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 

Just applying properties of determinants, we have:
$$\begin{vmatrix}aa_{11}+bb_{11} & aa_{12}+bb_{12}\\ aa_{21}+bb_{21}& aa_{22}+bb_{22}\end{vmatrix}- \begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}- \begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}$$
$$=\begin{vmatrix}aa_{11} & aa_{12}+bb_{12}\\ aa_{21}& aa_{22}+bb_{22}\end{vmatrix}+\begin{vmatrix}bb_{11} & aa_{12}+bb_{12}\\ bb_{21}& aa_{22}+bb_{22}\end{vmatrix}- \begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}- \begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}$$
$$=\begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}+\begin{vmatrix}aa_{11} & bb_{12}\\ aa_{21}& bb_{22}\end{vmatrix}+\begin{vmatrix}bb_{11} & aa_{12}\\ bb_{21}& aa_{22}\end{vmatrix}+\begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}- \begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}- \begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}$$
$$=\begin{vmatrix}aa_{11} & bb_{12}\\ aa_{21}& ba_{22}\end{vmatrix}+\begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& ba_{22}\end{vmatrix}$$
$$=ab\left(\begin{vmatrix}a_{11} & b_{12}\\ a_{21}& b_{22}\end{vmatrix}+\begin{vmatrix}b_{11} & a_{12}\\ b_{21}& a_{22}\end{vmatrix}\right)$$
$$=ab\left(\begin{vmatrix}a_{11}+b_{11} & a_{12}+b_{12}\\ a_{21}+b_{21}& a_{22}+b_{22}\end{vmatrix}- \begin{vmatrix}a_{11} & a_{12}\\ a_{21}& a_{22}\end{vmatrix}- \begin{vmatrix}b_{11} & b_{12}\\ b_{21}& b_{22}\end{vmatrix}\right)$$
