iii) multiply eqn. 2 by  and add to eqn. 3; replace eqn. 3.
We have eliminated x1 and x2 from eqn.3 and x1 from eqn. 2.
back substitution iv) solve eqn. 3 for x3, substitute in eqn. 2 & 1.
solve eqn. 2 for x2, substitute in eqn. 1.
solve eqn. 1 for x1.
b. Pivoting
Due to the finite number of digits carried along by the machine, we have to worry about the relative magnitudes of the matrix elements, especially the diagonal elements. In other words, the inverse matrix,  may be effectively singular even if not actually so. To minimize this possibility, we commonly rearrange the set of equations to place the largest coefficients on the diagonal, to the extent possible. This process is called pivoting.
e.g.
37x2 – 3x3 = 4
19x1 – 2x2 + 48x3 = 99
7x1 + 0.6x2 +15x3 = -9
rearrange
19x1 – 2x2 + 48x3 = 99
37x2 – 3x3 = 4
7x1 + 0.6x2 +15x3 = -9
or
7x1 + 0.6x2 +15x3 = -9
37x2 – 3x3 = 4
19x1 – 2x2 + 48x3 = 99
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