Prove $\det(P+Q+R)=\det(P+Q)+\det(Q+R)+\det(R+P)-\det(P)-\det(Q)-\det(R)$ 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
 A: 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
$$\begin{align}\det(P+Q+R) 
= & \det(P+Q) + \det(R) + \langle P+Q, R \rangle\\
= & \det(P+Q) + \det(R) + \color{red}{\langle P, R \rangle} + \color{blue}{\langle Q, R \rangle}\\
= & \det(P+Q) + \det(R) 
   +\left(\color{red}{\det(P+R) - \det(P) - \det(R)}\right)\\
  & \quad + \left(\color{blue}{\det(Q+R) - \det(Q) - \det(R)}\right)\\
= & \det(P+Q) + \det(P+R) + \det(Q+R) - \det(P) - \det(Q) - \det(R)
\end{align}
$$
A: HINT:
$\det S$ is a homogenous polynomial of degree $2$ in the entries of $S$, that is, a quadratic form. It is easier to prove the general fact: 
If $q\colon V \to W$ is a quadratic form then 
$$q(P+Q+R) - q(P+Q)- q(P+R) -q(Q+R) + q(P) + q(Q) + q(R)=0$$
The thing about quadratic forms, they come from bilinear forms ( symmetric too, if $char \ne 2$, but that is not important for this). So 
$$q(S) = b(S,S)$$ Substitute in the above, use bilinearity,  and observe that all of the terms cancel out. 
$\bf{Added:}$ One should try to guess what is the equality that holds for cubic forms, or, more generally, for forms of degree $n$. It is also related to "polarization". 
A: If $\det(P) = p_{11}p_{22}-p_{21}p_{12}$, thrn
$$\det(P+Q+R) = (p_{11}+q_{11}+r_{11})(p_{22}+q_{22}+r_{22})-(p_{21}+q_{21}+r_{21})(p_{12}+q_{12}+r_{12})$$
You can start in that way. 
A: Note that $(a_1 + a_2 + a_3) (d_1+d_2 + d_3)$ can be expanded as
$$(a_1 + a_2 + a_3) (d_1+d_2 + d_3) \\= (a_1 + a_2 )(d_1+d_2) + (a_2 + a_3)(d_2 + d_3) + (a_1 + a_3)(d_3+ d_1) - a_1 d_1 - a_2 d_2 -a_3 d_3.\tag 1$$
Similarly,
$$(b_1 + b_2 + b_3) (c_1+c_2 + c_3) \\= (b_1 + b_2 )(c_1+c_2) + (b_2 + b_3)(c_2 + c_3) + (b_1 + b_3)(c_3+ c_1) - b_1 c_1 - b_2 c_2 -b_3 c_3.\tag 2$$
Therefore, if $$P = \begin{bmatrix}a_1 & b_1 \\ c_1 & d_1\end{bmatrix},$$
$$Q = \begin{bmatrix}a_2 & b_2 \\ c_2 & d_2\end{bmatrix},$$
and
$$R = \begin{bmatrix}a_3 & b_3 \\ c_3 & d_3\end{bmatrix},$$
then using $(1)$ and $(2)$, we can show that
$$\det(P+Q+R) = (a_1 + a_2 + a_3) (d_1+d_2 + d_3)-(b_1 + b_2 + b_3) (c_1+c_2 + c_3)\\=\det(P+Q) + \det(Q+R) + \det(R+P) - \det(P) -\det(Q) -\det(R).$$
A: Regard the determinate as a mapping $\det: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$ which is bilinear and a matrix $A\in \mathbb{R}^{2\times 2}$ as two vectors
\begin{align}
\begin{pmatrix}
a_{1,1} & a_{1,2} \\
a_{2,1} & a_{2,2}
\end{pmatrix}
\cong
\begin{pmatrix}
a_{1,1} \\
a_{2,1} 
\end{pmatrix},
\begin{pmatrix}
 a_{1,2} \\
 a_{2,2}
\end{pmatrix}
=
a_1
,a_2
\end{align}
where $a_1,a_2$ are vectors in $\mathbb{R}^2$.
So we have $P \cong p_1,p_2$, $Q\cong q_1,q_2$ and $R \cong r_1, r_2$ and we start with
\begin{align}
\det(P+Q+R) = \det(p_1+q_1+r_1,p_2+q_2+r_2)
\end{align}
note that one property of bilinearity is $\det (a_1+b_1,a_2) = \det (a_1,a_2)+\det (b_1,a_2)$ and analogue in the second argument. Using this holds
\begin{align}
\det(p_1+q_1+r_1,p_2+q_2+r_2) &= \det(p_1+q_1,p_2+q_2+r_2) + \det(r_1,p_2+q_2+r_2)\\
&= \underbrace{\det(p_1+q_1,p_2+q_2)}_{\det(P+Q)} + \det(p_1+q_1,r_2) + \det(r_1,p_2+q_2+r_2)
\end{align}
Let us regard the remaining parts separately


*

*\begin{align}\begin{split}
\det(p_1+q_1,r_2) &= \det(q_1+r_1,r_2) + \det(p_1-r_1,r_2)\\
&= \underbrace{\det(q_1+r_1,q_2+r_2)}_{\det(Q+R)} + \det(q_1+r_1,-q_2) + \det(p_1-r_1,r_2)
\end{split}\tag{1}
\end{align}

*\begin{align}\begin{split}
\det(r_1,p_2+q_2+r_2) &= \det(r_1,p_2+r_2) + \det(r_1,q_2) \\
&= \underbrace{\det(p_1+r_1,p_2+r_2)}_{\det(P+R)} + \det(-p_1,p_2+r_2) + \det(r_1,q_2)
\end{split}\tag{2}
\end{align}
Note that from the bilinearity we have that $\det(a_1,-a_2) =  -\det(a_1,a_2)$
adding the remainders of $(1)$ and $(2)$ together yields
\begin{align}
\color{red}{\det(q_1+r_1,-q_2)} + \color{blue}{\det(p_1-r_1,r_2)}+\color{blue}{\det(-p_1,p_2+r_2)} + \color{red}{\det(r_1,q_2)}
\\= \color{red}{-\det(q_1,q_2)} + (\color{blue}{-\det(p_1,p_2) - \det(r_1,r_2)})
= -\det(P) - \det(Q)-\det(R)
\end{align}
Altogether this proves the desired identity.
