Gauss Elimination Method

DEFINITION 2.2.10 (Forward/Gauss Elimination Method)   Gaussian elimination is a method of solving a linear system $A {\mathbf x}= {\mathbf b}$ (consisting of $m$ equations in $n$ unknowns) by bringing the augmented matrix

$$[A \;\; {\mathbf b}] = \left[\begin{array}{cccc\vert c} a_{11} & ... ... \vdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} & b_m \end{array}\right]$$

to an upper triangular form

$$\left[\begin{array}{cccc\vert c} c_{11} & c_{12} & \cdots & c_{1n... ... \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & c_{mn} & d_m \end{array}\right].$$

This elimination process is also called the forward elimination method.

The following examples illustrate the Gauss elimination procedure.

EXAMPLE 2.2.11   Solve the linear system by Gauss elimination method.

$$y + z = 2$$

$$2 x + 3 z = 5$$

$$x+ y + z = 3$$

Solution: In this case, the augmented matrix is $\begin{bmatrix}0 & 1 & 1 & 2 \\ 2 & 0 & 3 & 5 \\ 1 & 1 & 1 & 3 \end{bmatrix}.$ The method proceeds along the following steps.

1. Interchange $1^{\mbox{st}}$ and $2^{\mbox{nd}}$ equation (or $R_{12}$ ).
   $$\begin{array}{cr} 2 x + 3 z &= 5 \\ y + z &= 2 \\ x + y + z &= 3 ... ...} \begin{bmatrix}2 & 0 & 3 & 5 \\ 0 & 1 & 1 & 2 \\ 1 & 1 & 1 & 3 \end{bmatrix}.$$

2. Divide the $1^{\mbox{st}}$ equation by $2$ (or $R_1(1/2)$ ).
   $$\begin{array}{cr} x + \frac{3}{2} z &= \frac{5}{2} \\ y + z &= 2... ...0 & \frac{3}{2} & \frac{5}{2} \\ 0 & 1 & 1 & 2 \\ 1 & 1 & 1 & 3 \end{bmatrix}.$$

3. Add $-1$ times the $1^{\mbox{st}}$ equation to the $3^{\mbox{rd}}$ equation (or $R_{31}(-1)$ ).
   $$\begin{array}{cr} x + \frac{3}{2} z &= \frac{5}{2} \\ y + z &... ...1 & 1 & 2 \\ 0 & 1 & -\frac{1}{2} & \frac{1}{2} \end{bmatrix}.$$

4. Add $-1$ times the $2^{\mbox{nd}}$ equation to the $3^{\mbox{rd}}$ equation (or $R_{32}(-1)$ ).
   $$\begin{array}{cr} x + \frac{3}{2} z &= \frac{5}{2} \\ y + z &= 2 ... ...ac{5}{2} \\ 0 & 1 & 1 & 2 \\ 0 & 0 & -\frac{3}{2} & -\frac{3}{2} \end{bmatrix}.$$

5. Multiply the $3^{\mbox{rd}}$ equation by $\frac{-2}{3}$ (or $R_3(-\frac{2}{3})$ ).
   $$\begin{array}{cr} x + \frac{3}{2} z &= \frac{5}{2} \\ y + z &= 2 ... ... 0 & \frac{3}{2} & \frac{5}{2} \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & 1 \end{bmatrix}.$$

The last equation gives $z=1,$ the second equation now gives $y=1.$ Finally the first equation gives $x = 1.$ Hence the set of solutions is $(x, y, z)^t = (1, 1, 1)^t,$ A UNIQUE SOLUTION.

EXAMPLE 2.2.12   Solve the linear system by Gauss elimination method.

$$x+ y + z = 3$$

$$x + 2 y + 2 z = 5$$

$$3 x + 4 y + 4 z = 11$$

Solution: In this case, the augmented matrix is $\begin{bmatrix}1 & 1 & 1 & 3 \\ 1 & 2 & 2 & 5 \\ 3 & 4 & 4 & 11 \end{bmatrix}$ and the method proceeds as follows:

1. Add $-1$ times the first equation to the second equation.
   $$\begin{array}{cr} x + y + z &= 3 \\ y + z &= 2 \\ 3 x +4 y +4 z ... ...begin{bmatrix} 1 & 1 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 3 & 4 & 4 & 11 \end{bmatrix}.$$

2. Add $-3$ times the first equation to the third equation.
   $$\begin{array}{cr} x + y + z &= 3 \\ y + z &= 2 \\ y + z &= 2 \en... ...\begin{bmatrix} 1 & 1 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 1 & 1 & 2 \end{bmatrix}.$$

3. Add $-1$ times the second equation to the third equation
   $$\begin{array}{cr} x + y+ z &= 3 \\ y + z &= 2 \end{array} \hspace... ... \begin{bmatrix}1 & 1 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix}.$$

Thus, the set of solutions is $(x, y, z)^t = (1, 2-z, z)^t = (1, 2, 0)^t + z (0, -1, 1)^t,$ with $z$ arbitrary. In other words, the system has INFINITE NUMBER OF SOLUTIONS.

EXAMPLE 2.2.13   Solve the linear system by Gauss elimination method.

$$x+ y + z = 3$$

$$x + 2 y + 2 z = 5$$

$$3 x + 4 y + 4 z = 12$$

Solution: In this case, the augmented matrix is $\begin{bmatrix}1 & 1 & 1 & 3 \\ 1 & 2 & 2 & 5 \\ 3 & 4 & 4 & 12 \end{bmatrix}$ and the method proceeds as follows:

1. Add $-1$ times the first equation to the second equation.
   $$\begin{array}{cr} x + y + z &= 3 \\ y + z &= 2 \\ 3 x +4 y +4 z ... ...begin{bmatrix} 1 & 1 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 3 & 4 & 4 & 12 \end{bmatrix}.$$

2. Add $-3$ times the first equation to the third equation.
   $$\begin{array}{cr} x + y + z &= 3 \\ y + z &= 2 \\ y + z &= 3 \en... ...\begin{bmatrix} 1 & 1 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 1 & 1 & 3 \end{bmatrix}.$$

3. Add $-1$ times the second equation to the third equation
   $$\begin{array}{cr} x + y+ z &= 3 \\ y + z &= 2 \\ 0 & = 1 \end{arr... ... \begin{bmatrix}1 & 1 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix}.$$

The third equation in the last step is

$$0 x + 0 y + 0 z = 1.$$

This can never hold for any value of $x, y, z.$ Hence, the system has NO SOLUTION.

Remark 2.2.14   Note that to solve a linear system, $A {\mathbf x}= {\mathbf b},$ one needs to apply only the elementary row operations to the augmented matrix $[ A \; \; {\mathbf b}].$