I'm trying to learn about barycentric coordinates so after a Google search found this PDF file which didn't look too scary. I'm only on page 3 and getting confused so hope I can get help here...
A triangle has vertices $p_1$, $p_2$, $p_3$ and a barycentric combination of the three points takes the form $$p = up_1 + vp_2 + wp_3$$ where $$u + v + w = 1$$ It says p can be rewritten $$p=up_1 + vp_2 + (1 - u - v)p_3$$
Then it asks "How can we find the barycentric coordinates of a given point p?" The claim is that there are 3 equations and the following linear system is setup:
$$\begin{bmatrix}p_1&p_2&p_3\\1&1&1\end{bmatrix}\begin{bmatrix}u\\v\\w\end{bmatrix} = \begin{bmatrix}p\\1\end{bmatrix}$$ for the unknown $u,v,w$. The system is then solved using Cramer's rule:
$$A = \begin{vmatrix}p_1&p_2&p_3\\1&1&1\end{vmatrix}$$ $$A_1 = \begin{vmatrix}p&p_2&p_3\\1&1&1\end{vmatrix}$$ $$A_2 = \begin{vmatrix}p_1&p&p_3\\1&1&1\end{vmatrix}$$ $$A_3 = \begin{vmatrix}p_1&p_2&p\\1&1&1\end{vmatrix}$$
My 2 points of confusion: 1) Is there 3 equations as claimed? I can see $p = \dots$ and $u + v + w = 1$ which is 2 equations. 2) Cramer's rule uses determinants, i.e. a value linked to a square matrix. The coefficient matrix here is 2-by-3, not square. Can someone help me see what the author is doing here please?