I'm brushing up on some high school maths and I'm currently revisiting determinants, specifically the factorisation of determinants. I'm working my way through a problem set and I keep getting stuck at the same step on every question.
The example question is as follows. Factorise:
$A =\begin{bmatrix}a & b & c\\a^2 & b^2 & c^2\\a^3 & b^3 & c^3\end{bmatrix}$
There are multiple steps from this point, but we arrive at:
$A = abc(b-a)(c-a)\begin{bmatrix}1 & 0 & 0\\a & 1 & 1\\a^2 & b + a & c + a\end{bmatrix}$
$Det(A) = abc(b-a)(c-a)[(c+a)-(b+a)]$
$Det(A) = abc(b-a)(c-a)(c-b)$
I understand everything up until this point, however the particular step where I get stuck is this last step. I don't understand how the book arrives at this equation from the previous equation:
$Det(A) = abc(a-b)(b-c)(c-a)$
Is there a rule that I am unaware of, that allows for the values within linear factors to be moved around? Or have they simply been expanded out and refactored?
Any help would be very much appreciated.
EDIT:
Thank you for your answer Paul Sinclair, if I had enough rep I would up-vote and mark as answered. I never would have though to have multiplied through twice by negative one, this has made things a whole lot clearer!
I have also modified my question to remove the ambiguity with regards to determinants and matrices.
Determinants are numbers, not matrices. Given a square matrix, $A$, the determinant of $A$ is simply a real number derived from a certain formula on the matrix elements.
Multiplication of real numbers is commutative ($xy = yx)$ and associative ($x(yz) = (xy)z$), so it doesn't matter in which order they are multiplied. Also $(b - a) = -1(a - b)$, and $(c - b) = -1(b - c)$. So, $$\begin{align}\det(A) &= abc(b - a)(c - a)(c - b)\\ &=abc(-1)(a-b)(c-a)(-1)(b-c)\\ &=abc(a-b)(b-c)(c-a)(-1)(-1)\\ &=abc(a-b)(b-c)(c-a)1\\ &=abc(a-b)(b-c)(c-a)\end{align}$$