Solution set of this linear system in reduced row echelon form. 
So we have this matrix:
$\begin{bmatrix}
1 & 0 & 2 & 1\\
 0 & 1 & -1 & 2
\end{bmatrix}$
Does this matrix have infinitely many solution sets? How do I know? Is it because there are more variables than equations?
Here's my solution. Is this right?
let $ z = t$
so $y = 2 + t$
$x = 1 - 2t$
Is this right?
 A: We want to find all solutions to $x+2z=1$ and $y-z=2$. We can re-write this as $x=1-2z$ and $y=2+z$. Now it is clear that we can pick and number for $z$ and it will give us unique numbers for $x$ and $y$. Thus there are infinitely many solutions of the form
$$\begin{bmatrix}x\\y\\z \end{bmatrix}=\begin{bmatrix}1-2t\\2+t\\t \end{bmatrix} $$
for $t\in\Bbb R$. So yes, your answer is correct.
In general an $m\times n$ matrix is a map $\Bbb R^n\to \Bbb R^m$. The rank tells you the dimension of the image, and the number of columns minus the rank tells you the dimension of the kernel. The dimension of the kernel is exactly the dimension of the number of solutions to a particular equation $Mx=b$, so in this case the solutions are $1$-dimensional, so there are infinitely many.
A: Let's solve the augmented matrix
$$
A=
\begin{bmatrix}
1 & 0 & 2 & 1\\
0 & 1 & -1 & 2
\end{bmatrix}$$
i.e. 
$$\begin{bmatrix}
1 & 0 & 2 & 1\\
0 & 1 & -1 & 2
\end{bmatrix} 
\begin{bmatrix}
x \\
y \\
z \\
-1
\end{bmatrix} =
0$$
That is 
$$\begin{cases}
x + 2z -1 &= 0 \\
y -z -2 &= 0
\end{cases}$$
Since this is a system of two equations in three unknowns, we have to parametrize in terms of one unknown, let's say $z = t$, then we get from the first equation
$$x = 1 -2t$$
and from the second equation
$$y = t+2$$
So the solution set is 
$$\lbrace 
t \in \mathbb{R}: \
\begin{bmatrix}
1 -2t \\
t+2\\
t
\end{bmatrix}
\rbrace$$
Hence, your answer is correct.
