Question:
Let $A=\{1,2,3,4,5, \ldots, 10\}$ and $f: A \rightarrow A$ be an invertible function. Then, $\sum_{r=1}^{10}\left(f^{-1} o f\right)(r)=$ ___________.
Solution:
Given: f : A → A is an invertible function, where A = {1, 2, 3, 4, 5, ..., 10}
Since, $f$ is invertible
Therefore, $f^{-1} o f(x)=x \quad \ldots(1)$
Now,
$\sum_{r=1}^{10}\left(f^{-1} o f\right)(r)=f^{-1} o f(1)+f^{-1} o f(2)+f^{-1} o f(3)+\ldots .+f^{-1} o f(10)$ $\left(\because 1+2+3+\ldots+n=\frac{n(n+1)}{2}\right)$
$=1+2+3+\ldots+10 \quad($ From $(1))$
$=\frac{10(10+1)}{2}$
$=5(11)$
$=55$
Hence, $\sum_{r=1}^{10}\left(f^{-1} o f\right)(r)=\underline{55}$.