Linear transformations and triangles I was solving my linear algebra homework and the activity says:
"Find a linear transformation $T: \mathbb{R}^2   \rightarrow     \mathbb{R}^2    $ that transforms the triangle with vertices $A(0;0)$, $B(3;0)$ and $C(5;6)$ into a triangle with vertices $P(1;3)$, $Q(0;0)$ and $S(1;0)$."
I don't know how to find the transformation, what should I do? Solving an equation system, or replacing points or what? Because I just know how to do the opposite thing
 A: We know that $T$ will map $A$ to $Q$ since $T$ must leave the origin fixed.  So we have two choices:
$$TB=P;\ TC=S$$ or $$TB=S;\ TC=P$$
Now let the matrix of $T$ in terms of the standard basis be $$\pmatrix{a&b\\c&d}$$  Pick one of the two possibilities for $T$ and write down the equations.  This will give you four equations in the four unknowns $a,b,c,d.$ 
It's best to just start out with the equations for $TB$.  The $y$ value of $0$ makes it trivial to solve for $a$ and $c$, and then you can plug these into the second set of equations. 
EDIT
Another way of looking at it, prompted by the OP's comment.  If we write 
$$\pmatrix{a&b\\c&d}\pmatrix{3\\0}=\pmatrix{3\\1}\\\pmatrix{a&b\\c&d}\pmatrix{5\\6}=\pmatrix{1\\0} $$ that comes to the same thing as writing
$$\pmatrix{a&b\\c&d}\pmatrix{3&5\\0&6}=\pmatrix{3&1\\1&0}$$  Since $$\pmatrix{3&5\\0&6}$$ is invertible, we can solve for $a,b,c,d$ no matter what values the matrix on the right-hand side is.  So it doesn't matter which point you decide to map $B$ to, you'll get a valid answer.  
In this case, there are two correct answers to the question.  You should consider giving them both.     
A: I think about what the tranformation does do the basis vectors.
$B =  3\begin{bmatrix} 1\\ 0 \end{bmatrix}+0\begin{bmatrix} 0\\1\end{bmatrix}.$
$T(B) = 3T\left(\begin{bmatrix} 1\\ 0 \end{bmatrix}\right)= \begin{bmatrix} 1\\ 0 \end{bmatrix}\\
T\left(\begin{bmatrix} 1\\ 0 \end{bmatrix}\right) =\begin{bmatrix} \frac 13\\ 0 \end{bmatrix}$
That gives us what is happening to one basis vector.
$C  = 5\begin{bmatrix} 1\\ 0 \end{bmatrix}+6\begin{bmatrix} 0\\1\end{bmatrix}\\
T(C) = 5 T\left(\begin{bmatrix} 1\\ 0 \end{bmatrix}\right)+6T\left(\begin{bmatrix} 0\\ 1 \end{bmatrix}\right)=\begin{bmatrix} \frac 53 \\0\end {bmatrix} + 6T\left(\begin{bmatrix} 0\\ 1 \end{bmatrix}\right) = \begin{bmatrix} 1 \\3\end {bmatrix}\\
6T\left(\begin{bmatrix} 0\\ 1 \end{bmatrix}\right) = \begin{bmatrix} 1 \\3\end {bmatrix} - \begin{bmatrix} \frac 53 \\ 0\end {bmatrix}\\
T\left(\begin{bmatrix} 0\\ 1 \end{bmatrix}\right) = \begin{bmatrix} -\frac 19 \\ \frac 12\end {bmatrix}$
Now that we know what is happening to the basis vectors, this defines our transformation.
$\begin{bmatrix} \frac 13 &-\frac 19 \\ 0&\frac 12\end {bmatrix}\begin{bmatrix} 3 &5\\0&6 \end {bmatrix} = \begin{bmatrix} 1 &1\\0&3 \end {bmatrix}$
