 Solution

ID: 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

B

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 $B^{- 1}$ is upper triangular. This deduction uses both the inverse-transpose identity and the lower-triangular result.

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