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and using backward substitution yields the solution.
Note that there exists the possibility that the set of equations
has no solution, or that the prior procedure will fail to find it.
During the triangularization step, if a zero is encountered on the
diagonal, we cannot use that row to eliminate coefficients below
that zero element. However, in that case, we can continue by
interchanging rows and eventually achieve an upper triangular
matrix of coefficients. The real stumbling block is finding a zero
on the diagonal after we have triangularized. If that occurs, it
means that the determinat is zero and there is no solution. Let us
now state what we mean by elementary row operations, that we have
used to solve the above system.
There are three of these
operations:
1. We may multiply any row of the augmented matrix by a
constant.
2. We can add a multiple of one row to a multiple of any other
row.
3. We can inter change the order of any two rows.
It is intuitively obvious that all the three above operations do
not change the solution of the system.