Backward substitution solves a matrix equation in the form Ux = b
by an iterative process for an upper triangular matrix U.
The process is so called because for an upper triangular matrix, one first computes
then substitutes that backward into the next equation to solve for xn-1,
and repeats until x1.
Note that some diagonal entries in U can be 0s, provided that the system of equations is
1 & 2 & 3\\
0 & 0 & 5\\
0 & 0 & 0