How to Describe the Solutions of Linear Systems: Linear Algebra

What is a Solution Set?

When it comes to operations on linear systems, the purpose of our efforts are often motivated by a need to solve the system. For this to be true, we must identify a solution. The solutions of a linear systems denote the values which when substituted into the equation, returns a true value. In some cases, the number of solutions may be finite, whereas in others, they may be infinite. This article investigates the methods of identifying the solution, as well as different manners of modeling these solutions.

Firstly, the previous article presented a thorough investigation into the Gaussian Method and its ability to solve a system for its solution. If you have not read it, it can be found here; it will be helpful for understanding the manner in which solution sets are described. Briefly, for a linear system of 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

Then the following tuple is a solution to the system:

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

In this case, the n-tuple of ‘s’ is a solution to a set of linear equations if it solves all of the equations.

Understanding Solutions

Linear systems with unique solutions have solution sets with one or more elements. Linear systems that lack a solution, however, possess empty solution sets. Firstly, let us consider the linear system that follows:

2x\enspace\enspace\enspace +z=3 \newline x-y-z=1 \newline 3x-y\enspace\enspace\enspace =4

Furthermore, as we look at this system, we notice that we must first get this system into echelon form. If you have not read our previous article on echelon form, you may find it here. Firstly, we modify equations 2 and 3 by scaling these by equation 1. As a result, this creates the modified system which follows:

2x\enspace\enspace\enspace\enspace +z=3 \newline \enspace\enspace\enspace\enspace-y-\frac{3}{2}z=-\frac{1}{2}\newline \cdot\newline \enspace\enspace\enspace\enspace-y-\frac{3}{2}z=-\frac{1}{2}

Subsequently, adding equation 2 to equation 3 yields:

2x\enspace\enspace\enspace\enspace\enspace\enspace\enspace +z=3 \newline \enspace\enspace\enspace\enspace\enspace\enspace-y-\frac{3}{2}z=-\frac{1}{2}\newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace 0=0

With this system now in echelon form, we can confirm that the system actually has many solutions. This is due to the fact that not all of the variables are leading variables. Therefore, we may not directly solve for z in a single instant; however, x and y have leading variables. Therefore, we can describe the system of equations with another system which takes the form:

{(x,y,z)\enspace |\enspace 2x + z =3\enspace and\enspace -y – \frac{3x}{2} = -\frac{1}{2}}

While this description is slightly more simplified than the system itself, it still retains some rather confusing properties. To simplify this, we may rewrite these equations with respect to z. Thus we may write:

{[(\frac{3}{2}-\frac{1}{2}z)\enspace , (\frac{1}{2}-\frac{3}{2}z)\enspace, z]\enspace |\enspace z \in R}

Expectantly, this greatly improves our solution to the system, as now it has been modified with respect to z which can take on any real number. In this case, we call x and y leading variables, while z is a free variable. In this case, free variables are those which are not leading variables.

Solutions to Linear Systems with Two Free Variables

In some cases, there may be multiple free variables as opposed to just one. For example, consider the system below:

\enspace\enspace x+y+z-w=1\newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace y-z+w=-1\newline 3x\enspace\enspace\enspace +\enspace\enspace 6z-6w=6\newline \enspace\enspace\enspace\enspace-y+z-w=1

We may use the Gaussian method to put this system into echelon form. As a result, this action produces the system as follows:

x+y+z+w=1\newline \enspace\enspace\enspace\enspace\enspace\enspace y-z+w=-1 \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace0=0\newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace0=0

We can observe in this system that both variables z and w are not leading variables. For that reason, z and w are free variables. As we did in the previous example, we should put equations 1 and 2 in terms of the free variables z and w. We can put the equation 2 in terms of z and w, which provides us with a value for y. We can then substitute this in to equation 1. This puts both equations in terms of z and w. This produces the following solution:

{[(2-2z+2w),(-1+z-w),z,w)]\enspace |\enspace z,w \in R}

Solution Parameters

Variables that we use to describe a system of linear equations are known as parameters. For example, in our previous system, z and w are parameters of the solution. However, do not confuse the parameters of a solution with the free variables, although they are often the same. Free variables describes the state of the variables in the system of linear equations (they are not leading variables in echelon form). Parameters are the variables used to describe the set of solutions for a linear system. In most cases, if not all, the parameters of the solution are in fact free variables. In parameterizing a solution set, wherein the parameters can take on any real number, this provides an indication that the system as infinitely many solutions.

Matrix Notation of Linear Systems

It is sometimes neater to organize a system of linear equations in the form of a matrix. A given matrix of the form ‘m’ by ‘n’ is comprised of ‘m’ rows and ‘n’ columns. The number of values in a matrix is denoted by multiplying the number of rows by the number of columns. Each value in the matrix is called an entry. We denote a matrix as an upper case letter. For example, consider the matrix below:

A =\begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{pmatrix}

Here, the matrix ‘A’ is a 2×3 matrix with six entries. Its important to remember that when it comes to matrices, the number of rows is indicated first, followed by the columns. An entry in a matrix is denoted by its position; more precisely, its row and position. For example, the entry b2=A2,2. If you need insights into matrix algebra, check out this article.

When it comes to using the Gaussian method with matrices, the same rules apply, simply with respect to matrix notation. The leading entry of a particular row is its first non-zero value. In the same manner as before, we utilize the same operations to put the matrix in echelon form. For this to be true, the leading variable of a row must be to the right of the leading variable in the row above.

Converting a Linear System to Matrix Form

Let us attempt to convert a system of linear equations into a representative matrix. Firstly, consider the system below:

x+2y\enspace\enspace\enspace\enspace =4\newline \enspace\enspace\enspace\enspace y-z=0\newline x\enspace\enspace\enspace\enspace +2z=4

Importantly, for equations where the variable is not included, fill that position with a 0. Because of this, the matrix of the system appears as follows:

\begin{pmatrix} 1 & 2 & \enspace 0 | &4\\ 0 & 1 & -1 | & 0\\ 1 & 0 & \enspace 2 | & 4\\ \end{pmatrix}

The bar signifies the separation between the terms and the constant.

Solutions of Linear Systems With Vectors

A column vector is a matrix with just a single column. Alternatively, a matrix with a single row is a row vector. Here, we present an example of a vector:

\overrightarrow{v} = \begin{pmatrix} 1 \\ 3 \\ 7 \\ \end{pmatrix}

Note that vectors are denoted with lower case symbols along with an over-right arrow. Again, if you have not had previous experience in working with vectors, check out our previous article on vector calculus which discusses this topic in depth.

Recall from the previous article on the Gaussian method that a solution to a system of linear equations takes the form:

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

When using a linear system defined as a matrix, however, we may represent the solution as a vector. For such a system, the solution takes the form of the following vector:

\overrightarrow{s}=\begin{pmatrix} s_1 \\ s_2 \\ \cdot \\ \cdot \\ s_n \end{pmatrix}

This vector constitutes the solution to our linear system.

Summary of Solutions of Linear Systems

It’s crucial to hold on to the idea that solutions of linear systems may be described by the parameters of the system. These parameters consist of free variables which do not constitute leading variables in the system Furthermore, these systems tend to have infinitely many solutions. In addition to this idea, take note of the fact that we may model our linear systems in terms of matrices, which makes them significantly easier to work with. In this capacity, we may additionally utilize vectors which allow us to model the solution linearly.

If some of these concepts do not make sense, certainly return to our previous article on the Gaussian method, as this serves as an essential foundation to linear algebra. Furthermore, for the purpose of comprehending solutions to linear solutions, it is incumbent upon a mathematician to understand the use of vectors. For this purpose, return to our previous series on vector calculus. This series will provide great insight into the mechanics and operations of vectors, along with their modeling.

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