If $B$ is an invertible <upper triangular matrix>, $B^T$ is invertible and lower triangular. Part (ii) makes $(B^T)^{-1}$ lower triangular, while part (i) identifies it with $(B^{-1})^T$. Transposing gives that \b[$B^{-1}$ is upper triangular]. This deduction uses both the inverse-transpose identity and the lower-triangular result.

Solution

Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Digital systems, linear algebra, introduction to formal proof / 5 / b / iii / Solution

 Solution (source code)