Let K be a finite field with only the two elements 0 and 1, where 1+1=0.

(i) How many 2 x 2 matrices with entries in K are there?

Ok so for this I expect the answer to be the number of different ways one can write a 4 digit combination of the numbers 1 and 0 and so my answer is 16. But it just seems so trivial...does it seem right?

(ii) How many of these are non-singular?

Now for this I know that a matrix is non-singular if it is invertible, which is the case when the determinant of the matrix is not equal to 0. But the thing is, I got quite confused about the part where 1+1=0. How do I factor this in?

Now all the possible matrices are $\displaystyle \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ (det=0), $\displaystyle \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ (det=0), $\displaystyle \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ (det=0), $\displaystyle \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ (det=0), $\displaystyle \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ (det=0), $\displaystyle \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$ (det=0), $\displaystyle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ (det=-1), $\displaystyle \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}$ (det=0), $\displaystyle \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ (det=1), $\displaystyle \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}$ (det=0), $\displaystyle \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$ (det=0), $\displaystyle \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ (det=-1), $\displaystyle \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$ (det=-1), $\displaystyle \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ (det=1), $\displaystyle \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ (det=1), $\displaystyle \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ (det=1-1).

now which ones are non-singular, considering that 1+1=0?