Using the equation $ A^{-1} + B^{-1} = (A+B)^{-1} $ to find relations between det(A) and det(B) We are given a condition, 
$$
A^{-1} + B^{-1} =  (A+B)^{-1}
$$
Further, $|A| =4$ and we are asked to find the value of |B|. 
I tried to simplify the LHS 
$$
A^{-1} + B^{-1}= B^{-1}BA^{-1} + B^{-1}AA^{-1}
$$
$$
A^{-1} + B^{-1}= B^{-1}(A+B)A^{-1}
$$
That only lead me to get
$$
|A+B|^2=|A||B|
$$
Am stuck after this. Any help would be appreciated! Thanks. 
 A: If $A,B$ have sizes $n$ with $n > 2$, then it is impossible to determine $|B|$.  
If $A,B$ have real entries and $n$ is odd, then it is impossible for $A,B$ to satisfy the given condition.
If $n = 2$ and $A,B$ have real entries, then $|B| = |A|$ must hold.

We can write
$$
A^{-1} + B^{-1} = (A + B)^{-1} \iff\\
A(A^{-1} + B^{-1}) = A(A + B)^{-1} \iff\\
AA^{-1} + AB^{-1} = ((A + B)A^{-1})^{-1} \iff\\
I + AB^{-1} = (I + BA^{-1})^{-1}.
$$
Writing $M = BA^{-1}$ (or equivalently $B = MA$) lets us rewrite this as
$$
I + M^{-1} = (I + M)^{-1} \iff\\
(I + M)(I + M^{-1}) = I \iff\\
M + 2I + M^{-1} = I \iff\\
M + I + M^{-1} = 0 \iff\\
M^2 + M + I = 0.
$$
In other words: invertible matrices $A,B$ will satisfy the given equation if and only if $B = MA$ for some (invertible) matrix $M$ with $M^2 + M + I = 0$. $M$ will satisfy this equation if and only if it is diagonalizable and its eigenvalues satisfy $\lambda^2 + \lambda + 1 = 0$. In other words, $M$ will satisfy the equation iff its eigenvalues are all equal to either of the two possibilities $\lambda = -\frac 12 \pm i\frac{\sqrt{3}}2$.  
For a $2 \times 2$ real matrix, this implies that $|M| = 1$, so that $|B| = |MA| = |A|\cdot |M| = |A|$.  
A: From $(A+B)(A^{-1}+B^{-1})=I$ you get
$$
I+AB^{-1}+BA^{-1}+I=I \tag{1}
$$
and therefore
$$
I+AB^{-1}+BA^{-1}=0 \tag{2}
$$
Multiply $(2)$ on the right by $A$:
$$
A+AB^{-1}A+B=0 \tag{3}
$$
Multiply $(2)$ on the right by $B$:
$$
B+A+BA^{-1}B=0 \tag{4}
$$
Therefore, comparing $(3)$ and $(4)$,
$$
AB^{-1}A=BA^{-1}B\tag{5}
$$
If $a=\det A$ and $b=\det B$, we get from $(5)$, using Binet's theorem, 
$$
a^2b^{-1}=a^{-1}b^2 \tag{6}
$$
Therefore $b^3=a^3$ and, assuming matrices with real coefficients, $b=a$.
Notes. 


*

*This answer doesn't touch the problem of existence of the given matrices. Other answers give conditions about this.

*If the matrices are allowed to have complex entries, the only possible conclusion is that $b/a$ is a cube root of unity.
A: Hi and welcome to MSE.
Your condition is not sufficient to determine an exact value of $\det(B)$. For, take the vector space $\mathbb C$. A linear application $T_c:\mathbb C \rightarrow \mathbb C$ is just a multiplication for an element $c$ of $\mathbb C$ and the determinant is exactly $c$. So the matrix that represent $T_c$ is just the $1\times 1$ matrix $(c)$.
Take now the linear application $A=T_4$ (that is an application with determinant equal to $4$). Let $B=T_b$; thanks to your condition we have:
\begin{gather}
A^{-1}+B^{-1} = (A+B)^{-1}\\
T_4^{-1}+T_b^{-1} = (T_4+T_b)^{-1}\\
\end{gather}
Using now the fact that $T_c^{-1}=T_{\frac{1}{c}}$ and $T_c + T_d = T_{c+d}$ we obtain the following equation:
\begin{equation}
\frac{1}{4} + \frac{1}{b} = \frac{1}{4+b}
\end{equation}
that has two solutions:
\begin{gather}
b_1 = -2 + 2i\sqrt 3\\
b_2 = -2 - 2i\sqrt 3
\end{gather}
Then, you have two linear application $B_1=T_{b_1}$ and $B_2 = T_{b_2}$ that satisfie the initial condition but they have different determinant $\det(B_1)=b_1\neq b_2=\det (B_2)$. So the initial conditions are not sufficient to determine an exact value of $\det(B)$.
