Editing Barycentric coordinate system (section)

==== Vertex approach ====
Another way to solve the conversion from Cartesian to barycentric coordinates is to write the relation in the [[Matrix (mathematics)|matrix]] form <math display="block">
\mathbf{R} \boldsymbol{\lambda} = \mathbf{r}</math>with <math>\mathbf{R} = \left(\, \mathbf{r}_1 \,|\, \mathbf{r}_2 \,|\, \mathbf{r}_3 \right)</math> and <math>\boldsymbol{\lambda} = \left(\lambda_1,\lambda_2,\lambda_3\right)^\top,</math> i.e.<math display="block">
\begin{pmatrix}
x_1 & x_2 & x_3\\
y_1 & y_2 & y_3
\end{pmatrix}
\begin{pmatrix}
\lambda_1 \\ \lambda_2 \\ \lambda_3
\end{pmatrix} =
\begin{pmatrix}x\\y\end{pmatrix}
</math>To get the unique normalized solution we need to add the condition <math>\lambda_1 + \lambda_2 + \lambda_3 = 1</math>. The barycentric coordinates are thus the solution of the [[System of linear equations|linear system]]<math display="block">
\left(\begin{matrix}
1 & 1 & 1 \\
x_1 & x_2 & x_3\\
y_1 & y_2 & y_3
\end{matrix}\right)
\begin{pmatrix}
\lambda_1 \\ \lambda_2 \\ \lambda_3
\end{pmatrix} =
\left(\begin{matrix}
1\\x\\y
\end{matrix}\right)
</math>which is<math display="block">
\begin{pmatrix}
\lambda_1 \\ \lambda_2 \\ \lambda_3
\end{pmatrix} = \frac{1}{2A}
\begin{pmatrix}
x_2y_3-x_3y_2 & y_2-y_3 & x_3-x_2 \\
x_3y_1-x_1y_3 & y_3-y_1 & x_1-x_3 \\
x_1y_2-x_2y_1 & y_1-y_2 & x_2-x_1 
\end{pmatrix}\begin{pmatrix}
1\\x\\y
\end{pmatrix}
</math>where <math display="block">
2A = \det(1|R) = x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)</math>is twice the signed area of the triangle. The area interpretation of the barycentric coordinates can be recovered by applying [[Cramer's rule]] to this linear system.