Find two $2\times2$ real matrix $A$ and $B$ such that $A$, $B$ , $A+B$ are all invertible with $(A+B)^{-1}=A^{-1}+B^{-1}$ Find two $2\times2$ real matrices $A$ and $B$ such that $A$, $B$ , $A+B$ are all invertible with $(A+B)^{-1}=A^{-1}+B^{-1}$

Tried to write the matrices as
$$A=\pmatrix {a&b\\c&d},B=\pmatrix {e&f\\g&h}$$
and solve $(A+B)^{-1}=A^{-1}+B^{-1}$, But make it too complex. Any more convenient ways?
 A: You get $I = (A^{-1}+B^{-1})(A+B) = I + A^{-1}B + B^{-1}A + I$. Thus we get:
$$(A^{-1}B) + (A^{-1}B)^{-1} = -I$$
$$(A^{-1}B)^2 + I = -(A^{-1}B)$$
So $A^{-1}B$ satisfies the polynomial $x^2 + x + 1 = 0$. Take any matrix satisfying this polynomial; for example you can take 
$$A^{-1}B = \begin{bmatrix} -1 &1 \\ -1&0 \end{bmatrix}$$
$$B = A\begin{bmatrix} -1 &1 \\ -1&0 \end{bmatrix}$$
Hence you can take any invertible $A$ and produce $B$ of the wanted form.
A: From $(A+B)^{-1}=A^{-1}+B^{-1}$ we can get 
$$A^{-1}B+B^{-1}A=-I \quad (*)$$
now let's chose $A=I$ so (*) become 
$$B+B^{-1}=-I$$ If we choice $B$ a diagonale matrix, then 
$$B=\pmatrix {\alpha & 0\\0&\beta};B^{-1}=\pmatrix {\frac{1}{\alpha} & 0\\0&\ \frac{1}{\beta}} $$
Then every thing is done by solving  $$\alpha + \frac{1}{\alpha}=-1$$
A: Using the formula from 
Inverse of the sum of matrices
we can take, for arbitrary $a,b$ with $a+b\neq 0$,
$$
A=\begin{pmatrix} -1 & 1 \cr 0 & 2 \end{pmatrix},\quad 
B=\begin{pmatrix} a & b \cr \frac{2(a^2-a+1)}{a+b} &  \frac{-2(b-ab+1)}{a+b} \end{pmatrix}.
$$
All of $A$, $B$ and $A+B$ have determinant $-2$. It is easy to check that $(A+B)^{-1}=A^{-1}+B^{-1}$. So your way is not too "complex".
