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

Transpose both inverse identities, using reversal of multiplication under matrix transposition:Thus $(A^T{)}^{-1}=(A^{-1}{)}^T$, by uniqueness of the matrix inverse.

Let column $j$ of $A^{-1}$ be $x$, so $Ax=e_j$. Every diagonal entry of the invertible lower triangular matrix $A$ is nonzero, since its determinant is the product of those entries. In row one through row $j-1$, the right-hand side is zero. Forward substitution first gives $x_1=0$, then $x_2=0$, and successively $x_i=0$ for all $i<j$. These are precisely the entries above the diagonal in column $j$ of $A^{-1}$. Every column satisfies this, proving that $A^{-1}$ is lower triangular. This is triangularity of an inverse matrix.

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