Row echelon form to solve this system I want to use the row echelon form to solve this system:
$$
\begin{cases}
4x +2 -z= 0\\
y^2 -4 =0\\
2z -x =0
\end{cases}.
$$
I am confused by the second equation: if it was $y-4$ instead of $y^2 -4$, the matrix to reduce in row echelon form should have been
$$
\begin{pmatrix}
4 & 0 & -1 &-2\\
0 &1 & 0 & 4\\
-1& 0 & 2& 0
\end{pmatrix}
$$
but how to proceed when there is a quadratic term? Could anyone please help me?
Thank you in advance!
 A: This started out as a comment, but then it got too long.

Okay, so the comments pretty much answer the specific question directly, so allow me to address the general case, for posterity's sake.

What do we do when we see a nonlinear term in a system of linear equations?

Well, it helps to consider that "linear" is a relative term; really we ought to say that an equation is linear with respect to some particular variable. For instance $x^2-2y=-3$ is linear in $x^2$, but not $x$.
With this in mind let's look at an example. Consider the following system of equations:
$$\begin{cases}
5x + 2y^2 = 3\\
2x-4z = 0\\
-3x + y^2 +z = 0
\end{cases}$$
Notice that the variable $y$ only ever appears in the form of $y^2$, so we can treat $y^2$ as its own variable (feel free to replace '$y^2$' with '$u$', if it helps). Doing so, we get the augmented matrix
$$
\begin{bmatrix}
5&2&0&|&3\\
2&0&-4&|&0\\
-3&1&1&|&0
\end{bmatrix}
$$
Which we can reduce to get
$$
\begin{bmatrix}
1&0&0&|&\frac3{10}\\
0&1&0&|&\frac34\\
0&0&1&|&\frac3{20}
\end{bmatrix}
$$
So $x=\frac3{10}$, $y^2=\frac34$, and $z=\frac3{20}$, whence $y=\pm\sqrt\frac34$.
What's really going to bake your noodle later on is: could we still do this with a system of polynomial equations?
A: Since $y$ appears in only one equation, the fact that we have $y^2$ is not a problem. You can just note $Y=y^2$, and then you get the system :
$$\begin{cases} 4x +2 -z= 0\\ Y -4 =0\\ 2z -x =0
\end{cases}$$
