Showing $\det(aM+bN)-\det(aM)-\det(bN)=ab\left[\det(M+N)-\det(M)-\det(N)\right]$ 
Why is $\det(aM+bN)-\det(aM)-\det(bN)=ab\left[\det(M+N)-\det(M)-\det(N)\right]$

$M,N$ are $2\times 2$ matrices, $a,b\in K$ for some field $K$. If I write $M=(m_{ij})_{i,j}$ for $\ 1\le i,j\le2$ then it is true, but a a long computation. Is there a shorter way to verify it 
 A: Just applying properties of determinants, we have:
$$\begin{vmatrix}aa_{11}+bb_{11} & aa_{12}+bb_{12}\\ aa_{21}+bb_{21}& aa_{22}+bb_{22}\end{vmatrix}- \begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}- \begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}$$
$$=\begin{vmatrix}aa_{11} & aa_{12}+bb_{12}\\ aa_{21}& aa_{22}+bb_{22}\end{vmatrix}+\begin{vmatrix}bb_{11} & aa_{12}+bb_{12}\\ bb_{21}& aa_{22}+bb_{22}\end{vmatrix}- \begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}- \begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}$$
$$=\begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}+\begin{vmatrix}aa_{11} & bb_{12}\\ aa_{21}& bb_{22}\end{vmatrix}+\begin{vmatrix}bb_{11} & aa_{12}\\ bb_{21}& aa_{22}\end{vmatrix}+\begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}- \begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}- \begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}$$
$$=\begin{vmatrix}aa_{11} & bb_{12}\\ aa_{21}& ba_{22}\end{vmatrix}+\begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& ba_{22}\end{vmatrix}$$
$$=ab\left(\begin{vmatrix}a_{11} & b_{12}\\ a_{21}& b_{22}\end{vmatrix}+\begin{vmatrix}b_{11} & a_{12}\\ b_{21}& a_{22}\end{vmatrix}\right)$$
$$=ab\left(\begin{vmatrix}a_{11}+b_{11} & a_{12}+b_{12}\\ a_{21}+b_{21}& a_{22}+b_{22}\end{vmatrix}- \begin{vmatrix}a_{11} & a_{12}\\ a_{21}& a_{22}\end{vmatrix}- \begin{vmatrix}b_{11} & b_{12}\\ b_{21}& b_{22}\end{vmatrix}\right)$$
A: Fix $2 \times 2$ real matrices $M$ and $N$. The function
$$
f(a, b) = \det(aM + bN) - \det(aM) - \det(bN) - ab[\det(M + N) - \det M - \det N]
$$
is clearly a homogeneous quadratic polynomial, hence has the form
$$
f(a, b) = C_{11}a^{2} + 2C_{12} ab + C_{22} b^{2}
$$
for some elements $C_{ij}$ of $K$.
The facts that $f(a, 0) = f(0, b) = 0$ imply $C_{11} = C_{22} = 0$, after which the fact that $f(1, 1) = 0$ implies $C_{12} = 0$.
A: Let $M_1$ and $M_2$ be two columns of $M$, and $N_1$ and $N_2$ be two columns of $N$. Then the LHS evaluates to
$$
[aM_1+bN_1,aM_2+bN_2]-[aM_1,aM_2]-[bN_1,bN_2]=
ab([M_1,N_2]+[M_2,N_1])
$$
by bilinearity.
Similarly, in the RHS, the term $\det(M+N)-\det(M)-\det(N)$ evaluates to
$$
[M_1+N_1,M_2+N_2]-[M_1,M_2]-[N_1,N_2]=
[M_1,N_2]+[M_2,N_1]
$$
as wished.
