Symmetric bilinear form over $Mat_{2x2}(\mathbb{R})$ Let $V = Mat_{2x2}(\mathbb{R})$. Define $\varphi(A,B) := det(A+B) - det(A) - det(B)$. 
I have to show that $\varphi(A,B)$ is a symmetric bilinear form. 
It is easy to see that $\varphi$ is symmetric. My problem is to show the bilinearity. 
I know i have to show that: 
$\varphi(\lambda(A)+\mu(C),B) = \lambda \varphi(A,B) + \mu \varphi(C,B)$.
And the same for the second argument. 
But I don't know how to get to the equality. 
 A: We can do the computaion in coordinate. Using the notation below
\begin{equation}
X= 
\left[
\begin{matrix}
x_1 & x_2\\
x_3 & x_4
\end{matrix}
\right]
\end{equation}
we have:
\begin{multline}
\varphi(A,B) = [(a_1+b_1)(a_4+b_4)-(a_2+b_2)(a_3+b_3)] - [a_1a_4 -a_2a_3] - [b_1b_4-b_2b_3] =\\
a_1b_4+b_1a_4-a_2b_3-b_2a_3
\end{multline}
At this step you have only to do the computation in order to find the bilinearity. 
But I prefer another approach. Under the isomorphism of coordinates:
\begin{equation}
f:Mat_{2\times 2}(\mathbb R)\longrightarrow \mathbb R^4, \quad f(X) = (x_1,x_2,x_3,x_4)^T
\end{equation}
you can see that $\varphi$ corresponds to the function:
\begin{gather}
\tilde \varphi:\mathbb R^4 \times \mathbb R^4 \longrightarrow \mathbb R\\
(a,b) \longmapsto a^T M b
\end{gather}
where $a=(a_1,a_2,a_3,a_4)^T$, $b=(b_1,b_2,b_3,b_4)^T$ and 
\begin{equation}
M= 
\left[
\begin{matrix}
0 & 0 & 0 &1\\
0 & 0 & -1 & 0\\
0 & -1 & 0 & 0\\
1 & 0 & 0 &0
\end{matrix}
\right]
\end{equation}
and this is obviously a scalar product.
