REVISITED$^1$: How is the determinant of a matrix $A\in M_{2\times 2}(\mathbb{R})$ considered a bilinear form? I'm trying to prove that $B(X,Y)=\det (X+Y) - \det (X) - \det (Y)$ is a blinear form on the vector space $A$ is from, and also trying to determine if it is an inner product space. I think if I know how $\det$ is a bilinear form, then this will become easier.

EDIT$^1$:
I was thinking, if I let $\det : \mathbb{R}^2\times\mathbb{R}^2\rightarrow\mathbb{R}$ be a bilinear form defined by
$$ \det\pmatrix{\pmatrix{a_{11}\\a_{21}}, \pmatrix{a_{12}\\a_{22}}} =a_{11}a_{22}-a_{21}a_{12},$$
then if I can show that $\det$ is a bilinear form, then this will suffice to show that $B$ is a bilinear form as the sum of bilinear forms is yet again a bilinear form, as is remarked in Friedberg's Linear Algebra (page 423).
 A: I think what your textbook means is something like this:

If $B(X,Y)$ is a finite sum of bilinear forms in $X$ and $Y$, then $B$ is also a bilinear form in $X$ and $Y$. 

If this is the case, there is no hope to apply this statement to your question directly, because neither $\det(X+Y),\,\det(X)$ nor $\det(Y)$ is a bilinear form in $X$ and $Y$.

Adding to Will Jagy's comment, if you can show that some $q(x)$ is a quadratic form in $x$, then $b(x,y):=\frac{q(x+y)-q(x)-q(y)}2$ is a (symmetric) bilinear form in $x$ and $y$ with $q(x)\equiv b(x,x)$. In your case, if you can show that $q(X):=2\det(X)$ is a quadratic form in $X$, then you are done. $q(X)$ is obviously a polynomial in the entries of $X$. Can you show that $q(kX)=k^2q(X)$ for all scalar $k$?

Alternatively, let $X=\pmatrix{a&b\\ c&d}$. Can you find a symmetric matrix $M$ such that $2\det(X) = (a,b,c,d)M(a,b,c,d)^T$? If you can, then for $Y=\pmatrix{x&y\\ z&w}$, we can express the (symmetric) bilinear form $B(X,Y)$ as $(a,b,c,d)M(x,y,z,w)^T$.

You may also work from the first principle and evaluate $B(X,Y)$ in terms of the entries of $X$ and $Y$:
\begin{align*}
&\det\pmatrix{a+x&b+y\\ c+z&d+w} - \det\pmatrix{a&b\\ c&d} - \det\pmatrix{x&y\\ z&w}\\
=&(a+x)(d+w) - (b+y)(c+z) - (ad-bc) - (xw-yz).\tag{1}
\end{align*}
Can you simplify $(1)$ and show that it is a bilinear form?

Finally, if $B(X,X)$ defines an inner product, $B(X,X)$ should be positive for all $X\neq0$. Is that true?
