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Solving Systems with Gaussian Elimination

Write the augmented matrix of a system of equations.

Solving Systems with Gaussian Elimination

Use row operations on a matrix (IA 4.5.2)
Solve systems of equations using matrices (IA 4.5.3)

In the last section, we learned how to write the augmented matrix for a system of equations.

Once a system of equations is in its augmented matrix form, we will solve by elimination by performing operations on the rows that will lead us to the solution. Our goal will be to get 1 on the diagonal of the matrix and all entries below the diagonal must be zeros.

In a matrix, the following operations can be performed on any row and the resulting matrix will be equivalent to the original matrix.

Interchange any two rows.
Multiply a row by any real number except 0.
Add a nonzero multiple of one row to another row.

These actions are called row operations and will help us use the matrix to solve a system of equations.

Example

Try it.

Use the indicated row operations on the augmented matrix:

ⓐ Interchange rows 2 and 3.
ⓑ Multiply row 2 by 5.
ⓒ Multiply row 3 by −2 and add to row 1.

$[\begin{array}{lll}6 & -5 & 2 \\ 1 & 1 & -4 \\ 3 & -3 & 1\end{array}|\begin{array}{l}3 \\ 5 \\ -1\end{array}]$

Solution

ⓐ Interchange rows 2 and 3.
ⓑ Multiply row 2 by 5.
ⓒ Multiply row 3 by −2 and add to row 1.

Example

Try it.

Use the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: $[\begin{array}{llllll}1 & & & -1 & & \\ 4 & & & -8 & & \end{array}|\begin{array}{lll} & & 2 \\ & & 0\end{array}].$

Solution

To make the 4 a 0, we could multiply row 1 by $-4$ and then add it to row 2.

Try it.

Use the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix

$\left[\begin{array}{lll}1 & -1 & 2 \\ 3 & -6 & 2\end{array}\right]$

Condensed — the full section is in OpenStax College Algebra 2e.

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 $2\times 2$ system of equations.

\[\begin{array}{l}3x+4y=7 \\ 4x-2y=5\end{array}\]

We can write this system as an augmented matrix:

\[[\begin{array}{ll}3 & 4 \\ 4 & -2\end{array}\ |\ \begin{array}{l}7 \\ 5\end{array}]\]

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

\[[\begin{array}{ll}3 & 4 \\ 4 & -2\end{array}]\]

A three-by-three system of equations such as

\[\begin{array}{l}3x-y-z=0 \\ x+y=5 \\ 2x-3z=2\end{array}\]

has a coefficient matrix

\[[\begin{array}{lll}3 & -1 & -1 \\ 1 & 1 & 0 \\ 2 & 0 & -3\end{array}]\]

and is represented by the augmented matrix

\[[\begin{array}{lll}3 & -1 & -1 \\ 1 & 1 & 0 \\ 2 & 0 & -3\end{array}\ |\ \begin{array}{l}0 \\ 5 \\ 2\end{array}]\]

Example

Try it.

Write the augmented matrix for the given system of equations.

\[\begin{array}{l}\ x+2y-z=3 \\ 2x-y+2z=6 \\ x-3y+3z=4\end{array}\]

Solution

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

\[[\begin{array}{lll}1 & 2 & -1 \\ 2 & -1 & 2 \\ 1 & -3 & 3\end{array}\ |\ \begin{array}{l}3 \\ 6 \\ 4\end{array}]\]

Condensed — the full section is in OpenStax College Algebra 2e.

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.

Example

Try it.

Find the system of equations from the augmented matrix.

\[[\begin{array}{lll}1 & -3 & -5 \\ 2 & -5 & -4 \\ -3 & 5 & 4\end{array}\ |\ \begin{array}{l}-2 \\ 5 \\ 6\end{array}]\]

Solution

When the columns represent the variables $x,$ $y,$ and $z,$

\[[\begin{array}{lll}1 & -3 & -5 \\ 2 & -5 & -4 \\ -3 & 5 & 4\end{array}\ |\ \begin{array}{l}-2 \\ 5 \\ 6\end{array}]\to \begin{array}{l}x-3y-5z=-2 \\ 2x-5y-4z=5 \\ -3x+5y+4z=6\end{array}\]

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.

\[\begin{array}{lllllll}\text{Row-echelon form} \\ [\begin{array}{lll}1 & a & b \\ 0 & 1 & d \\ 0 & 0 & 1\end{array}]\end{array}\]

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: ${R}_{i}\ ↔\ {R}_{j}$ )
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.

Condensed — the full section is in OpenStax College Algebra 2e.

Solving a System of Linear Equations Using Matrices

We have seen how to write a system of equations with an augmented matrix, and then how to use row operations and back-substitution to obtain row-echelon form. Now, we will take row-echelon form a step farther to solve a 3 by 3 system of linear equations. The general idea is to eliminate all but one variable using row operations and then back-substitute to solve for the other variables.

Example

Try it.

Solve the system of linear equations using matrices.

\[\begin{array}{l}\begin{array}{l} \\ \\ x-y+z=8\end{array} \\ 2x+3y-z=-2 \\ 3x-2y-9z=9\end{array}\]

Solution

First, we write the augmented matrix.

\[[\begin{array}{lll}1 & -1 & 1 \\ 2 & 3 & -1 \\ 3 & -2 & -9\end{array}\ |\ \begin{array}{l}8 \\ -2 \\ 9\end{array}]\]

Next, we perform row operations to obtain row-echelon form.

\[\begin{array}{lllllllllllllllllllllllllllllllllllllllll}-2{R}_{1}+{R}_{2}={R}_{2}\to [\begin{array}{llllll}1 & & -1 & & 1 & \\ 0 & & 5 & & -3 & \\ 3 & & -2 & & -9 & \end{array}|\begin{array}{ll} & 8 \\ & -18 \\ & 9\end{array}] & & & & -3{R}_{1}+{R}_{3}={R}_{3}\to [\begin{array}{llllll}1 & & -1 & & 1 & \\ 0 & & 5 & & -3 & \\ 0 & & 1 & & -12 & \end{array}|\begin{array}{ll} & 8 \\ & -18 \\ & -15\end{array}]\end{array}\]

The easiest way to obtain a 1 in row 2, column 2 is to interchange ${R}_{2}$ and ${R}_{3}.$

\[\text{Interchange}\ {R}_{2}\ \text{and}\ {R}_{3}\to [\begin{array}{lllllll}1 & & -1 & & 1 & & 8 \\ 0 & & 1 & & -12 & & -15 \\ 0 & & 5 & & -3 & & -18\end{array}]\]

Then

\[\begin{array}{lllllllllllllllllllllllllllllllllllllllll} \\ \begin{array}{lllllllllllllllllllllllllllllllllllllllll}-5{R}_{2}+{R}_{3}={R}_{3}\to [\begin{array}{llllll}1 & & -1 & & 1 & \\ 0 & & 1 & & -12 & \\ 0 & & 0 & & 57 & \end{array}|\begin{array}{ll} & 8 \\ & -15 \\ & 57\end{array}] & & & & -\frac{1}{57}{R}_{3}={R}_{3}\to [\begin{array}{llllll}1 & & -1 & & 1 & \\ 0 & & 1 & & -12 & \\ 0 & & 0 & & 1 & \end{array}|\begin{array}{ll} & 8 \\ & -15 \\ & 1\end{array}]\end{array}\end{array}\]

The last matrix represents the equivalent system.

\[\begin{array}{l}\ x-y+z=8 \\ y-12z=-15 \\ z=1\end{array}\]

Using back-substitution, we obtain the solution as $(4,-3,1).$

Condensed — the full section is in OpenStax College Algebra 2e.

Key Concepts

An augmented matrix is one that contains the coefficients and constants of a system of equations. See .
A matrix augmented with the constant column can be represented as the original system of equations. See .
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. See , , and .
Row operations are performed on matrices to obtain row-echelon form. See .
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. See and .
A calculator can be used to solve systems of equations using matrices. See .
Many real-world problems can be solved using augmented matrices. See and .

Solving Systems with Gaussian Elimination

Use row operations on a matrix (IA 4.5.2)
Solve systems of equations using matrices (IA 4.5.3)

In the last section, we learned how to write the augmented matrix for a system of equations.

In a matrix, the following operations can be performed on any row and the resulting matrix will be equivalent to the original matrix.

Interchange any two rows.
Multiply a row by any real number except 0.
Add a nonzero multiple of one row to another row.

These actions are called row operations and will help us use the matrix to solve a system of equations.

Example

Try it.

Use the indicated row operations on the augmented matrix:

ⓐ Interchange rows 2 and 3.
ⓑ Multiply row 2 by 5.
ⓒ Multiply row 3 by −2 and add to row 1.

$[\begin{array}{lll}6 & -5 & 2 \\ 1 & 1 & -4 \\ 3 & -3 & 1\end{array}|\begin{array}{l}3 \\ 5 \\ -1\end{array}]$

Solution

ⓐ Interchange rows 2 and 3.
ⓑ Multiply row 2 by 5.
ⓒ Multiply row 3 by −2 and add to row 1.

Example

Try it.

Solution

To make the 4 a 0, we could multiply row 1 by $-4$ and then add it to row 2.

Try it.

Use the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix

$\left[\begin{array}{lll}1 & -1 & 2 \\ 3 & -6 & 2\end{array}\right]$

Condensed — the full section is in OpenStax Precalculus 2e.

Writing the Augmented Matrix of a System of Equations

For example, consider the following $2\times 2$ system of equations.

\[\begin{array}{l}3x+4y=7 \\ 4x-2y=5\end{array}\]

We can write this system as an augmented matrix:

\[[\begin{array}{ll}3 & 4 \\ 4 & -2\end{array}\ |\ \begin{array}{l}7 \\ 5\end{array}]\]

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

\[[\begin{array}{ll}3 & 4 \\ 4 & -2\end{array}]\]

A three-by-three system of equations such as

\[\begin{array}{l}3x-y-z=0 \\ x+y=5 \\ 2x-3z=2\end{array}\]

has a coefficient matrix

\[[\begin{array}{lll}3 & -1 & -1 \\ 1 & 1 & 0 \\ 2 & 0 & -3\end{array}]\]

and is represented by the augmented matrix

\[[\begin{array}{lll}3 & -1 & -1 \\ 1 & 1 & 0 \\ 2 & 0 & -3\end{array}\ |\ \begin{array}{l}0 \\ 5 \\ 2\end{array}]\]

Example

Try it.

Write the augmented matrix for the given system of equations.

\[\begin{array}{l}\ x+2y-z=3 \\ 2x-y+2z=6 \\ x-3y+3z=4\end{array}\]

Solution

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

\[[\begin{array}{lll}1 & 2 & -1 \\ 2 & -1 & 2 \\ 1 & -3 & 3\end{array}\ |\ \begin{array}{l}3 \\ 6 \\ 4\end{array}]\]

Condensed — the full section is in OpenStax Precalculus 2e.

Writing a System of Equations from an Augmented Matrix

Example

Try it.

Find the system of equations from the augmented matrix.

\[[\begin{array}{lll}1 & -3 & -5 \\ 2 & -5 & -4 \\ -3 & 5 & 4\end{array}\ |\ \begin{array}{l}-2 \\ 5 \\ 6\end{array}]\]

Solution

When the columns represent the variables $x,$ $y,$ and $z,$

\[[\begin{array}{lll}1 & -3 & -5 \\ 2 & -5 & -4 \\ -3 & 5 & 4\end{array}\ |\ \begin{array}{l}-2 \\ 5 \\ 6\end{array}]\to \begin{array}{l}x-3y-5z=-2 \\ 2x-5y-4z=5 \\ -3x+5y+4z=6\end{array}\]

Performing Row Operations on a Matrix

\[\begin{array}{lllllll}\text{Row-echelon form} \\ [\begin{array}{lll}1 & a & b \\ 0 & 1 & d \\ 0 & 0 & 1\end{array}]\end{array}\]

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

Condensed — the full section is in OpenStax Precalculus 2e.

Solving a System of Linear Equations Using Matrices

Example

Try it.

Solve the system of linear equations using matrices.

\[\begin{array}{l}\begin{array}{l} \\ \\ x-y+z=8\end{array} \\ 2x+3y-z=-2 \\ 3x-2y-9z=9\end{array}\]

Solution

First, we write the augmented matrix.

\[[\begin{array}{lll}1 & -1 & 1 \\ 2 & 3 & -1 \\ 3 & -2 & -9\end{array}\ |\ \begin{array}{l}8 \\ -2 \\ 9\end{array}]\]

Next, we perform row operations to obtain row-echelon form.

The easiest way to obtain a 1 in row 2, column 2 is to interchange ${R}_{2}$ and ${R}_{3}.$

\[\text{Interchange}\ {R}_{2}\ \text{and}\ {R}_{3}\to [\begin{array}{lllllll}1 & & -1 & & 1 & & 8 \\ 0 & & 1 & & -12 & & -15 \\ 0 & & 5 & & -3 & & -18\end{array}]\]

Then

The last matrix represents the equivalent system.

\[\begin{array}{l}\ x-y+z=8 \\ y-12z=-15 \\ z=1\end{array}\]

Using back-substitution, we obtain the solution as $(4,-3,1).$

Condensed — the full section is in OpenStax Precalculus 2e.

Key Concepts

An augmented matrix is one that contains the coefficients and constants of a system of equations. See .
A matrix augmented with the constant column can be represented as the original system of equations. See .
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. See , , and .
Row operations are performed on matrices to obtain row-echelon form. See .
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. See and .
A calculator can be used to solve systems of equations using matrices. See .
Many real-world problems can be solved using augmented matrices. See and .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Use the indicated row operations on the augmented matrix:
1. ⓐ Interchange rows 2 and 3.
2. ⓑ Multiply row 2 by 5.
3. ⓒ Multiply row 3 by −2 and add to row 1.
$[\begin{array}{lll}6 & -5 & 2 \\ 1 & 1 & -4 \\ 3 & -3 & 1\end{array}|\begin{array}{l}3 \\ 5 \\ -1\end{array}]$
答えを明らかにしろ

ⓐ Interchange rows 2 and 3.
ⓑ Multiply row 2 by 5.
ⓒ Multiply row 3 by −2 and add to row 1.
Use the indicated row operations on the augmented matrix:
- ⓐ Interchange rows 1 and 3.
- ⓑ Multiply row 3 by 3.
- ⓒ Multiply row 3 by 2 and add to row 2.
$\left[\begin{array}{llll}5 & -2 & -2 & -2 \\ 4 & -1 & -4 & 4 \\ -2 & 3 & 0 & -1\end{array}\right]$
Use the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: $[\begin{array}{llllll}1 & & & -1 & & \\ 4 & & & -8 & & \end{array}|\begin{array}{lll} & & 2 \\ & & 0\end{array}].$

答えを明らかにしろ

To make the 4 a 0, we could multiply row 1 by $-4$ and then add it to row 2.
Use the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix
$\left[\begin{array}{lll}1 & -1 & 2 \\ 3 & -6 & 2\end{array}\right]$

Solve the system of equations using matrices

$\{\begin{array}{l}3x+8y+2z=-5 \\ 2x+5y-3z=0 \\ x+2y-2z=-1\end{array}$

答えを明らかにしろ


Write the augmented matrix for the system of equations.
Interchange row 1 and row 3 to get a 1 in the first row and first column.
Using row operations, get zeros in column 1 below the 1

The entry in row 2, column 2 is now 1.
Continue the process until the matrix is in row-echelon form.

The matrix is now in row-echelon form.
Write the corresponding system of equations.
Use substitution to find the remaining variables.

Write the solution as an ordered pair or triple.
Check that the solution makes the original equations true.

Solve the system of equations using matrices
$\left\{\begin{array}{l}x-y-z=1 \\ -x+2y-3z=-4 \\ 3x-2y-7z=0\end{array}\right$
Write the augmented matrix for the given system of equations.
\[\begin{array}{l}\ x+2y-z=3 \\ 2x-y+2z=6 \\ x-3y+3z=4\end{array}\]

答えを明らかにしろ

The augmented matrix displays the coefficients of the variables, and an additional column for the constants.
\[[\begin{array}{lll}1 & 2 & -1 \\ 2 & -1 & 2 \\ 1 & -3 & 3\end{array}\ |\ \begin{array}{l}3 \\ 6 \\ 4\end{array}]\]
Write the augmented matrix of the given system of equations.
\[\begin{array}{l}4x-3y=11 \\ 3x+2y=4\end{array}\]

答えを明らかにしろ

$[\begin{array}{ll}4 & -3 \\ 3 & 2\end{array}|\begin{array}{l}11 \\ 4\end{array}]$
Find the system of equations from the augmented matrix.
\[[\begin{array}{lll}1 & -3 & -5 \\ 2 & -5 & -4 \\ -3 & 5 & 4\end{array}\ |\ \begin{array}{l}-2 \\ 5 \\ 6\end{array}]\]

答えを明らかにしろ

When the columns represent the variables $x,$ $y,$ and $z,$
\[[\begin{array}{lll}1 & -3 & -5 \\ 2 & -5 & -4 \\ -3 & 5 & 4\end{array}\ |\ \begin{array}{l}-2 \\ 5 \\ 6\end{array}]\to \begin{array}{l}x-3y-5z=-2 \\ 2x-5y-4z=5 \\ -3x+5y+4z=6\end{array}\]
Write the system of equations from the augmented matrix.
\[[\begin{array}{lll}1 & -1 & 1 \\ 2 & -1 & 3 \\ 0 & 1 & 1\end{array}|\begin{array}{l}5 \\ 1 \\ -9\end{array}]\]

答えを明らかにしろ

$\begin{array}{l}x-y+z=5 \\ 2x-y+3z=1 \\ y+z=-9\end{array}$
Solve the given system by Gaussian elimination.
\[\begin{array}{l}2x+3y=6 \\ x-y=\frac{1}{2}\end{array}\]

答えを明らかにしろ

First, we write this as an augmented matrix.
\[[\begin{array}{ll}2 & 3 \\ 1 & -1\end{array}\ |\ \begin{array}{l}6 \\ \frac{1}{2}\end{array}]\]
We want a 1 in row 1, column 1. This can be accomplished by interchanging row 1 and row 2.
\[{R}_{1}↔{R}_{2}\to [\begin{array}{lll}1 & -1 & \\ 2 & 3 & \end{array}|\begin{array}{ll} & \frac{1}{2} \\ & 6\end{array}]\]
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.
\[-2{R}_{1}+{R}_{2}={R}_{2}\to [\begin{array}{lll}1 & -1 & \\ 0 & 5 & \end{array}|\begin{array}{ll} & \frac{1}{2} \\ & 5\end{array}]\]
We only have one more step, to multiply row 2 by $\frac{1}{5}.$
\[\frac{1}{5}{R}_{2}={R}_{2}\to [\begin{array}{lll}1 & -1 & \\ 0 & 1 & \end{array}|\begin{array}{ll} & \frac{1}{2} \\ & 1\end{array}]\]
Use back-substitution. The second row of the matrix represents $y=1.$ Back-substitute $y=1$ into the first equation.
\[\begin{array}{l}x-(1)=\frac{1}{2} \\ x=\frac{3}{2}\end{array}\]
The solution is the point $(\frac{3}{2},1).$
Solve the given system by Gaussian elimination.
\[\begin{array}{l}4x+3y=11 \\ \ x-3y=-1\end{array}\]

答えを明らかにしろ

$(2,\ 1)$
Use Gaussian elimination to solve the given $2\times 2$ system of equations.
\[\begin{array}{l}\ 2x+y=1 \\ 4x+2y=6\end{array}\]

答えを明らかにしろ

Write the system as an augmented matrix.
\[[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\ |\ \begin{array}{l}1 \\ 6\end{array}]\]
Obtain a 1 in row 1, column 1. This can be accomplished by multiplying the first row by $\frac{1}{2}.$
\[\frac{1}{2}{R}_{1}={R}_{1}\to [\begin{array}{ll}1 & \frac{1}{2} \\ 4 & 2\end{array}\ |\ \begin{array}{l}\frac{1}{2} \\ 6\end{array}]\]
Next, we want a 0 in row 2, column 1. Multiply row 1 by $-4$ and add row 1 to row 2.
\[-4{R}_{1}+{R}_{2}={R}_{2}\to [\begin{array}{ll}1 & \frac{1}{2} \\ 0 & 0\end{array}\ |\ \begin{array}{l}\frac{1}{2} \\ 4\end{array}]\]
The second row represents the equation $0=4.$ Therefore, the system is inconsistent and has no solution.
Solve the system of equations.
\[\begin{array}{l}3x+4y=12 \\ 6x+8y=24\end{array}\]

答えを明らかにしろ

Perform row operations on the augmented matrix to try and achieve row-echelon form.
\[A=[\begin{array}{llll}3 & & 4 & \\ 6 & & 8 & \end{array}|\begin{array}{ll} & 12 \\ & 24\end{array}]\]\[\begin{array}{lllllllllllllllll} \\ \begin{array}{lllllllll}-\frac{1}{2}{R}_{2}+{R}_{1}={R}_{1}\to [\begin{array}{llll}0 & & 0 & \\ 6 & & 8 & \end{array}|\begin{array}{ll} & 0 \\ & 24\end{array}] \\ {R}_{1}↔{R}_{2}\to [\begin{array}{llll}6 & & 8 & \\ 0 & & 0 & \end{array}|\begin{array}{ll} & 24 \\ & 0\end{array}]\end{array}\end{array}\]
The matrix ends up with all zeros in the last row: $0y=0.$ 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 $y.$
\[\begin{array}{l}3x+4y=12 \\ 4y=12-3x \\ y=3-\frac{3}{4}x\end{array}\]
So the solution to this system is $(x,3-\frac{3}{4}x).$
Perform row operations on the given matrix to obtain row-echelon form.
\[[\begin{array}{lll}1 & -3 & 4 \\ 2 & -5 & 6 \\ -3 & 3 & 4\end{array}\ |\ \begin{array}{l}3 \\ 6 \\ 6\end{array}]\]

答えを明らかにしろ

The first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by $-2$ and add it to row 2. Then replace row 2 with the result.
\[-2{R}_{1}+{R}_{2}={R}_{2}\to [\begin{array}{llllll}1 & & -3 & & 4 & \\ 0 & & 1 & & -2 & \\ -3 & & 3 & & 4 & \end{array}|\begin{array}{ll} & 3 \\ & 0 \\ & 6\end{array}]\]
Next, obtain a zero in row 3, column 1.
\[3{R}_{1}+{R}_{3}={R}_{3}\to [\begin{array}{llllll}1 & & -3 & & 4 & \\ 0 & & 1 & & -2 & \\ 0 & & -6 & & 16 & \end{array}|\begin{array}{ll} & 3 \\ & 0 \\ & 15\end{array}]\]
Next, obtain a zero in row 3, column 2.
\[6{R}_{2}+{R}_{3}={R}_{3}\to [\begin{array}{llllll}1 & & -3 & & 4 & \\ 0 & & 1 & & -2 & \\ 0 & & 0 & & 4 & \end{array}|\begin{array}{ll} & 3 \\ & 0 \\ & 15\end{array}]\]
The last step is to obtain a 1 in row 3, column 3.
\[\frac{1}{4}{R}_{3}={R}_{3}\to [\begin{array}{lll}1 & -3 & 4 \\ 0 & 1 & -2 \\ 0 & 0 & 1\end{array}\ |\ \begin{array}{l}3 \\ 0 \\ \frac{15}{4}\end{array}]\]
Write the system of equations in row-echelon form.
\[\begin{array}{l}\ x-2y+3z=9 \\ -x+3y=-4 \\ 2x-5y+5z=17\end{array}\]

答えを明らかにしろ

$[\begin{array}{lll}1 & -\frac{5}{2} & \frac{5}{2} \\ \text{}0 & 1 & 5 \\ 0 & 0 & 1\end{array}|\begin{array}{l}\frac{17}{2} \\ 9 \\ 2\end{array}]$
Solve the system of linear equations using matrices.
\[\begin{array}{l}\begin{array}{l} \\ \\ x-y+z=8\end{array} \\ 2x+3y-z=-2 \\ 3x-2y-9z=9\end{array}\]

答えを明らかにしろ

First, we write the augmented matrix.
\[[\begin{array}{lll}1 & -1 & 1 \\ 2 & 3 & -1 \\ 3 & -2 & -9\end{array}\ |\ \begin{array}{l}8 \\ -2 \\ 9\end{array}]\]
Next, we perform row operations to obtain row-echelon form.
\[\begin{array}{lllllllllllllllllllllllllllllllllllllllll}-2{R}_{1}+{R}_{2}={R}_{2}\to [\begin{array}{llllll}1 & & -1 & & 1 & \\ 0 & & 5 & & -3 & \\ 3 & & -2 & & -9 & \end{array}|\begin{array}{ll} & 8 \\ & -18 \\ & 9\end{array}] & & & & -3{R}_{1}+{R}_{3}={R}_{3}\to [\begin{array}{llllll}1 & & -1 & & 1 & \\ 0 & & 5 & & -3 & \\ 0 & & 1 & & -12 & \end{array}|\begin{array}{ll} & 8 \\ & -18 \\ & -15\end{array}]\end{array}\]
The easiest way to obtain a 1 in row 2, column 2 is to interchange ${R}_{2}$ and ${R}_{3}.$
\[\text{Interchange}\ {R}_{2}\ \text{and}\ {R}_{3}\to [\begin{array}{lllllll}1 & & -1 & & 1 & & 8 \\ 0 & & 1 & & -12 & & -15 \\ 0 & & 5 & & -3 & & -18\end{array}]\]
Then
\[\begin{array}{lllllllllllllllllllllllllllllllllllllllll} \\ \begin{array}{lllllllllllllllllllllllllllllllllllllllll}-5{R}_{2}+{R}_{3}={R}_{3}\to [\begin{array}{llllll}1 & & -1 & & 1 & \\ 0 & & 1 & & -12 & \\ 0 & & 0 & & 57 & \end{array}|\begin{array}{ll} & 8 \\ & -15 \\ & 57\end{array}] & & & & -\frac{1}{57}{R}_{3}={R}_{3}\to [\begin{array}{llllll}1 & & -1 & & 1 & \\ 0 & & 1 & & -12 & \\ 0 & & 0 & & 1 & \end{array}|\begin{array}{ll} & 8 \\ & -15 \\ & 1\end{array}]\end{array}\end{array}\]
The last matrix represents the equivalent system.
\[\begin{array}{l}\ x-y+z=8 \\ y-12z=-15 \\ z=1\end{array}\]
Using back-substitution, we obtain the solution as $(4,-3,1).$
Solve the following system of linear equations using matrices.
\[\begin{array}{l}-x-2y+z=-1 \\ 2x+3y=2 \\ y-2z=0\end{array}\]

答えを明らかにしろ

Write the augmented matrix.
\[[\begin{array}{lll}-1 & -2 & 1 \\ 2 & 3 & 0 \\ 0 & 1 & -2\end{array}\ |\ \begin{array}{l}-1 \\ 2 \\ 0\end{array}]\]
First, multiply row 1 by $-1$ to get a 1 in row 1, column 1. Then, perform row operations to obtain row-echelon form.
\[-{R}_{1}\to [\begin{array}{lll}1 & 2 & -1 \\ 2 & 3 & 0 \\ 0 & 1 & -2\end{array}\ |\ \begin{array}{l}1 \\ 2 \\ 0\end{array}]\]\[{R}_{2}↔{R}_{3}\to [\begin{array}{lllll}1 & & 2 & & -1 \\ 0 & & 1 & & -2 \\ 2 & & 3 & & 0\end{array}\ |\begin{array}{ll} & 1 \\ & 0 \\ & 2\end{array}]\]\[-2{R}_{1}+{R}_{3}={R}_{3}\to [\begin{array}{llllll}1 & & 2 & & -1 & \\ 0 & & 1 & & -2 & \\ 0 & & -1 & & 2 & \end{array}|\begin{array}{ll} & 1 \\ & 0 \\ & 0\end{array}]\]\[{R}_{2}+{R}_{3}={R}_{3}\to [\begin{array}{llllll}1 & & 2 & & -1 & \\ 0 & & 1 & & -2 & \\ 0 & & 0 & & 0 & \end{array}|\begin{array}{ll} & 1 \\ & 1 \\ & 0\end{array}]\]
The last matrix represents the following system.
\[\begin{array}{l}\ x+2y-z=1 \\ y-2z=0 \\ 0=0\end{array}\]
We see by the identity $0=0$ that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the second equation for $y$ and substituting it into the first equation we can solve for $z$ in terms of $x.$
\[\begin{array}{l}x+2y-z=1 \\ y=2z \\ \\ x+2(2z)-z=1 \\ x+3z=1 \\ z=\frac{1-x}{3}\end{array}\]
Now we substitute the expression for $z$ into the second equation to solve for $y$ in terms of $x.$
\[\begin{array}{l}y-2z=0 \\ z=\frac{1-x}{3} \\ \\ y-2(\frac{1-x}{3})=0 \\ y=\frac{2-2x}{3}\end{array}\]
The generic solution is $(x,\frac{2-2x}{3},\frac{1-x}{3}).$
Solve the system using matrices.
\[\begin{array}{l}x+4y-z=4 \\ 2x+5y+8z=15 \\ x+3y-3z=1\end{array}\]

答えを明らかにしろ

$(1,\ 1,\ 1)$
Solve the system of equations.
\[\begin{array}{l}5x+3y+9z=-1 \\ -2x+3y-z=-2 \\ -x-4y+5z=1\end{array}\]

答えを明らかにしろ

Write the augmented matrix for the system of equations.
\[[\begin{array}{lll}5 & 3 & 9 \\ -2 & 3 & -1 \\ -1 & -4 & 5\end{array}\ |\ \begin{array}{l}-1 \\ -2 \\ -1\end{array}]\]
On the matrix page of the calculator, enter the augmented matrix above as the matrix variable $[A].$
\[[A]=[\begin{array}{lllllll}5 & & 3 & & 9 & & -1 \\ -2 & & 3 & & -1 & & -2 \\ -1 & & -4 & & 5 & & 1\end{array}]\]
Use the ref( function in the calculator, calling up the matrix variable $[A].$
\[\text{ref}([A])\]
Evaluate.
\[\begin{array}{llllllllll} \\ [\begin{array}{llll}1 & \ \frac{3}{5} & \ \frac{9}{5} & -\frac{1}{5} \\ 0 & \ 1 & \ \frac{13}{21} & -\frac{4}{7} \\ 0 & \ 0 & \ 1 & -\frac{24}{187}\end{array}]\to \begin{array}{l}x+\frac{3}{5}y+\frac{9}{5}z=-\frac{1}{5} \\ y+\frac{13}{21}z=-\frac{4}{7} \\ z=-\frac{24}{187}\end{array}\end{array}\]
Using back-substitution, the solution is $(\frac{61}{187},-\frac{92}{187},-\frac{24}{187}).$
Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate?

答えを明らかにしろ

We have a system of two equations in two variables. Let $x=$ the amount invested at 10.5% interest, and $y=$ the amount invested at 12% interest.
\[\begin{array}{l}\ x+y=12,000 \\ 0.105x+0.12y=1,335\end{array}\]
As a matrix, we have
\[[\begin{array}{ll}1 & 1 \\ 0.105 & 0.12\end{array}\ |\ \begin{array}{l}12,000 \\ 1,335\end{array}]\]
Multiply row 1 by $-0.105$ and add the result to row 2.
\[[\begin{array}{ll}1 & 1 \\ 0 & 0.015\end{array}\ |\ \begin{array}{l}12,000 \\ 75\end{array}]\]
Then,
\[\begin{array}{l}0.015y=75 \\ y=5,000\end{array}\]
So $12,000-5,000=7,000.$
Thus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest.
Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate?

答えを明らかにしろ

We have a system of three equations in three variables. Let $x$ be the amount invested at 5% interest, let $y$ be the amount invested at 8% interest, and let $z$ be the amount invested at 9% interest. Thus,
\[\begin{array}{l}\ x+y+z=10,000 \\ 0.05x+0.08y+0.09z=770 \\ 2x-z=0\end{array}\]
As a matrix, we have
\[[\begin{array}{lll}1 & 1 & 1 \\ 0.05 & 0.08 & 0.09 \\ 2 & 0 & -1\end{array}\ |\ \begin{array}{l}10,000 \\ 770 \\ 0\end{array}]\]
Now, we perform Gaussian elimination to achieve row-echelon form.
\[\begin{array}{lllllllllllllllllll}\begin{array}{lllllllllllllllllll} \\ -0.05{R}_{1}+{R}_{2}={R}_{2}\to [\begin{array}{llllll}1 & & 1 & & 1 & \\ 0 & & 0.03 & & 0.04 & \\ 2 & & 0 & & -1 & \end{array}|\begin{array}{ll} & 10,000 \\ & 270 \\ & 0\end{array}]\end{array} \\ -2{R}_{1}+{R}_{3}={R}_{3}\to [\begin{array}{llllll}1 & & 1 & & 1 & \\ 0 & & 0.03 & & 0.04 & \\ 0 & & -2 & & -3 & \end{array}|\begin{array}{ll} & 10,000 \\ & 270 \\ & -20,000\end{array}] \\ \frac{1}{0.03}{R}_{2}={R}_{2}\to [\begin{array}{llllll}0 & & 1 & & 1 & \\ 0 & & 1 & & \frac{4}{3} & \\ 0 & & -2 & & -3 & \end{array}|\begin{array}{ll} & 10,000 \\ & 9,000 \\ & -20,000\end{array}] \\ 2{R}_{2}+{R}_{3}={R}_{3}\to [\begin{array}{llllll}1 & & 1 & & 1 & \\ 0 & & 1 & & \frac{4}{3} & \\ 0 & & 0 & & -\frac{1}{3} & \end{array}|\begin{array}{ll} & 10,000 \\ & 9,000 \\ & -2,000\end{array}]\end{array}\]
The third row tells us $-\frac{1}{3}z=-2,000;$ thus $z=6,000.$
The second row tells us $y+\frac{4}{3}z=9,000.$ Substituting $z=6,000,$ we get
\[\begin{array}{l}y+\frac{4}{3}(6,000)=9,000 \\ y+8,000=9,000 \\ y=1,000\end{array}\]
The first row tells us $x+y+z=10,000.$ Substituting $y=1,000$ and $z=6,000,$ we get
\[\begin{array}{l}x+1,000+6,000=10,000 \\ x=3,000\end{array}\]
The answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.
A small shoe company took out a loan of $1,500,000 to expand their inventory. Part of the money was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the annual interest on all three loans was $130,500. Use matrices to find the amount borrowed at each rate.

答えを明らかにしろ

$150,000 at 7%, $750,000 at 8%, $600,000 at 10%
Can any system of linear equations be written as an augmented matrix? Explain why or why not. Explain how to write that augmented matrix.

答えを明らかにしろ

Yes. For each row, the coefficients of the variables are written across the corresponding row, and a vertical bar is placed; then the constants are placed to the right of the vertical bar.
Can any matrix be written as a system of linear equations? Explain why or why not. Explain how to write that system of equations.
Is there only one correct method of using row operations on a matrix? Try to explain two different row operations possible to solve the augmented matrix $[\begin{array}{ll}9 & 3 \\ 1 & -2\end{array}\ |\ \begin{array}{l}0 \\ 6\end{array}].$

答えを明らかにしろ

No, there are numerous correct methods of using row operations on a matrix. Two possible ways are the following: (1) Interchange rows 1 and 2. Then ${R}_{2}={R}_{2}-9{R}_{1}.$ (2) ${R}_{2}={R}_{1}-9{R}_{2}.$ Then divide row 1 by 9.
Can a matrix whose entry is 0 on the diagonal be solved? Explain why or why not. What would you do to remedy the situation?
Can a matrix that has 0 entries for an entire row have one solution? Explain why or why not.

答えを明らかにしろ

No. A matrix with 0 entries for an entire row would have either zero or infinitely many solutions.
$\begin{array}{l}8x-37y=8 \\ 2x+12y=3\end{array}$
$\begin{array}{l}\ 16y=4 \\ 9x-y=2\end{array}$

答えを明らかにしろ

$[\begin{array}{llll}0 & & 16 & \\ 9 & & -1 & \end{array}|\begin{array}{ll} & 4 \\ & 2\end{array}]$
$\begin{array}{l}\ 3x+2y+10z=3 \\ -6x+2y+5z=13 \\ 4x+z=18\end{array}$
$\begin{array}{l} \\ x+5y+8z=19 \\ 12x+3y=4 \\ 3x+4y+9z=-7\end{array}$

答えを明らかにしろ

$[\begin{array}{llllll}1 & & 5 & & 8 & \\ 12 & & 3 & & 0 & \\ 3 & & 4 & & 9 & \end{array}|\begin{array}{ll} & 19 \\ & 4 \\ & -7\end{array}]$
$\begin{array}{l}6x+12y+16z=4 \\ 19x-5y+3z=-9 \\ x+2y=-8\end{array}$
$[\begin{array}{ll}-2 & 5 \\ 6 & -18\end{array}\ |\ \begin{array}{l}5 \\ 26\end{array}]$

答えを明らかにしろ

$\begin{array}{l}-2x+5y=5 \\ 6x-18y=26\end{array}$
$[\begin{array}{ll}3 & 4 \\ 10 & 17\end{array}\ |\ \begin{array}{l}10 \\ 439\end{array}]$
$[\begin{array}{lll}3 & 2 & 0 \\ -1 & -9 & 4 \\ 8 & 5 & 7\end{array}\ |\ \begin{array}{l}3 \\ -1 \\ 8\end{array}]$

答えを明らかにしろ

$\begin{array}{l}3x+2y=3 \\ -x-9y+4z=-1 \\ 8x+5y+7z=8\end{array}$
$[\begin{array}{lll}8 & 29 & 1 \\ -1 & 7 & 5 \\ 0 & 0 & 3\end{array}\ |\ \begin{array}{l}43 \\ 38 \\ 10\end{array}]$
$[\begin{array}{lll}4 & 5 & -2 \\ 0 & 1 & 58 \\ 8 & 7 & -3\end{array}\ |\ \begin{array}{l}12 \\ 2 \\ -5\end{array}]$

答えを明らかにしろ

$\begin{array}{l}4x+5y-2z=12 \\ y+58z=2 \\ 8x+7y-3z=-5\end{array}$
$[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\ |\ \begin{array}{l}3 \\ 0\end{array}]$
$[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\ |\ \begin{array}{l}1 \\ 2\end{array}]$

答えを明らかにしろ

No solutions

Symbols used here

imaginary unit
i² = −1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

matrix
A rectangular array of numbers; a linear map.

\det A,\ |A|

determinant
Scaling factor of area/volume under A; zero means singular.

A^{-1},\ A^{T}

inverse, transpose
The matrix that undoes A; A with rows and columns swapped.

\lambda

lambda (eigenvalue)
The factor by which an eigenvector is stretched: Av = λv.

\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|

dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Solving Systems with Gaussian Elimination

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.
Use row operations on a matrix (IA 4.5.2)
Solve systems of equations using matrices (IA 4.5.3)
Interchange any two rows.
Multiply a row by any real number except 0.

Questions people ask

What does a determinant mean geometrically?

It is the factor by which the matrix scales area (2×2) or volume (3×3), with a negative sign if orientation flips. Zero means the matrix flattens space and cannot be undone.

What is an eigenvector?

A direction the matrix does not turn — it only stretches it by the eigenvalue. Along eigenvectors a complicated matrix acts like multiplication by a number.

Why is matrix multiplication not commutative?

Because a matrix is a transformation and AB means "do B, then A". Rotating then reflecting is not the same as reflecting then rotating.

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Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ここに Linear Algebra

Determinants Matrix inverse Eigenvalues and eigenvectors Matrix multiplication Row reduction Vector spaces, span and linear independence Orthogonality, projections and least squares Diagonalisation and matrix powers Linear transformations and change of basis