# Hint for finding all of the units of the ring $M_{2} (\mathbb{Z}_2)$

I want to find all of the units of the ring $$M_{2} (\mathbb{Z}_2)$$ (the set of $$2\times 2$$ matrices with entires from $$\mathbb{Z}_2$$)

I have tried to solve it by taking two matrices and labeling the entries $$a,b,c,d$$ and $$w,x,y,z$$ respectively. Then if you multiply them together you get:

$$\begin{array}{cc} aw+by & ax+bz \\ cw+dy & cx+dz \\ \end{array}$$

this then gives me the following requirements:

$$aw+by\equiv cx+dz \equiv 1$$ and $$zx+bz \equiv cw+dy \equiv 0$$ (mod 2)

From this I can show any matrix of the form

$$\begin{array}{cc} 0 & 0\\ \alpha & \beta \\ \end{array}$$

or

$$\begin{array}{cc} \alpha & \beta \\ 0 & 0 \\ \end{array}$$

cannot be a unit $$\forall \alpha, \beta \in \mathbb{Z}_2$$

Do I just brute force the rest of the possible matrices? Or am I missing something?

I would really just like a hint not the solution.

• Do you know how to solve this in a more typical setting? For example, do you know which matrices in $M_2(\Bbb R)$ are units? Commented Oct 11, 2022 at 18:02
• Also, do you know the formula for the inverse of a $2\times 2$ matrix? Commented Oct 11, 2022 at 18:04
• Alternatively: a matrix is invertible if and only if its rows are linearly independent. There aren't that many nonzero vectors in $(\mathbb{Z}_2)^2$, and since two vectors are linearly independent if and only if neither a scalar multiple of the other, you can easily list all the answers. Commented Oct 11, 2022 at 18:07

A matrix in $$M_2(\mathbb Z_2)$$ is a unit if and only if its determinant is equal to $$1$$. This should give a very simple way to find all units (there is not much to check).
(1) Ignoring whether or not the matrix is invertible, what would the general $$2\times 2$$ matrix $$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ look like, if $$a,b,c,d$$ are REAL numbers?
(3) Same questions as (1) and (2), but this time over $$\mathbb{Z}_2$$, that is, find a general form of what would be the inverse if it were to exist, then find the obstruction (HINT: division by 0 in $$\mathbb{Z}_2$$), then deduce that if this obstruction doesn't hold, the matrix is invertible.