Solving Systems with Gaussian Elimination using Matrices

Matthew Schurmann

Learning Objectives

In this section, you will:

Write the augmented matrix of a system of equations.
Write the system of equations from an augmented matrix.
Perform row operations on a matrix.
Solve a system of linear equations using matrices.

**Figure 1.** German mathematician Carl Friedrich Gauss (1777–1855).

Carl Friedrich Gauss lived during the late 18th century and early 19th century, but he is still considered one of the most prolific mathematicians in history. His contributions to the science of mathematics and physics span fields such as algebra, number theory, analysis, differential geometry, astronomy, and optics, among others. His discoveries regarding matrix theory changed the way mathematicians have worked for the last two centuries.

We first encountered Gaussian elimination in Systems of Linear Equations: Two Variables. In this section, we will revisit this technique for solving systems, this time using matrices.

Writing the Augmented Matrix of a System of Equations

A matrix can serve as a device for representing and solving a system of equations. To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an augmented matrix.

For example, consider the following $2 \text{x} 2$ system of equations.

$\systeme*{3x+4y=7, 4x-2y=5}$

We can write this system as an augmented matrix:

$\begin{align*} \left[ \begin{array}{rr|r} 3 & 4 & 7 \\ 4 & -2 & 5 \end{array} \right] \end{align*}$

We can also write a matrix containing just the coefficients. This is called the coefficient matrix.

$\begin{align*} \begin{bmatrix} 3 & 4 \\ 4 & -2 \end{bmatrix} \end{align*}$

We will focus on two-by-two system of equations but it is worth noting that this process can be applied to a three-by-three system of equations such as

$\systeme*{3x-y-z=0, x+y=5, 2x-3z=2}$

has a coefficient matrix

$\begin{align*} \begin{bmatrix} 3 & -1 & -1 \\ 1 & 1 & 0 \\ 2 & 0 & -3 \end{bmatrix} \end{align*}$

and is represented by the augmented matrix

$\begin{align*} \left[ \begin{array}{rrr|r} 3 & -1 & -1 & 0 \\ 1 & 1 & 0 & 5 \\ 2 & 0 & -3 & 2 \end{array} \right] \end{align*}$

Notice that the matrix is written so that the variables line up in their own columns: x-terms go in the first column and y-terms in the second column. This can be extended to more variables with each variable being placed in it’s own column. It is very important that each equation is written in standard form $ax+by=c$ so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0.

How To

Given a system of equations in two variables, write an augmented matrix.

Write the coefficients of the x-terms as the numbers down the first column.
Write the coefficients of the y-terms as the numbers down the second column.
Draw a vertical line and write the constants to the right of the line.

Writing the Augmented Matrix for a System of Equations

Write the augmented matrix for the given system of equations.

$\systeme*{x+2y=3, 2x-y=6}$

Show Solution

The augmented matrix displays the coefficients of the variables, and an additional column for the constants.

$\begin{align*} \left[ \begin{array}{rr|r} 1 & 2 & 3 \\ 2 & -1 & 6 \end{array} \right] \end{align*}$

Try It

Write the augmented matrix of the given system of equations.

$\systeme*{4x-3y=11, 3x+2y=4}$

Show Solution

$\begin{align*} \left[ \begin{array}{rr|r} 4 & -3 & 11 \\ 3 & 2 & 4 \end{array} \right] \end{align*}$

Writing a System of Equations from an Augmented Matrix

We can use augmented matrices to help us solve systems of equations because they simplify operations when the systems are not encumbered by the variables. However, it is important to understand how to move back and forth between formats in order to make finding solutions smoother and more intuitive. Here, we will use the information in an augmented matrix to write the system of equations in standard form.

Writing a System of Equations from an Augmented Matrix Form

Find the system of equations from the augmented matrix.

$\begin{align*} \left[ \begin{array}{rr|r} 3 & -2 & 1 \\ 2 & -3 & 7 \end{array} \right] \end{align*}$

Show Solution

The columns represent the variables $x$ and $y$ , and the constant. The first row represents the first equation and the second row represents the second equation.

$\systeme*{3x -2y = 1, 2x -3y = 7}$

Try It

Write the system of equations from the augmented matrix.

$\begin{align*} \left[ \begin{array}{rr|r} -3 & -1 & 17 \\ 5 & 6 & -27 \end{array} \right] \end{align*}$

Show Solution

$\systeme*{-3x -y = 17, 5x =6y = -27}$

Performing Row Operations on a Matrix

Now that we can write systems of equations in augmented matrix form, we will examine the various row operations that can be performed on a matrix, such as addition, multiplication by a constant, and interchanging rows.

Performing row operations on a matrix is the method we use for solving a system of equations. In order to solve the system of equations, we want to convert the matrix to row-echelon form, in which there are ones down the main diagonal from the upper left corner to the lower right corner, and zeros in every position below the main diagonal as shown.

$\begin{align*} \left[ \begin{array}{rr|r} 1 & a & b \\ 0 & 1 & c \end{array} \right] \end{align*}$

We use row operations corresponding to equation operations to obtain a new matrix that is row-equivalent in a simpler form. Here are the guidelines to obtaining row-echelon form.

In any nonzero row, the first nonzero number is a 1. It is called a leading 1.
Any all-zero rows are placed at the bottom on the matrix.
Any leading 1 is below and to the right of a previous leading 1.
Any column containing a leading 1 has zeros in all other positions in the column.

To solve a system of equations we can perform the following row operations to convert the coefficient matrix to row-echelon form and do back-substitution to find the solution.

Interchange rows. (Notation: ${R}_{i} \leftrightarrow {R}_{j}$ )
Multiply a row by a constant. (Notation: $c{R}_{i}$ )
Add the product of a row multiplied by a constant to another row. (Notation: ${R}_{i}+c{R}_{j})$

Each of the row operations corresponds to the operations we have already learned to solve systems of equations in three variables. With these operations, there are some key moves that will quickly achieve the goal of writing a matrix in row-echelon form. To obtain a matrix in row-echelon form for finding solutions, we use Gaussian elimination, a method that uses row operations to obtain a 1 as the first entry so that row 1 can be used to convert the remaining rows.

Gaussian Elimination

The Gaussian elimination method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write a new matrix with the number 1 as the entry down the main diagonal and have all zeros below.

$\begin{align*} \begin{bmatrix} a_{11} & a_{12} &| & a_{13} \\ a_{21} & a_{22} & | & a_{23} \end{bmatrix} \quad \xrightarrow{\text{After Gaussian elimination}} \quad \begin{bmatrix} 1 & b_{12} & | &b_{13} \\ 0 & 1 & | & b_{23} \end{bmatrix} \end{align*}$

The first step of the Gaussian strategy includes obtaining a 1 as the first entry, so that row 1 may be used to alter the rows below.

How To

Given an augmented matrix, perform row operations to achieve row-echelon form.

The first equation should have a leading coefficient of 1. Interchange rows or multiply by a constant, if necessary.
Use row operations to obtain zeros down the first column below the first entry of 1.
Use row operations to obtain a 1 in row 2, column 2.
Use row operations to obtain zeros down column 2, below the entry of 1.
Continue this process for all rows until there is a 1 in every entry down the main diagonal and there are only zeros below.
If any rows contain all zeros, place them at the bottom.

Solving a $2×2$ System by Gaussian Elimination

Solve the given system by Gaussian elimination.

$\systeme*{2x+3y=6, x-y=\dfrac{1}{2}}$

Show Solution

First, we write this as an augmented matrix.

$\begin{align*} \left[ \begin{array}{rr|r} 2 & 3 & 6 \\ 1 & -1 & \dfrac{1}{2} \end{array} \right] \end{align*}$

We want a 1 in row 1, column 1. This can be accomplished by interchanging row 1 and row 2.

${R}_{1} \leftrightarrow {R}_{2}$

$\begin{align*} \left[ \begin{array}{rrr|r} 1 & -1 & & \dfrac{1}{2} \\ 2 & 3 & & 6 \end{array} \right] \end{align*}$

We now have a 1 as the first entry in row 1, column 1. Now let’s obtain a 0 in row 2, column 1. This can be accomplished by multiplying row 1 by $-2$ and then adding the result to row 2.

$-2{R}_{1}+{R}_{2}={R}_{2}$

$\begin{align*} \left[ \begin{array}{rrr|r} 1 & -1 & & \dfrac{1}{2} \\ 0 & 5 & & 5 \end{array} \right] \end{align*}$

We only have one more step, to multiply row 2 by $\dfrac{1}{5}$ .

$\dfrac{1}{5}{R}_{2}={R}_{2} to$

$\begin{align*} \left[ \begin{array}{rrr|r} 1 & -1 & & \dfrac{1}{2} \\ 0 & 1 & & 1 \end{array} \right] \end{align*}$

Use back-substitution. The second row of the matrix represents $y=1$ . Back-substitute $y=1$ into the first equation.

$\begin{align*} x - \left(1\right) &= \frac{1}{2} \\ x &= \frac{3}{2} \end{align*}$

The solution is the point $\left(\dfrac{3}{2},1\right)$ .

Try It

Solve the given system by Gaussian elimination.

$\systeme*{4x+3y=11, x-3y=-1}$

Show Solution

$\left(2,1\right)$

Solving an Inconsistent System

Use Gaussian elimination to solve the given $2 × 2$
system of equations.
$\systeme*{2x+y=1, 4x+2y=6}$

Show Solution

Write the system as an augmented matrix.

$\begin{align*} \left[ \begin{array}{ll|l} 2 & 1 & 1 \\ 4 & 2 & 6 \end{array} \right] \end{align*}$

Obtain a 1 in row 1, column 1. This can be accomplished by multiplying the first row by $\dfrac{1}{2}$ .

$\dfrac{1}{2}{R}_{1}={R}_{1}$

$\begin{align*} \left[ \begin{array}{cc|c} 1 & \dfrac{1}{2} & \dfrac{1}{2} \\ 4 & 2 & 6 \end{array} \right] \end{align*}$

Next, we want a 0 in row 2, column 1. Multiply row 1 by $-4$ and add row 1 to row 2.

$-4{R}_{1}+{R}_{2}={R}_{2}$

$\begin{align*} \left[ \begin{array}{cc|c} 1 & \dfrac{1}{2} & \dfrac{1}{2} \\ 0 & 0 & 4 \end{array} \right] \end{align*}$

The second row represents the equation $0=4$ . Therefore, the system is inconsistent and has no solution.

Solving a Dependent System

Solve the system of equations.
$\systeme*{3x+4y=12, 6x+8y=24}$

Show Solution

Perform row operations on the augmented matrix to try and achieve row-echelon form.

$\begin{align*} \left[ \begin{array}{cccc|c} 3 & & 4 & & 12 \\ 6 & & 8 & & 24 \end{array} \right] \end{align*}$

$-\dfrac{1}{2}{R}_{2}+{R}_{1}={R}_{1}$

$\begin{align*} \left[ \begin{array}{ll|l} 0 & 0 & 0 \\ 6 & 8 & 24 \end{array} \right] \end{align*}$

${R}_{1}\leftrightarrow{R}_{2}$

$\begin{align*} \left[ \begin{array}{ll|l} 6 & 8 & 24 \\ 0 & 0 & 0 \end{array} \right] \end{align*}$

The matrix ends up with all zeros in the last row: $0x+0y=0$ . Thus, there are an infinite number of solutions, and the system is classified as dependent. To find the generic solution, return to one of the original equations and solve for $y$ .

$\begin{align*} 3x + 4y &= 12 \\ 4y &= 12 - 3x \\ y &= 3 - \frac{3}{4}x \end{align*}$

So the solution to this system is $\left(x,3-\dfrac{3}{4}x\right)$ .

Can any system of linear equations be solved by Gaussian elimination?

Yes, a system of linear equations of any size can be solved by Gaussian elimination. We are focusing on a fourth method to solve a system of equation in two variables but these techniques can be applied to any size system of equations.

Access these online resources for additional instruction and practice with solving systems of linear equations using Gaussian elimination.

Solve a System of Two Equations Using an Augmented Matrix

Key Concepts

An augmented matrix is one that contains the coefficients and constants of a system of equations.
A matrix augmented with the constant column can be represented as the original system of equations.
Row operations include multiplying a row by a constant, adding one row to another row, and interchanging rows.
We can use Gaussian elimination to solve a system of equations.
Row operations are performed on matrices to obtain row-echelon form.
To solve a system of equations, write it in augmented matrix form. Perform row operations to obtain row-echelon form. Back-substitute to find the solutions.

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Learning Objectives

Writing the Augmented Matrix of a System of Equations

How To

Writing the Augmented Matrix for a System of Equations

Try It

Writing a System of Equations from an Augmented Matrix

Writing a System of Equations from an Augmented Matrix Form

Try It

Performing Row Operations on a Matrix

Gaussian Elimination

How To

Solving a System by Gaussian Elimination

Try It

Solving an Inconsistent System

Solving a Dependent System

Key Concepts

License

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Solving a $2×2$ System by Gaussian Elimination