Question

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?

Answer 1

Substracting the first column from the second and third, you get

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

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

$=(b-a)\cdot (c-a)\cdot (c+a-b-a)=(b-a)\cdot (c-a)\cdot(c-b).$

Answer 2

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}