- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.8 Exercises
In Exercises 22 through 25, $S$ and $T$ denote functions with domain $V$ and values in $V.$ In general, $ST \neq TS.$ If $ST = TS,$ we say that $S$ and $T$ commute.
23.$\quad$ If $S$ and $T$ are invertible, prove that $ST$ is also invertible and that $(ST)^{-1} = T^{-1} S^{-1}.$ In other words, the inverse of $ST$ is the composition of inverses, taken in reverse order.
Solution. $\quad$ Let $S^{-1}$ and $T^{-1}$ denote the inverses of $S$ and $T,$ respectively. Applying $S^{-1}$ to $ST$ and using the associative property of composition, we get \begin{align*} S^{-1}(ST) &= (S^{-1}S)T \\ &= IT \\ &= T \end{align*} Then, applying $T^{-1},$ we get \begin{align*} T^{-1}S^{-1}(ST) &= T^{-1}T \\ &= I \end{align*} Thus, we have shown that $ST$ is invertible with $(ST)^{-1} = T^{-1}S^{-1}. \quad \blacksquare$