= Solution
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 \b[$A^{-1}$ is lower triangular]. This is <triangularity of an inverse matrix>.
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