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Paul Dale · (This post was last modified: 11-13-2019 12:42 AM by Paul Dale.)

I think you'll find the pivots are zero based not one based, they are also not necessarily a permutation. An unpivoted matrix would then end up with pivots 012 -> 12. Your 112 is saying to use the second row first and then the rest of the updated matrix in order (no more pivoting).

You are correct about the forward and back substitution process to solve systems of equations.
Unlike the 15C, the 34S has no space to remember that a matrix is in LU form (the 15C stored pivots in the matrix descriptor I believe).

Pauli