matrix practice problems pdf

7.5 Matrices and Matrix Operations

Learning objectives.

In this section, you will:

• Find the sum and difference of two matrices.
• Find scalar multiples of a matrix.
• Find the product of two matrices.

Two club soccer teams, the Wildcats and the Mud Cats, are hoping to obtain new equipment for an upcoming season. Table 1 shows the needs of both teams.

A goal costs $300; a ball costs $10; and a jersey costs $30. How can we find the total cost for the equipment needed for each team? In this section, we discover a method in which the data in the soccer equipment table can be displayed and used for calculating other information. Then, we will be able to calculate the cost of the equipment.

Finding the Sum and Difference of Two Matrices

To solve a problem like the one described for the soccer teams, we can use a matrix , which is a rectangular array of numbers. A row in a matrix is a set of numbers that are aligned horizontally. A column in a matrix is a set of numbers that are aligned vertically. Each number is an entry , sometimes called an element, of the matrix. Matrices (plural) are enclosed in [ ] or ( ), and are usually named with capital letters. For example, three matrices named A , B , A , B , and C C are shown below.

Describing Matrices

A matrix is often referred to by its size or dimensions: m × n m × n indicating m m rows and n n columns. Matrix entries are defined first by row and then by column. For example, to locate the entry in matrix A A identified as a i j , a i j , we look for the entry in row i , i , column j . j . In matrix A ,   A ,   shown below, the entry in row 2, column 3 is a 23 . a 23 .

A square matrix is a matrix with dimensions n × n , n × n , meaning that it has the same number of rows as columns. The 3 × 3 3 × 3 matrix above is an example of a square matrix.

A row matrix is a matrix consisting of one row with dimensions 1 × n . 1 × n .

A column matrix is a matrix consisting of one column with dimensions m × 1. m × 1.

A matrix may be used to represent a system of equations. In these cases, the numbers represent the coefficients of the variables in the system. Matrices often make solving systems of equations easier because they are not encumbered with variables. We will investigate this idea further in the next section, but first we will look at basic matrix operations .

A matrix is a rectangular array of numbers that is usually named by a capital letter: A , B , C , A , B , C , and so on. Each entry in a matrix is referred to as a i j , a i j , such that i i represents the row and j j represents the column. Matrices are often referred to by their dimensions: m × n m × n indicating m m rows and n n columns.

Finding the Dimensions of the Given Matrix and Locating Entries

Given matrix A : A :

• ⓐ What are the dimensions of matrix A ? A ?
• ⓑ What are the entries at a 31 a 31 and a 22 ? a 22 ? A = [ 2 1 0 2 4 7 3 1 − 2 ] A = [ 2 1 0 2 4 7 3 1 − 2 ]
• ⓐ The dimensions are 3 × 3 3 × 3 because there are three rows and three columns.
• ⓑ Entry a 31 a 31 is the number at row 3, column 1, which is 3. The entry a 22 a 22 is the number at row 2, column 2, which is 4. Remember, the row comes first, then the column.

Adding and Subtracting Matrices

We use matrices to list data or to represent systems. Because the entries are numbers, we can perform operations on matrices. We add or subtract matrices by adding or subtracting corresponding entries.

In order to do this, the entries must correspond. Therefore, addition and subtraction of matrices is only possible when the matrices have the same dimensions . We can add or subtract a 3 × 3 3 × 3 matrix and another 3 × 3 3 × 3 matrix, but we cannot add or subtract a 2 × 3 2 × 3 matrix and a 3 × 3 3 × 3 matrix because some entries in one matrix will not have a corresponding entry in the other matrix.

Given matrices A A and B B of like dimensions, addition and subtraction of A A and B B will produce matrix C C or matrix D D of the same dimension.

Matrix addition is commutative.

It is also associative.

Finding the Sum of Matrices

Find the sum of A A and B , B , given

Add corresponding entries.

Adding Matrix A and Matrix B

Find the sum of A A and B . B .

Add corresponding entries. Add the entry in row 1, column 1, a 11 , a 11 , of matrix A A to the entry in row 1, column 1, b 11 , b 11 , of B . B . Continue the pattern until all entries have been added.

Finding the Difference of Two Matrices

Find the difference of A A and B . B .

We subtract the corresponding entries of each matrix.

Finding the Sum and Difference of Two 3 x 3 Matrices

Given A A and B : B :

• ⓐ Find the sum.
• ⓑ Find the difference.
• ⓐ Add the corresponding entries. A + B = [ 2 − 10 − 2 14 12 10 4 − 2 2 ] + [ 6 10 − 2 0 − 12 − 4 − 5 2 − 2 ] = [ 2 + 6 − 10 + 10 − 2 − 2 14 + 0 12 − 12 10 − 4 4 − 5 − 2 + 2 2 − 2 ] = [ 8 0 − 4 14 0 6 − 1 0 0 ] A + B = [ 2 − 10 − 2 14 12 10 4 − 2 2 ] + [ 6 10 − 2 0 − 12 − 4 − 5 2 − 2 ] = [ 2 + 6 − 10 + 10 − 2 − 2 14 + 0 12 − 12 10 − 4 4 − 5 − 2 + 2 2 − 2 ] = [ 8 0 − 4 14 0 6 − 1 0 0 ]
• ⓑ Subtract the corresponding entries. A − B = [ 2 −10 −2 14 12 10 4 −2 2 ] − [ 6 10 −2 0 −12 −4 −5 2 −2 ] = [ 2 − 6 −10 − 10 −2 + 2 14 − 0 12 + 12 10 + 4 4 + 5 −2 − 2 2 + 2 ] = [ −4 −20 0 14 24 14 9 −4 4 ] A − B = [ 2 −10 −2 14 12 10 4 −2 2 ] − [ 6 10 −2 0 −12 −4 −5 2 −2 ] = [ 2 − 6 −10 − 10 −2 + 2 14 − 0 12 + 12 10 + 4 4 + 5 −2 − 2 2 + 2 ] = [ −4 −20 0 14 24 14 9 −4 4 ]

Add matrix A A and matrix B . B .

Finding Scalar Multiples of a Matrix

Besides adding and subtracting whole matrices, there are many situations in which we need to multiply a matrix by a constant called a scalar. Recall that a scalar is a real number quantity that has magnitude, but not direction. For example, time, temperature, and distance are scalar quantities. The process of scalar multiplication involves multiplying each entry in a matrix by a scalar. A scalar multiple is any entry of a matrix that results from scalar multiplication.

Consider a real-world scenario in which a university needs to add to its inventory of computers, computer tables, and chairs in two of the campus labs due to increased enrollment. They estimate that 15% more equipment is needed in both labs. The school’s current inventory is displayed in Table 2 .

Converting the data to a matrix, we have

To calculate how much computer equipment will be needed, we multiply all entries in matrix C C by 0.15.

We must round up to the next integer, so the amount of new equipment needed is

Adding the two matrices as shown below, we see the new inventory amounts.

Thus, Lab A will have 18 computers, 19 computer tables, and 19 chairs; Lab B will have 32 computers, 40 computer tables, and 40 chairs.

Scalar Multiplication

Scalar multiplication involves finding the product of a constant by each entry in the matrix. Given

the scalar multiple c A c A is

Scalar multiplication is distributive. For the matrices A , B , A , B , and C C with scalars a a and b , b ,

Multiplying the Matrix by a Scalar

Multiply matrix A A by the scalar 3.

Multiply each entry in A A by the scalar 3.

Given matrix B , B , find −2 B −2 B where

Finding the Sum of Scalar Multiples

Find the sum 3 A + 2 B . 3 A + 2 B .

First, find 3 A , 3 A , then 2 B . 2 B .

Now, add 3 A + 2 B . 3 A + 2 B .

Finding the Product of Two Matrices

In addition to multiplying a matrix by a scalar, we can multiply two matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. If A A is an m × r m × r matrix and B B is an r × n r × n matrix, then the product matrix A B A B is an m × n m × n matrix. For example, the product A B A B is possible because the number of columns in A A is the same as the number of rows in B . B . If the inner dimensions do not match, the product is not defined.

We multiply entries of A A with entries of B B according to a specific pattern as outlined below. The process of matrix multiplication becomes clearer when working a problem with real numbers.

To obtain the entries in row i i of A B , A B , we multiply the entries in row i i of A A by column j j in B B and add. For example, given matrices A A and B , B , where the dimensions of A A are 2 × 3 2 × 3 and the dimensions of B B are 3 × 3 , 3 × 3 , the product of A B A B will be a 2 × 3 2 × 3 matrix.

Multiply and add as follows to obtain the first entry of the product matrix A B . A B .

• To obtain the entry in row 1, column 1 of A B , A B , multiply the first row in A A by the first column in B , B , and add. [ a 11 a 12 a 13 ] [ b 11 b 21 b 31 ] = a 11 ⋅ b 11 + a 12 ⋅ b 21 + a 13 ⋅ b 31 [ a 11 a 12 a 13 ] [ b 11 b 21 b 31 ] = a 11 ⋅ b 11 + a 12 ⋅ b 21 + a 13 ⋅ b 31
• To obtain the entry in row 1, column 2 of A B , A B , multiply the first row of A A by the second column in B , B , and add. [ a 11 a 12 a 13 ] [ b 12 b 22 b 32 ] = a 11 ⋅ b 12 + a 12 ⋅ b 22 + a 13 ⋅ b 32 [ a 11 a 12 a 13 ] [ b 12 b 22 b 32 ] = a 11 ⋅ b 12 + a 12 ⋅ b 22 + a 13 ⋅ b 32
• To obtain the entry in row 1, column 3 of A B , A B , multiply the first row of A A by the third column in B , B , and add. [ a 11 a 12 a 13 ] [ b 13 b 23 b 33 ] = a 11 ⋅ b 13 + a 12 ⋅ b 23 + a 13 ⋅ b 33 [ a 11 a 12 a 13 ] [ b 13 b 23 b 33 ] = a 11 ⋅ b 13 + a 12 ⋅ b 23 + a 13 ⋅ b 33

We proceed the same way to obtain the second row of A B . A B . In other words, row 2 of A A times column 1 of B ; B ; row 2 of A A times column 2 of B ; B ; row 2 of A A times column 3 of B . B . When complete, the product matrix will be

Properties of Matrix Multiplication

For the matrices A , B , A , B , and C C the following properties hold.

• Matrix multiplication is associative: ( A B ) C = A ( B C ) . ( A B ) C = A ( B C ) .
• Matrix multiplication is distributive: C ( A + B ) = C A + C B , ( A + B ) C = A C + B C . C ( A + B ) = C A + C B , ( A + B ) C = A C + B C .

Note that matrix multiplication is not commutative.

Multiplying Two Matrices

Multiply matrix A A and matrix B . B .

First, we check the dimensions of the matrices. Matrix A A has dimensions 2 × 2 2 × 2 and matrix B B has dimensions 2 × 2. 2 × 2. The inner dimensions are the same so we can perform the multiplication. The product will have the dimensions 2 × 2. 2 × 2.

We perform the operations outlined previously.

• ⓐ Find A B . A B .
• ⓑ Find B A . B A .
• ⓐ As the dimensions of A A are 2 × 3 2 × 3 and the dimensions of B B are 3 × 2 , 3 × 2 , these matrices can be multiplied together because the number of columns in A A matches the number of rows in B . B . The resulting product will be a 2 × 2 2 × 2 matrix, the number of rows in A A by the number of columns in B . B . A B = [ −1 2 3 4 0 5 ]    [ 5 −1 − 4 0 2 3 ] = [ −1 ( 5 ) + 2 ( −4 ) + 3 ( 2 ) −1 ( −1 ) + 2 ( 0 ) + 3 ( 3 ) 4 ( 5 ) + 0 ( −4 ) + 5 ( 2 ) 4 ( −1 ) + 0 ( 0 ) + 5 ( 3 ) ] = [ −7 10 30 11 ] A B = [ −1 2 3 4 0 5 ]    [ 5 −1 − 4 0 2 3 ] = [ −1 ( 5 ) + 2 ( −4 ) + 3 ( 2 ) −1 ( −1 ) + 2 ( 0 ) + 3 ( 3 ) 4 ( 5 ) + 0 ( −4 ) + 5 ( 2 ) 4 ( −1 ) + 0 ( 0 ) + 5 ( 3 ) ] = [ −7 10 30 11 ]
• ⓑ The dimensions of B B are 3 × 2 3 × 2 and the dimensions of A A are 2 × 3. 2 × 3. The inner dimensions match so the product is defined and will be a 3 × 3 3 × 3 matrix. B A = [ 5 −1 −4 0 2 3 ]    [ −1 2 3 4 0 5 ] = [ 5 ( −1 ) + −1 ( 4 ) 5 ( 2 ) + −1 ( 0 ) 5 ( 3 ) + −1 ( 5 ) −4 ( −1 ) + 0 ( 4 ) −4 ( 2 ) + 0 ( 0 ) −4 ( 3 ) + 0 ( 5 ) 2 ( −1 ) + 3 ( 4 ) 2 ( 2 ) + 3 ( 0 ) 2 ( 3 ) + 3 ( 5 ) ] = [ −9 10 10 4 −8 −12 10 4 21 ] B A = [ 5 −1 −4 0 2 3 ]    [ −1 2 3 4 0 5 ] = [ 5 ( −1 ) + −1 ( 4 ) 5 ( 2 ) + −1 ( 0 ) 5 ( 3 ) + −1 ( 5 ) −4 ( −1 ) + 0 ( 4 ) −4 ( 2 ) + 0 ( 0 ) −4 ( 3 ) + 0 ( 5 ) 2 ( −1 ) + 3 ( 4 ) 2 ( 2 ) + 3 ( 0 ) 2 ( 3 ) + 3 ( 5 ) ] = [ −9 10 10 4 −8 −12 10 4 21 ]

Notice that the products A B A B and B A B A are not equal.

This illustrates the fact that matrix multiplication is not commutative.

Is it possible for AB to be defined but not BA ?

Yes, consider a matrix A with dimension 3 × 4 3 × 4 and matrix B with dimension 4 × 2. 4 × 2. For the product AB the inner dimensions are 4 and the product is defined, but for the product BA the inner dimensions are 2 and 3 so the product is undefined.

Using Matrices in Real-World Problems

Let’s return to the problem presented at the opening of this section. We have Table 3 , representing the equipment needs of two soccer teams.

We are also given the prices of the equipment, as shown in Table 4 .

We will convert the data to matrices. Thus, the equipment need matrix is written as

The cost matrix is written as

We perform matrix multiplication to obtain costs for the equipment.

The total cost for equipment for the Wildcats is $2,520, and the total cost for equipment for the Mud Cats is $3,840.

Given a matrix operation, evaluate using a calculator.

• Save each matrix as a matrix variable [ A ] , [ B ] , [ C ] , ... [ A ] , [ B ] , [ C ] , ...
• Enter the operation into the calculator, calling up each matrix variable as needed.
• If the operation is defined, the calculator will present the solution matrix; if the operation is undefined, it will display an error message.

Using a Calculator to Perform Matrix Operations

Find A B − C A B − C given

On the matrix page of the calculator, we enter matrix A A above as the matrix variable [ A ] , [ A ] , matrix B B above as the matrix variable [ B ] , [ B ] , and matrix C C above as the matrix variable [ C ] . [ C ] .

On the home screen of the calculator, we type in the problem and call up each matrix variable as needed.

The calculator gives us the following matrix.

Access these online resources for additional instruction and practice with matrices and matrix operations.

• Dimensions of a Matrix
• Matrix Addition and Subtraction
• Matrix Operations
• Matrix Multiplication

7.5 Section Exercises

Can we add any two matrices together? If so, explain why; if not, explain why not and give an example of two matrices that cannot be added together.

Can we multiply any column matrix by any row matrix? Explain why or why not.

Can both the products A B A B and B A B A be defined? If so, explain how; if not, explain why.

Can any two matrices of the same size be multiplied? If so, explain why, and if not, explain why not and give an example of two matrices of the same size that cannot be multiplied together.

Does matrix multiplication commute? That is, does A B = B A ? A B = B A ? If so, prove why it does. If not, explain why it does not.

For the following exercises, use the matrices below and perform the matrix addition or subtraction. Indicate if the operation is undefined.

A + B A + B

C + D C + D

A + C A + C

B − E B − E

C + F C + F

D − B D − B

For the following exercises, use the matrices below to perform scalar multiplication.

1 2 C 1 2 C

100 D 100 D

For the following exercises, use the matrices below to perform matrix multiplication.

For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed.

A + B − C A + B − C

4 A + 5 D 4 A + 5 D

2 C + B 2 C + B

3 D + 4 E 3 D + 4 E

C −0.5 D C −0.5 D

100 D −10 E 100 D −10 E

For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. (Hint: A 2 = A ⋅ A A 2 = A ⋅ A )

B 2 A 2 B 2 A 2

A 2 B 2 A 2 B 2

( A B ) 2 ( A B ) 2

( B A ) 2 ( B A ) 2

( A B ) C ( A B ) C

A ( B C ) A ( B C )

For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. Use a calculator to verify your solution.

A B C A B C

For the following exercises, use the matrix below to perform the indicated operation on the given matrix.

Using the above questions, find a formula for B n . B n . Test the formula for B 201 B 201 and B 202 , B 202 , using a calculator.

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9.6: Solving Systems with Gaussian Elimination

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Learning Objectives

• 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.

Carl Friedrich Gauss lived during the late \(18^{th}\) century and early \(19^{th}\) 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 \(2 × 2\) system of equations.

\[\begin{align*} 3x+4y&= 7\\ 4x-2y&= 5 \end{align*}\]

We can write this system as an augmented matrix:

\(\left[ \begin{array}{cc|c} 3&4&7\\4&-2&5\end{array} \right]\)

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

\(\begin{bmatrix}3&4\\4&−2\end{bmatrix}\)

A three-by-three system of equations such as

\[\begin{align*} 3x-y-z&= 0\\ x+y&= 5\\ 2x-3z&= 2 \end{align*}\]

has a coefficient matrix

\(\begin{bmatrix}3&−1&−1\\1&1&0\\2&0&−3\end{bmatrix}\)

and is represented by the augmented matrix

\(\left[ \begin{array}{ccc|c}3&−1&−1&0\\1&1&0&5\\2&0&−3&2\end{array} \right]\)

Notice that the matrix is written so that the variables line up in their own columns: \(x\)-terms go in the first column, \(y\)-terms in the second column, and \(z\)-terms in the third column. It is very important that each equation is written in standard form \(ax+by+cz=d\) 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, 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.
• If there are \(z\)-terms, write the coefficients as the numbers down the third column.
• Draw a vertical line and write the constants to the right of the line.

Example \(\PageIndex{1}\): Writing the Augmented Matrix for a System of Equations

Write the augmented matrix for the given system of equations.

\[\begin{align*} x+2y-z&= 3\\ 2x-y+2z&= 6\\ x-3y+3z&= 4 \end{align*}\]

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

\(\left[ \begin{array}{ccc|c}1&2&−1&3\\2&−1&2&6\\1&−3&3&4\end{array} \right]\)

Exercise \(\PageIndex{1}\)

Write the augmented matrix of the given system of equations.

\[\begin{align*} 4x-3y&= 11\\ 3x+2y&= 4 \end{align*}\]

\(\left[ \begin{array}{cc|c} 4&−3&11\\3&2&4\end{array} \right]\)

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 \(\PageIndex{2}\): Writing a System of Equations from an Augmented Matrix Form

Find the system of equations from the augmented matrix.

\(\left[ \begin{array}{ccc|c}1&−3&−5&-2\\2&−5&−4&5\\−3&5&4&6 \end{array} \right]\)

When the columns represent the variables \(x\), \(y\), and \(z\),

\[\left[ \begin{array}{ccc|c}1&-3&-5&-2\\2&-5&-4&5\\-3&5&4&6 \end{array} \right] \rightarrow \begin{align*} x-3y-5z&= -2\\ 2x-5y-4z&= 5\\ -3x+5y+4z&= 6 \end{align*}\]

Exercise \(\PageIndex{2}\)

Write the system of equations from the augmented matrix.

\(\left[ \begin{array}{ccc|c}1&−1& 1&5\\2&−1&3&1\\0&1&1&-9\end{array}\right]\)

\(\begin{align*} x-y+z&= 5\\ 2x-y+3z&= 1\\ y+z&= -9 \end{align*}\)

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.

Row-echelon form \(\begin{bmatrix}1&a&b\\0&1&d\\0&0&1\end{bmatrix}\)

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: \(cR_i\))
• Add the product of a row multiplied by a constant to another row. (Notation: \(R_i+cR_j\))

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 matrix \(A\) with the number \(1\) as the entry down the main diagonal and have all zeros below.

\(A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\xrightarrow{After\space Gaussian\space elimination} A=\begin{bmatrix}1&b_{12}& b_{13}\\0&1&b_{23}\\0&0&1\end{bmatrix}\)

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.
• Use row operations to obtain a \(1\) in row 3, column 3.
• 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.

Example \(\PageIndex{3}\): Solving a \(2×2\) System by Gaussian Elimination

Solve the given system by Gaussian elimination.

\[\begin{align*} 2x+3y&= 6\\ x-y&= \dfrac{1}{2} \end{align*}\]

First, we write this as an augmented matrix.

\(\left[ \begin{array}{cc|c} 2&3&6\\1&−1&12\end{array} \right]\)

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

\(R_1\leftrightarrow R_2\rightarrow \left[ \begin{array}{cc|c} 1&−1&12\\2&3&6\end{array} \right]\)

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.

\(-2R_1+R_2=R_2\rightarrow \left[ \begin{array}{cc|c} 1&−1&12\\0&5&5\end{array} \right]\)

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

\(\dfrac{1}{5}R_2=R_2\rightarrow \left[ \begin{array}{cc|c} 1&−1&12\\0&1&1\end{array} \right]\)

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

\[\begin{align*} x-(1)&= \dfrac{1}{2}\\ x&= \dfrac{3}{2} \end{align*}\]

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

Exercise \(\PageIndex{3}\)

\[\begin{align*} 4x+3y&= 11\\ x-3y&= -1 \end{align*}\]

Example \(\PageIndex{4}\): Using Gaussian Elimination to Solve a System of Equations

Use Gaussian elimination to solve the given \(2 × 2\) system of equations .

\[\begin{align*} 2x+y&= 1\\ 4x+2y&= 6 \end{align*}\]

Write the system as an augmented matrix .

\(\left[ \begin{array}{cc|c} 2&1&1\\4&2&6\end{array} \right]\)

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

\(\dfrac{1}{2} R_1=R_1\rightarrow \left[ \begin{array}{cc|c} 1&\dfrac{1}{2}&\dfrac{1}{2}\\4&2&6\end{array} \right]\)

Next, we want a \(0\) in row 2, column 1. Multiply row 1 by \(−4\) and add row 1 to row 2.

\(-4R_1+R_2=R_2\rightarrow \left[ \begin{array}{cc|c} 1&\dfrac{1}{2}&\dfrac{1}{2}\\0&0&4\end{array} \right]\)

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

Example \(\PageIndex{5}\): Solving a Dependent System

Solve the system of equations.

\[\begin{align*} 3x+4y&= 12\\ 6x+8y&= 24 \end{align*}\]

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

\(A=\left[ \begin{array}{cc|c} 3&4&12\\6&8&24\end{array} \right]\)

\(-\dfrac{1}{2}R_2+R_1=R_1\rightarrow \left[ \begin{array}{cc|c} 0&0&0\\6&8&24\end{array} \right]\)

\(R_1\leftrightarrow R_2=\left[ \begin{array}{cc|c} 6&8&24\\0&0&0\end{array} \right]\)

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 \(y\).

\[\begin{align*} 3x+4y&= 12\\ 4y&= 12-3x\\ y&= 3-\dfrac{3}{4}x \end{align*}\]

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

Example \(\PageIndex{6}\): Performing Row Operations on a \(3×3\) Augmented Matrix to Obtain Row-Echelon Form

Perform row operations on the given matrix to obtain row-echelon form.

\(\left[ \begin{array}{ccc|c} 1&-3&4&3\\2&-5&6&6\\-3&3&4&6\end{array} \right]\)

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.

\(-2R_1+R_2=R_2 \left[ \begin{array}{ccc|c} 1&-3&4&3\\0&1&-2&0\\-3&3&4&6\end{array} \right]\)

Next, obtain a zero in row 3, column 1.

\(3R_1+R_3=R_3 \left[ \begin{array}{ccc|c} 1&-3&4&3\\0&1&-2&0\\0&-6&16&15\end{array} \right]\)

Next, obtain a zero in row 3, column 2.

\(6R_2+R_3=R_3 \left[ \begin{array}{ccc|c} 1&-3&4&3\\0&1&-2&0\\0&0&4&15\end{array} \right]\)

The last step is to obtain a 1 in row 3, column 3.

\(\dfrac{1}{3}R_3=R_3 \left[ \begin{array}{ccc|c} 1&-3&4&3\\0&1&-2&0\\0&0&1&\dfrac{21}{2}\end{array} \right]\)

Exercise \(\PageIndex{4}\)

Write the system of equations in row-echelon form.

\[\begin{align*} x−2y+3z &= 9 \\ −x+3y &= −4 \\ 2x−5y+5z &= 17 \end{align*}\]

\(\left[ \begin{array}{ccc|c} 1&-\dfrac{5}{2}&\dfrac{5}{2}&\dfrac{17}{2}\\0&1&5&9\\0&0&1&2\end{array} \right]\)

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 \(\PageIndex{7}\): Solving a System of Linear Equations Using Matrices

Solve the system of linear equations using matrices.

\[\begin{align*} x-y+z&= 8\\ 2x+3y-z&= -2\\ 3x-2y-9z&= 9 \end{align*}\]

First, we write the augmented matrix.

\(\left[ \begin{array}{ccc|c} 1&-1&1&8\\2&3&-1&-2\\3&-2&-9&9\end{array} \right]\)

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

\(−2R_1+R_2=R_2\rightarrow \left[ \begin{array}{ccc|c} 1&-1&1&8\\0&5&-3&-18\\3&-2&-9&9\end{array} \right]\)

\(−3R_1+R_3=R_3\rightarrow \left[ \begin{array}{ccc|c} 1&-1&1&8\\0&5&-3&-18\\0&1&-12&-15\end{array} \right]\)

The easiest way to obtain a \(1\) in row 2 of column 1 is to interchange \(R_2\) and \(R_3\).

\(Interchange\space R_2\space and\space R_3\rightarrow\left[ \begin{array}{ccc|c} 1&-1&1&8\\0&1&-12&-15\\0&5&-3&-18\end{array} \right]\)

\(−5R_2+R_3=R_3\rightarrow\left[ \begin{array}{ccc|c} 1&-1&1&8\\0&1&-12&-15\\0&0&57&57\end{array} \right]\)

\(−\dfrac{1}{57}R_3=R_3\rightarrow\left[ \begin{array}{ccc|c} 1&-1&1&8\\0&1&-12&-15\\0&0&1&1\end{array} \right]\)

The last matrix represents the equivalent system.

\[\begin{align*} x−y+z &= 8 \\ y−12z &= −15 \\ z &= 1 \end{align*}\]

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

Example \(\PageIndex{8}\): Solving a Dependent System of Linear Equations Using Matrices

Solve the following system of linear equations using matrices.

\[\begin{align*} −x−2y+z &= −1 \\ 2x+3y &= 2 \\ y−2z &= 0 \end{align*}\]

Write the augmented matrix.

\(\left[ \begin{array}{ccc|c} -1&-2&1&-1\\2&3&0&2\\0&1&-2&0\end{array} \right]\)

First, multiply row 1 by \(−1\) to get a \(1\) in row 1, column 1. Then, perform row operations to obtain row-echelon form.

\(-R_1\rightarrow \left[ \begin{array}{ccc|c} 1&2&-1&1\\2&3&0&2\\0&1&-2&0\end{array} \right]\)

\(R_2\leftrightarrow R_3\rightarrow \left[ \begin{array}{ccc|c} 1&2&-1&1\\0&1&-2&0\\2&3&0&2\end{array} \right]\)

\(−2R_1+R_3=R_3\rightarrow \left[ \begin{array}{ccc|c} 1&2&-1&1\\0&1&-2&0\\0&-1&2&0\end{array} \right]\)

\(R_2+R_3=R_3\rightarrow \left[ \begin{array}{ccc|c} 1&2&-1&1\\0&1&-2&0\\0&0&0&0\end{array} \right]\)

The last matrix represents the following system.

\[\begin{align*} x+2y−z &= 1 \\ y−2z &= 0 \\ 0 &= 0 \end{align*}\]

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 \(x\).

\[\begin{align*} x+2y−z &= 1 \\ y &= 2z \\ x+2(2z)−z &= 1 \\ x+3z &= 1 \\ z &=\dfrac{1−x}{3} \end{align*}\]

Now we substitute the expression for \(z\) into the second equation to solve for \(y\) in terms of \(x\).

\[\begin{align*} y−2z &= 0 \\ z &= \dfrac{1−x}{3} \\ y−2\left(\dfrac{1−x}{3}\right) &= 0 \\ y &= \dfrac{2−2x}{3} \end{align*}\]

The generic solution is \(\left(x,\dfrac{2−2x}{3},\dfrac{1−x}{3}\right)\).

Exercise \(\PageIndex{5}\)

Solve the system using matrices.

\[\begin{align*} x+4y-z&= 4\\ 2x+5y+8z&= 1\\ 5x+3y-3z&= 1 \end{align*}\]

\((1,1,1)\)

Q&A: 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.

How to: Given a system of equations, solve with matrices using a calculator

• Save the augmented matrix as a matrix variable \([A], [B], [C], ….\)
• Use the ref( function in the calculator, calling up each matrix variable as needed.

Example \(\PageIndex{9A}\): Solving Systems of Equations with Matrices Using a Calculator

\[\begin{align*} 5x+3y+9z&= -1\\ -2x+3y-z&= -2\\ -x-4y+5z&= 1 \end{align*}\]

Write the augmented matrix for the system of equations.

\(\left[ \begin{array}{ccc|c} 5&3&9&-1\\-2&3&-1&-2\\-1&-4&5&1\end{array} \right]\)

On the matrix page of the calculator, enter the augmented matrix above as the matrix variable \([A]\).

\([A]=\left[ \begin{array}{ccc|c} 5&3&9&-1\\-2&3&-1&-2\\-1&-4&5&1\end{array} \right]\)

Use the ref( function in the calculator, calling up the matrix variable \([A]\).

\[\begin{array}{cc} {\left[ \begin{array}{ccc|c} 1&\dfrac{3}{5}&\dfrac{9}{5}&\dfrac{1}{5}\\0&1&\dfrac{13}{21}&-\dfrac{4}{7}\\0&0&1&-\dfrac{24}{187}\end{array} \right] \rightarrow} & {\begin{align*} x+\dfrac{3}{5}y+\dfrac{9}{5}z &= -\dfrac{1}{5} \\ y+\dfrac{13}{21}z &= -\dfrac{4}{7} \\ z &= -\dfrac{24}{187} \end{align*}} \end{array}\]

Using back-substitution, the solution is \(\left(\dfrac{61}{187},−\dfrac{92}{187},−\dfrac{24}{187}\right)\).

Example \(\PageIndex{9B}\): Applying \(2×2\) Matrices to Finance

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?

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.

\[\begin{align*} x+y&= 12,000\\ 0.105x+0.12y&= 1,335 \end{align*}\]

As a matrix, we have

\(\left[ \begin{array}{cc|c} 1&1&12,000\\0.105&0.12&1,335\end{array} \right]\)

Multiply row 1 by \(−0.105\) and add the result to row 2.

\(\left[ \begin{array}{cc|c} 1&1&12,000\\0&0.015&75\end{array} \right]\)

\[\begin{align*} 0.015y &= 75 \\ y &= 5,000 \end{align*}\]

So \(12,000−5,000=7,000\).

Thus, \($5,000\) was invested at \(12%\) interest and \($7,000\) at \(10.5%\) interest.

Example \(\PageIndex{10}\): Applying \(3×3\) Matrices to Finance

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?

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,

\[\begin{align*} x+y+z &= 10,000 \\ 0.05x+0.08y+0.09z &= 770 \\ 2x−z &= 0 \end{align*}\]

\(\left[ \begin{array}{ccc|c} 1&1&1&10,000\\0.05&0.08&0.09&770\\2&0&-1&0\end{array} \right]\)

Now, we perform Gaussian elimination to achieve row-echelon form.

\(−0.05R_1+R_2=R_2\rightarrow \left[ \begin{array}{ccc|c} 1&1&1&10,000\\0&0.03&0.04&270\\2&0&-1&0\end{array} \right]\)

\(−2R_1+R_3=R_3\rightarrow \left[ \begin{array}{ccc|c} 1&1&1&10,000\\0&0.03&0.04&270\\0&-2&-3&-20,000\end{array} \right]\)

\(\dfrac{1}{0.03}R_2=R_2\rightarrow \left[ \begin{array}{ccc|c} 1&1&1&10,000\\0&1&\dfrac{4}{3}&9,000\\0&-2&-3&-20,000\end{array} \right]\)

\(2R_2+R_3=R_3\rightarrow \left[ \begin{array}{ccc|c} 1&1&1&10,000\\0&1&\dfrac{4}{3}&9,000\\0&0&-\dfrac{1}{3}&-2,000\end{array} \right]\)

The third row tells us \(−\dfrac{1}{3}z=−2,000\); thus \(z=6,000\).

The second row tells us \(y+\dfrac{4}{3}z=9,000\). Substituting \(z=6,000\),we get

\[\begin{align*} y+\dfrac{4}{3}(6,000) &= 9,000 \\ y+8,000 &= 9,000 \\ y &= 1,000 \end{align*}\]

The first row tells us \(x+y+z=10,000\). Substituting \(y=1,000\) and \(z=6,000\),we get

\[\begin{align*} x+1,000+6,000 &= 10,000 \\ x &= 3,000 \end{align*}\]

The answer is \($3,000\) invested at \(5%\) interest, \($1,000\) invested at \(8%\), and \($6,000\) invested at \(9%\) interest.

Exercise \(\PageIndex{6}\)

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.

\($150,000\) at \(7%\), \($750,000\) at \(8%\), \($600,000\) at \(10%\)

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
• Solve a System of Three Equations Using an Augmented Matrix
• Augmented Matrices on the Calculator

Key Concepts

• An augmented matrix is one that contains the coefficients and constants of a system of equations. See Example \(\PageIndex{1}\).
• A matrix augmented with the constant column can be represented as the original system of equations. See Example \(\PageIndex{2}\).
• 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 Example \(\PageIndex{3}\), Example \(\PageIndex{4}\), and Example \(\PageIndex{5}\).
• Row operations are performed on matrices to obtain row-echelon form. See Example \(\PageIndex{6}\).
• 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 Example \(\PageIndex{7}\) and Example \(\PageIndex{8}\).
• A calculator can be used to solve systems of equations using matrices. See Example \(\PageIndex{9}\).
• Many real-world problems can be solved using augmented matrices. See Example \(\PageIndex{10}\) and Example \(\PageIndex{11}\).