Gaussian Method of Solving Linear Systems: Linear Algebra

Introducing the Gaussian Method and Linear Algebra

Linear algebra is a subject devoted to understanding complex equations that interact together to form systems. The Gaussian method is the premise of linear algebra, and provides tools which help discern the solution of a linear system. The Gaussian method is an algorithm of steps which constitutes meaningful operations that allow these systems to be manipulated, and thus solved. The following article explores linear systems, the Gaussian method, and provides examples for understanding how each operation is performed.

An Overview of Linear Systems

Linear combinations consist of variable sequences x1, . . . , xn which takes the form:

a_1x_1+a_2x_2+a_3x_3+\cdot \cdot \cdot +a_nx_n

The values of ‘a’ are real numbers and constitute the coefficients of the combination. These combinations may comprise a linear equation which takes the form:

a_1x_1+a_2x_2+a_3x_3+\cdot \cdot \cdot +a_nx_n=d

In this case, the parameter ‘d’ is a real number serving as a constant. Associated with such linear equations is an n-tuple of of the form:

(s_1, s_2, . . . , s_n) \in R

This n-tuple of ‘s’ represents the solution set which satisfies the linear equation when is substituted in. The substitution then takes the form:

a_1s_1+a_2s_2+\cdot \cdot \cdot +a_ns_n=d

Now, when it comes to linear equations, we may observe systems of linear equations. Systems of linear equations take the form:

a_1,_1x_1+a_1,_2x_2+\cdot\cdot\cdot+a_1,_nx_n=d_1\newline a_2,_1x_1+a_2,_2x_2+\cdot\cdot\cdot+a_2,_nx_n=d_2\newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\cdot\newline\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\cdot \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\cdot\newline a_m,_1x_1+a_m,_2x_2+\cdot \cdot \cdot + a_m,_nx_n = d_m

In this instance, the n-tuple (s1,s2, … , sn) may be a solution to the system if it satisfies each equation of the system.

Finding the Solution: The Gaussian Method


Identifying the set of solutions to a linear system is known as solving the system. The method of solving a linear system is defined by an algorithm known as the Gaussian method. The Gaussian method is a step-by-step elimination process wherein we treat equations as they relate to each other. In this manner, we seek to modify the equations in order to easily solve it. Consider the following system of linear equations:

\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace3x_3=9 \newline x_1+5x_2-2x_3=2 \newline \frac{1}{3}x_1+2x_2\enspace\enspace\enspace\enspace=3

Here, we employ the Gaussian method to solve the system. First of all, we must transform the system according to the leading variable of the equations. In this manner, equations who’s leading variable is x1 should be placed first while equations who’s leading variable is x3 should be placed last. So, let us first undertake our system rearrangement:

\frac{1}{3}x_1+2x_2 \enspace\enspace\enspace\enspace\enspace= 3\newline \enspace \enspace x_1+5x_2-2x_3=2\newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace3x_3=9

Followed by this rearrangement, we must modify the first equation to get it into an integer form, making it easier to work with. We can do this by multiplying row 1 by a value of three. This modification yields:

x_1+6x_2\enspace\enspace\enspace\enspace\enspace=9 \newline x_1+5x_2-2x_3=2\newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace3x_3=9

Now, we add the negative of equation 1 to the second equation in order to get rid of the x1 parameter. Doing so yields

x_1+6x_2\enspace\enspace=9 \newline -1x_2-2x_3=-7 \newline \enspace\enspace\enspace\enspace\enspace\enspace3x_3=9

Now that the equations are in this form, we may now solve the system In equation three that when solving for x3, it is equal to 3. If we plug 3 in to x3 in equation two and solve for x2, we find x2 to equal 1. Then if we plug 1 in to x2 in equation one and solve for x1, we find that x1 is equal to 3. Therefore, the solution to the system is the set {(3,1,3)}.

The Three Tenets of Gauss’s Method

The Gaussian Method relies upon three fundamental tenets. If a linear system is modified by any of the following tenets:

  1. Equations are swapped
  2. The equation is multiplied on both sides by the same constant
  3. The equation is replaced by its sum and the multiple of another equation

Then those two systems has the same set of solutions. These tenets constitute the elementary reduction operations of the Gaussian method. Furthermore, they are summed up by the names of swapping, scaling, and recombination.

When executing these operations, the syntax of it is such that we say ‘row i’ by ‘pi‘. This is to say we are executing the operation pi to the row i. A proof of the Gaussian method may be found here.

Comprehending Echelon Form

When working with linear systems, it is often essential to place the component equations in echelon form. We mentioned this earlier but neglected to give this operation its proper title. To put a system in echelon form means to ensure that the equations are listed in order of their leading variable from least to greatest. Echelon form contends that if each variable is to the right of the leading variable of the row above it, then the system is in echelon form. Let’s consider what this means by employing an example.

Consider the system which follows:

x-y \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace=0 \newline 2x-2y+z+2w=4 \newline \enspace\enspace\enspace\enspace\enspace y\enspace\enspace\enspace\enspace+w\enspace=0 \newline\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace 2z+w=5
Operation #1

Here, we have to execute the swap operation in order to get this system into row echelon form. It is not already in echelon form because the leading variable of equation 2 is not to the right of the leading variable in equation 1. To correct this, we add negative 2 times equation 1 to equation 2. This returns the system:

x-y \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace= 0 \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace z+2w=4\newline \enspace\enspace\enspace\enspace y\enspace\enspace\enspace\enspace\enspace\enspace\enspace +w=0\newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace2z+w=5
Operation #2

So, we can see that from this manipulation, we successfully properly oriented equation 2 with respect to equation 1. However, if we look again, we see that we have not completely finished. This is due to the fact that now, the leading variable of equation 3 is not to the right of the leading variable of equation 2. Therefore, we must transpose the positions of these two equations. Finally, this transformation produces the system that follows:

x-y\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace =0 \newline \enspace\enspace\enspace\enspace y\enspace\enspace\enspace\enspace\enspace\enspace\enspace +w=0 \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace z+2w=4 \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace 2z+w=5
Operation #3

We have almost successfully put this equation into echelon form; however, the leading variable of equation 4 is not to the right of equation 3. Therefore, we will again use the equation swap method. This will proceed by adding negative 2 times equation 3 to equation 4. This produces the following system:

<!– wp:katex/display-block –> <div class=”wp-block-katex-display-block katex-eq” data-katex-display=”true”>x-y\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace =0 \newline \enspace\enspace\enspace\enspace y\enspace\enspace\enspace\enspace\enspace\enspace\enspace +w=0 \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace z+2w=4 \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace -3w=3</div> <!– /wp:katex/display-block –>

Here we can see that we have successfully placed the system into echelon form. At this point, we can readily solve the system. Finally, we use the same mechanics of back substitution from the previous example, and observe that the solution set of the system as {(-1,-1,2,1)}.

Summary of the Gaussian Method

The purpose of putting a system in echelon form is to favorably set up the system for back substitution. We apply the tenets of the Gaussian method so that the system to provide efficient solutions. Furthermore, we put the system into echelon form which allows inference of several attributes of the solution. For example, if there are no contradictory equations and the system is in echelon form, then we may comfortably infer that the system has a unique solution. In our subsequent article, we will thoroughly examine the manner in which we describe the solutions of a Gaussian linear system of equations. If you have not had previous experience in working with vectors, it may be helpful to address our series on vector calculus before moving onward. Find the collection here.

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