Solving $Ax = b$ using least squares (minimize $||Ax -b||_2$) $$A = \begin{bmatrix}
    1       & 1 \\
    -1       & 1\\
    1   & 2
\end{bmatrix}, b = \begin{bmatrix}
    1       \\
    0  \\
    1   
\end{bmatrix}$$
My teacher said about solving this in lots of ways. 
I know that finding min $||Ax -b||_2$ is the same as doing a projection of something onto something, but I'm not quite sure of what onto what.
What are the possible ways I can solve this?
 A: You can use the normal equations, i.e. the least squares solution $x_0$ is found by solving (for $x_0$) $$A^TAx_0 = A^T b. \quad(\star)$$
So calculate the matrix $A^TA$ and the vector $A^T b$, and you are left with solving the system of linear equations $(\star)$, which you can do by standard methods, like Gaussian elimination.
A proof that $x_0$ is a least squares solution if and only if it satisfies $A^T Ax_0 = A^T b$ can be found here.

$\newcommand{\CS}{\mathcal{C}}\newcommand{\R}{\mathbb{R}}$If you don't want to use the normal equations, you can use the method in my comment below. (Using the normal equations should be easier though.) Here is an explanation of that method. (I.e. why for the solution, $Ax$ should be the projection of $b$ onto the column space of $A$.)
Let $\CS(A)$ denote the column space of $A$. As $x$ varies, $Ax$ will take on all values in $\CS(A)$, effectively by definition of column space. But remember $\left\| Ax-b\right\|_2$ is the (Euclidean) distance between $Ax$ and $b$. So to minimise $\left\| Ax-b\right\|_2$ over $x$, you want the $x$ such that $Ax$ is closest to $b$ (in Euclidean distance). In other words, for the solution, $Ax$ should be the element in $\CS(A)$ that is closest to $b$ (since $Ax\in\CS(A)$ always). 
Now, there is a theorem that if $W$ is a subspace of $\R^n$, the closest vector (using Euclidean distance) in $W$ to a given vector $b\in\R^n$ is the vector $\mathrm{proj}_W(b)$ (i.e. the projection of $b$ onto $W$). So for your problem, the closest vector in $\CS(A)$ to $b$ is the projection of $b$ onto $\CS(A)$, say $\widehat{b}$. Hence for the solution $x$, we must have $Ax=\widehat{b}$, since we said that for the solution, $Ax$ must be the closest vector to $b$ in $\CS(A)$.
