Determinant of $2\times 2$ matrix over $\mathbb{Z}/2\mathbb{Z}$

Question

I have to prove that for any square matrix that is in $M_2(\mathbb{Z}/2\mathbb{Z})$ it is invertible if and only if its determinant is not $0$.

Here are my thoughts: Since all entries are modulo $2$, the determinant can either be $0$, $1$, or $2$. But since $2$ is the same as $0$, the determinant is either $0$ or $1$. Since you cannot divide by $0$ the matrix is not invertible when the determinant is $0$ and you can always divide by $1$.

However, I feel my reasoning is insufficient because my professor gave a hint that it might be useful to consider the cases when $a=c$ and $a\neq c$ separately. He also suggested something about using row-reduction. Any ideas ?

Answer 1

Actually, what is true in any commutative ring is that a matrix is invertible if and only if the determinant is a unit in that ring. This is a consequence of Cramer's rule.

In this case, you have two options.

Either just compute all 2x2-matrices over $\mathbb Z/2$ and check if they are invertible (there are just $2^4=16$ of them, so this is a feasible task).

Or, for a generic matrix (let the coefficients be unknown), use row reduction to find the inverse matrix. You will have to use that the determinant is non-zero at some point.

Answer 2

If you compute the inverse matrix of $A=\begin{pmatrix} a & b \\\ c & d \end{pmatrix}$ over any given field using row reduction, then you will arrive at $$ A^{-1}=\frac{1}{ad - bc} \begin{pmatrix} d & -b \\\ -c & a \end{pmatrix}, $$ provided $ad-bc=\det(A)$ is nonzero. It is easy to see that this condition is necessary and sufficient for the existence of the inverse matrix.

Your reasoning about dividing by zero is perhaps indeed not clear enough.