Prove triangle area formula for barycentric coordinates

The area of a triangle whose vertices have cartesian coordinates $(x_i, y_i)$ is $$\frac 1 2 \begin{vmatrix} x_2 - x_1 & y_2 - y_1 \\ x_3 - x_1 & y_3 - y_1 \end{vmatrix} = \frac 1 2 \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}.$$ If the three points $(x_i, y_i)$ have normalized barycentric coordinates $(u_i, v_i, w_i)$, then $$\frac 1 2 \begin{pmatrix} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{pmatrix} \begin{pmatrix} x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1 \end{pmatrix} = \frac 1 2 \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix}.$$ The determinant of a matrix product is the product of the determinants.

Without loss of generality, let $\,ABC\,$ be the three basis vectors in a cartesian coordinate system and let $\,O\,$ be the origin. The triangle $\,\triangle ABC\,$ is the convex hull of $\,\{A,B,C\}\,$ and is the base of a tetrahedron with vertex at $\,O\,$. Any three points $\,\{P_1,P_2,P_3\}\,$ in the plane of $\,\triangle ABC\,$ also form the base of a tetrahedron with vertex at $\,O.\,$ It is well-known that the volume of such a tetrahedon is $\,1/6\,$ the area of the base times the altitude to that base, and also that the volume is $\,1/6\,$ the determinant of the matrix given by the coordinates of the three points. The requested result follows. The key fact needed is that relative length, area, or volume is an affine invariant.