I am working in the vector space $\mathbb F^{2\times 2}$, which denotes the space of $2\times 2$ matrices over the field $\mathbb F$ (the characteristic of $\mathbb F$ is not equal to $2$) and I would like to proof that: $\det:A\mapsto\det A$ is a quadratic form. Furthermore, I would like to show that $$\det(A+B) - \det(A) - \det(B) = tr(AB^\#)$$ whereas $tr$ denotes the trace of the matrix and $B^\#$ denotes the cofactor matrix of $B$.
My definition of a quadratic form $q$ is that $q(cx) = c^2q(x)$ and that the map $$(x,y) \mapsto q(x+y)-q(x)-q(y)$$ is a bilinear form. I could easily prove the first criterion, but I am really stuck with the second. I also tried to first show the equality above (the trace equality), but for some reason, for matrices $$A= \begin{pmatrix} a & b\\c & d \end{pmatrix} \qquad B = \begin{pmatrix} w & x\\y & z \end{pmatrix}$$
the left side gives me $az+dw-by-cx$ and the right side leaves me with $az+dw-bx-cy$
I am pretty sure that I computed the determinants and the cofactor matrix correctly, so I am genuinely confused where my mistake lies and I also do not know how to prove the bilinear form criterion. Any help is greatly appreciated!
Your computation of the left hand side, $az+dw-by-cx$ is correct. Note that this sum of products can be interpreted as trace of $$\tag1\begin{pmatrix}az-by&?\\?&dw-cx\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}z&-x\\-y&w\end{pmatrix}$$ So maybe you check again the definition of $B^\#$. At any rate, the second factor in $(1)$ is clearly linear in $B$, hence $(1)$ is linear in $A$ for fixed $B$ and linear in $B$ for fixed $A$ - in other words, it is bilinear, as desired.