Find the matrix $S$ of stretch by a factor of $3$. All mappings are from $\mathbb{R}^2$ to $\mathbb{R}^2$.
Find the matrix $S$ of stretch by a factor of $3$ in the $y$-direction and the matrix $S^{-1}$.
So the matrix $S$ is a $2\times2$ matrix.
So if we say that:
$$
S =\pmatrix{a & b  \\ c& d}
$$
Then, would this mean that $S$ stretched by a factor of $3$ in the $y$-direction would then be:
$$
\pmatrix{a &3b  \\
c &3d} ??
$$
And then for the inverse of $S$ you would just have to use the equation for a $2\times 2$ matrix?
I am having trouble with this question and I can't find how to do them anywhere in my notes or textbook. If anyone could help it'd be appreciated.
 A: I fear that you have misunderstood the question: they aren't asking you to stretch a matrix; they are asking you (unless I seriously misunderstand) to find the matrix which corresponds to the transformation "stretch in the $y$-direction by a factor of $3$".
To put it another way: you are told that you want to find the matrix corresponding to the transformation $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ defined by $T(x,y)=(x,3y)$.
This is a nice, linear transformation, and so it does indeed have a matrix representation. We want to find a matrix $A$ such that
$$
A\cdot\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\3y\end{bmatrix}.
$$
There is a general method for doing these sorts of things with linear transformations: check to see how $A$ will modify the standard basis. In this case, we will have 
$$
A\cdot\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}\qquad A\cdot\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}0\\3\end{bmatrix}.
$$
From this, we deduce that the matrix $A$ has first column and second column given by these results. (Why?) So, the desired matrix is
$$
A:=\begin{bmatrix}1 & 0\\0 & 3\end{bmatrix}
$$
