I have to do what described next:
Consider the planar triangle $[p_1,p_2,p_3]$ with vertices $ p_1=\left(\begin{array}[c]\,-2\\-1\end{array}\right), p_2=\left(\begin{array}[c]\,3\\-1\end{array}\right), p_1=\left(\begin{array}[c]\,1\\4\end{array}\right), $ compute the barycen-tric coordinates $\lambda_1,\lambda_2,\lambda_3$ of the point $p=\left(\begin{array}[c]\,2\\1\end{array}\right),$ and verify that $p$ can indeed be expressed as the convex combination $\lambda_1p_1+\lambda_2p_2+\lambda_3p_3$.
Repeat this computation for the point $p'=\left(\begin{array}[c]\,3\\3\end{array}\right).$ Finally, transform the triangle and the point p with the linear mapping $\phi:R^2 \to R^2, \phi\left(\begin{array}[c]\,x\\y\end{array}\right) =\left(\begin{array}[c]\,2x-1\\x-3y-2\end{array}\right), $ and compute the barycen-tric coordinates of the transformed point $q=\phi(p)$ with respect to the transformed triangle $[q_1,q_2,q_3]$, where $q_i=\phi(p_i)$, for $i=1,2,3$.
Consider the planar triangle $[p_1,p_2,p_3]$ with vertices $p_1=\begin{pmatrix}-2\\-1\end{pmatrix}$, $p_2=\begin{pmatrix}3\\-1\end{pmatrix}$, $p_3=\begin{pmatrix}1\\4\end{pmatrix}$, compute the barycentric coordinates of the point $p=\begin{pmatrix}2\\1\end{pmatrix}$, and verify that $p$ can indeed be expressed as the convex combination $\lambda_1p_1+\lambda_2p_2+\lambda_3p_3$. Repeat this computation with the point $p'=\begin{pmatrix}3\\3\end{pmatrix}$. Finally transform the triangle and the point $p$ with the linear mapping $\phi \colon \mathbb R^2\to\mathbb R^2$, $\phi\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2x-1\\x-3y+2\end{pmatrix}$, and compute the barycentric coordinates of the transformed point $q=\phi(p)$ with respect to the transformed triangle $[q_1,q_2,q_3]$, where $q_i=\phi(p_i)$, for $i=1,2,3$.
Any ideas on how to do this?