TITLE: Prove $\det(P+Q+R)=\det(P+Q)+\det(Q+R)+\det(R+P)-\det(P)-\det(Q)-\det(R)$
QUESTION [6 upvotes]: If $P,Q,R$ be three $2\times2$ matrices, then prove that 
$$\det(P+Q+R)=\det(P+Q)+\det(Q+R)+\det(R+P)-\det(P)-\det(Q)-\det(R).$$
My Attempt: I have no idea from where to start

REPLY [8 votes]: Notice for any $A \in M_2(\mathbb{C})$, the characteristic polynomial has the form
$$\chi_A(\lambda) = \det(\lambda I_2 - A)
= \lambda^2 - {\rm tr}(A)\lambda + \det(A)$$
Using Cayley Hamilton and taking trace again, we can express the determinant
in terms of the trace:
$$A^2 - {\rm tr}(A) A + \det(A) I_2 = 0
\quad\implies\quad \det(A) = \frac12\left({\rm tr}(A)^2 - {\rm tr}(A^2)\right)
$$
For any $A, B \in M_2(\mathbb{C})$, define a bracket between them by
$$\langle A, B \rangle \stackrel{def}{=} \det(A+B) - \det(A) - \det(B)$$
Using above expression of determinant, it is easy to verify:
$$\langle A, B \rangle = {\rm tr}(A){\rm tr}(B) - {\rm tr}(AB)$$
This means this sort of bracket is bilinear in terms of its arguments.
As a result, we obtain
$$\begin{align}\det(P+Q+R) 
= & \det(P+Q) + \det(R) + \langle P+Q, R \rangle\\
= & \det(P+Q) + \det(R) + \color{red}{\langle P, R \rangle} + \color{blue}{\langle Q, R \rangle}\\
= & \det(P+Q) + \det(R) 
   +\left(\color{red}{\det(P+R) - \det(P) - \det(R)}\right)\\
  & \quad + \left(\color{blue}{\det(Q+R) - \det(Q) - \det(R)}\right)\\
= & \det(P+Q) + \det(P+R) + \det(Q+R) - \det(P) - \det(Q) - \det(R)
\end{align}
$$