IMC 2015 · Day 1 · P1

Multiplying the equation by $(A + B)$ we get $$\begin{gather*} I = (A + B)(A + B)^{-1} = (A + B)(A^{-1} + B^{-1}) = \\ = A A^{-1} + A B^{-1} + B A^{-1} + B B^{-1} = I + A B^{-1} + B A^{-1} + I \\ A B^{-1} + B A^{-1} + I = 0. \end{gather*}$$ Let $X = A B^{-1}$; then $A = X B$ and $B A^{-1} = X^{-1}$, so we have $X + X^{-1} + I = 0$; multiplying by $(X - I) X$, $$0 = (X - I) X \cdot (X + X^{-1} + I) = (X - I) \cdot (X^2 + X + I) = X^3 - I.$$ Hence, $$\begin{gather*} X^3 = I \\ (\det X)^3 = \det(X^3) = \det I = 1 \\ \det X = 1 \\ \det A = \det(XB) = \det X \cdot \det B = \det B. \end{gather*}$$

In case of complex matrices the statement is false. Let $\omega = \frac{1}{2}(-1 + i\sqrt{3})$. Obviously $\omega \notin \mathbb{R}$ and $\omega^3 = 1$, so $0 = 1 + \omega + \omega^2 = 1 + \omega + \bar{\omega}$.

Let $A = I$ and let $B$ be a diagonal matrix with all entries along the diagonal equal to either $\omega$ or $\bar{\omega} = \omega^2$ such a way that $\det(B) \ne 1$ (if $n$ is not divisible by 3 then one may set $B = \omega I$). Then $A^{-1} = I$, $B^{-1} = \bar{B}$. Obviously $I + B + \bar{B} = 0$ and $$(A + B)^{-1} = (-\bar{B})^{-1} = -B = I + \bar{B} = A^{-1} + B^{-1}.$$ By the choice of $A$ and $B$, $\det A = 1 \ne \det B$.