4. The Elimination Method

Consider an upper-triangular system given by

$$5x_1+3x_2-2x_3=-3$$

$$6x_2+x_3=-1$$

$$2x_3=10$$

It is very easy to obtain to solution. From the last equation, we see that $x_3=5$ and substituting in the $2^{nd}$ equation gives $x_2=-1$. Finally, substituting these values in the first equation given $x_1=2$. Thus the solution is $x_1=2,x_2=-1, x_3=5.$
The first objective of the elimination method is to reduce the matrix of coefficients to an upper-triangular form. Consider this example of three equation:

$$4x_1-2x_2+x_3=15$$

$$-3x_1-x_2+4x_3=8$$

$$x_1-x_2+3x_3=13$$

We first eliminate $x_1$ from the $2^{nd}$ and $3^{rd}$ equation. This is done by performing the calculations as $4R_2+3R_1$ and $4R_3-R_1$ (where $R_i$ stands for the $i^{th}$ row), we get

$$4x_1-2x_2+x_3=15$$

$$-10x_2+19x_3=77$$

$$-2x_2+11x_3=37$$

We now eliminate $x_2$ from the third equation; this is done by performing the calculations as $-10R_3+2R_2$ to get

$$4x_1-2x_2+x_3=15$$

$$-10x_2+19x_3=77$$

$$-72x_3=-216$$

This yields solution, by backward substitution, as

$$x_3=3,x_2=-2,x_1=2$$

We now present the above problem, solved in exactly the same way, in matrix notation. We write the given system as

$$\left[ \begin{array}{ccc} 4 & -2 & 1 \\ -3 & -1 & 4 \\ ... ...begin{array}{c} 15 \\ 8\\ 13 \\ \end{array}% \right]$$

and form the augmented matrix

$$A/b= \left[% \begin{array}{ccc\vert c} 4 & -2 & 1 & 15 \\ ... ...3 & -1 & 4 & 8 \\ 1 & -1 & 3 & 13 \\ \end{array}% \right]$$

We carry out the elementary row transformations to convert A to upper triangular form. Using the transformation as given above, we put

$$\left[% \begin{array}{cccc} 4 & -2 & 1 & 15 \\ 0 & -10 & 19 & 77 \\ 0 & 0 & -72 & -216 \\ \end{array}% \right]$$

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 rows 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.