Here is an exercise I have to solve:
Consider the variety $X \subseteq \mathbb{A}_{w,x,y,z}^{4}$ with four defining equations given via the following matrix equations:
$$ \begin{pmatrix} w & x \\ y & z \end{pmatrix} \begin{pmatrix} w & x \\ y & z \end{pmatrix} = \begin{pmatrix} w & x\\ y & z \end{pmatrix} $$
Decompose $X$ as a union of finitely many irreducible varieties.
So far I have been able to come up with the system:
$$ \begin{cases} w^2 - w + xy = 0 \\ x(w+z-1) = 0 \\ y(w+z-1) = 0 \\ z^2 - z + xy = 0 \end{cases} $$
But I don't see where to go from there... Thanks!
In general this type of problem can be solved by a computer algebra system by computing the primary decomposition of the defining ideal. In your case you obtain that $X$ is the union of the varieties $X_1=\{z+w-1=xy+w^2-w=0\}$, $X_2=\{w-1=z-1=x=y=0\}$ and $X_3=\{x=y=z=w=0\}$. Not that $X_2$ and $X_3$ are points and $X_1$ is a surface.
Now, let us work this out by hand. From the second and third equation you get that either $x=y=0$ (call this case 1) or $w+z-1=0$ (call this case 2).
In case 1, the equations one and four become $w^2-w=w(w-1)=0$ and $z^2-z=z(z-1)=0$, respectively. This gives the possibilities $$z=w=0,\ z=w-1=0,\ w=z-1=0,\ z-1=w-1=0.$$ We'll come back to this later.
In case 2, $w+z-1=0$ means that we can substitute one of $z$ or $w$ in equation one or four. First, $w=1-z$ plugged into the first equation gives us $0=xy+z^2-z$. The same way, plugging $z=1-w$ into the fourth equation gives $0=xy+w^2-w$. To sum up, $$w+z-1=0,\ xy+w^2-w=xy+z^2-z=0.$$
Let us gather together all of the different cases. We were obtaining the following equations $$x=y=z=w=0,\\ x=y=z=w-1=0, \\ x=y=w=z-1=0,\\ x=y=z-1=w-1=0,\\ w+z-1=xy+w^2-w=xy+z^2-z=0. $$
We are close to being done, what is left is to perform the same procedure to the last equation. This then gives you even more equations and you then have to check whether there are some inclusions between the varieties defined by this equations. Can you proceed from here on?