how to solve 2 equations with matrix

Solving Systems of Linear Equations Using Matrices

Hi there! This page is only going to make sense when you know a little about Systems of Linear Equations and Matrices , so please go and learn about those if you don't know them already.

The Example

One of the last examples on Systems of Linear Equations was this one:

Example: Solve

• x + y + z = 6
• 2y + 5z = −4
• 2x + 5y − z = 27

We went on to solve it using "elimination", but we can also solve it using Matrices!

Using Matrices makes life easier because we can use a computer program (such as the Matrix Calculator ) to do all the "number crunching".

But first we need to write the question in Matrix form.

In Matrix Form?

OK. A Matrix is an array of numbers:

Well, think about the equations:

They could be turned into a table of numbers like this:

We could even separate the numbers before and after the "=" into:

Now it looks like we have 2 Matrices.

In fact we have a third one, which is [x y z] :

Why does [x y z] go there? Because when we Multiply Matrices we use the "Dot Product" like this:

Which is the first of our original equations above (you might like to check that). Here it is for the second line.

 Try the third line for yourself.

The Matrix Solution

We can shorten this:

• A is the 3x3 matrix of x, y and z coefficients
• X is x, y and z , and
• B is 6, −4 and 27

Then (as shown on the Inverse of a Matrix page) the solution is this:

What does that mean?

It means that we can find the X matrix (the values of x, y and z) by multiplying the inverse of the A matrix by the B matrix .

So let's go ahead and do that.

First, we need to find the inverse of the A matrix (assuming it exists!)

Using the Matrix Calculator we get this:

(I left the 1/determinant outside the matrix to make the numbers simpler)

Then multiply A -1 by B (we can use the Matrix Calculator again):

And we are done! The solution is:

x = 5 y = 3 z = −2

Just like on the Systems of Linear Equations page.

Quite neat and elegant, and the human does the thinking while the computer does the calculating.

Just For Fun ... Do It Again!

For fun (and to help you learn), let us do this all again, but put matrix "X" first.

I want to show you this way, because many people think the solution above is so neat it must be the only way.

So we will solve it like this:

And because of the way that matrices are multiplied we need to set up the matrices differently now. The rows and columns have to be switched over ("transposed"):

And XA = B looks like this:

Then (also shown on the Inverse of a Matrix page) the solution is this:

This is what we get for A -1 :

In fact it is just like the Inverse we got before, but Transposed (rows and columns swapped over).

Next we multiply B by A -1 :

And the solution is the same:

x = 5 , y = 3 and z = −2

It didn't look as neat as the previous solution, but it does show us that there is more than one way to set up and solve matrix equations. Just be careful about the rows and columns!

• 4.5 Solve Systems of Equations Using Matrices
• Introduction
• 1.1 Use the Language of Algebra
• 1.2 Integers
• 1.3 Fractions
• 1.4 Decimals
• 1.5 Properties of Real Numbers
• Key Concepts
• Review Exercises
• Practice Test
• 2.1 Use a General Strategy to Solve Linear Equations
• 2.2 Use a Problem Solving Strategy
• 2.3 Solve a Formula for a Specific Variable
• 2.4 Solve Mixture and Uniform Motion Applications
• 2.5 Solve Linear Inequalities
• 2.6 Solve Compound Inequalities
• 2.7 Solve Absolute Value Inequalities
• 3.1 Graph Linear Equations in Two Variables
• 3.2 Slope of a Line
• 3.3 Find the Equation of a Line
• 3.4 Graph Linear Inequalities in Two Variables
• 3.5 Relations and Functions
• 3.6 Graphs of Functions
• 4.1 Solve Systems of Linear Equations with Two Variables
• 4.2 Solve Applications with Systems of Equations
• 4.3 Solve Mixture Applications with Systems of Equations
• 4.4 Solve Systems of Equations with Three Variables
• 4.6 Solve Systems of Equations Using Determinants
• 4.7 Graphing Systems of Linear Inequalities
• 5.1 Add and Subtract Polynomials
• 5.2 Properties of Exponents and Scientific Notation
• 5.3 Multiply Polynomials
• 5.4 Dividing Polynomials
• Introduction to Factoring
• 6.1 Greatest Common Factor and Factor by Grouping
• 6.2 Factor Trinomials
• 6.3 Factor Special Products
• 6.4 General Strategy for Factoring Polynomials
• 6.5 Polynomial Equations
• 7.1 Multiply and Divide Rational Expressions
• 7.2 Add and Subtract Rational Expressions
• 7.3 Simplify Complex Rational Expressions
• 7.4 Solve Rational Equations
• 7.5 Solve Applications with Rational Equations
• 7.6 Solve Rational Inequalities
• 8.1 Simplify Expressions with Roots
• 8.2 Simplify Radical Expressions
• 8.3 Simplify Rational Exponents
• 8.4 Add, Subtract, and Multiply Radical Expressions
• 8.5 Divide Radical Expressions
• 8.6 Solve Radical Equations
• 8.7 Use Radicals in Functions
• 8.8 Use the Complex Number System
• 9.1 Solve Quadratic Equations Using the Square Root Property
• 9.2 Solve Quadratic Equations by Completing the Square
• 9.3 Solve Quadratic Equations Using the Quadratic Formula
• 9.4 Solve Equations in Quadratic Form
• 9.5 Solve Applications of Quadratic Equations
• 9.6 Graph Quadratic Functions Using Properties
• 9.7 Graph Quadratic Functions Using Transformations
• 9.8 Solve Quadratic Inequalities
• 10.1 Finding Composite and Inverse Functions
• 10.2 Evaluate and Graph Exponential Functions
• 10.3 Evaluate and Graph Logarithmic Functions
• 10.4 Use the Properties of Logarithms
• 10.5 Solve Exponential and Logarithmic Equations
• 11.1 Distance and Midpoint Formulas; Circles
• 11.2 Parabolas
• 11.3 Ellipses
• 11.4 Hyperbolas
• 11.5 Solve Systems of Nonlinear Equations
• 12.1 Sequences
• 12.2 Arithmetic Sequences
• 12.3 Geometric Sequences and Series
• 12.4 Binomial Theorem

Learning Objectives

By the end of this section, you will be able to:

• Write the augmented matrix for a system of equations
• Use row operations on a matrix
• Solve systems of equations using matrices

Be Prepared 4.13

Before you get started, take this readiness quiz.

Solve: 3 ( x + 2 ) + 4 = 4 ( 2 x − 1 ) + 9 . 3 ( x + 2 ) + 4 = 4 ( 2 x − 1 ) + 9 . If you missed this problem, review Example 2.2 .

Be Prepared 4.14

Solve: 0.25 p + 0.25 ( p + 4 ) = 5.20 . 0.25 p + 0.25 ( p + 4 ) = 5.20 . If you missed this problem, review Example 2.13 .

Be Prepared 4.15

Evaluate when x = −2 x = −2 and y = 3 : 2 x 2 − x y + 3 y 2 . y = 3 : 2 x 2 − x y + 3 y 2 . If you missed this problem, review Example 1.21 .

Write the Augmented Matrix for a System of Equations

Solving a system of equations can be a tedious operation where a simple mistake can wreak havoc on finding the solution. An alternative method which uses the basic procedures of elimination but with notation that is simpler is available. The method involves using a matrix . A matrix is a rectangular array of numbers arranged in rows and columns.

A matrix is a rectangular array of numbers arranged in rows and columns.

A matrix with m rows and n columns has order m × n . m × n . The matrix on the left below has 2 rows and 3 columns and so it has order 2 × 3 . 2 × 3 . We say it is a 2 by 3 matrix.

Each number in the matrix is called an element or entry in the matrix.

We will use a matrix to represent a system of linear equations. We write each equation in standard form and the coefficients of the variables and the constant of each equation becomes a row in the matrix. Each column then would be the coefficients of one of the variables in the system or the constants. A vertical line replaces the equal signs. We call the resulting matrix the augmented matrix for the system of equations.

Notice the first column is made up of all the coefficients of x , the second column is the all the coefficients of y , and the third column is all the constants.

Example 4.37

Write each system of linear equations as an augmented matrix:

ⓐ { 5 x − 3 y = −1 y = 2 x − 2 { 5 x − 3 y = −1 y = 2 x − 2 ⓑ { 6 x − 5 y + 2 z = 3 2 x + y − 4 z = 5 3 x − 3 y + z = −1 { 6 x − 5 y + 2 z = 3 2 x + y − 4 z = 5 3 x − 3 y + z = −1

ⓐ The second equation is not in standard form. We rewrite the second equation in standard form.

We replace the second equation with its standard form. In the augmented matrix, the first equation gives us the first row and the second equation gives us the second row. The vertical line replaces the equal signs.

ⓑ All three equations are in standard form. In the augmented matrix the first equation gives us the first row, the second equation gives us the second row, and the third equation gives us the third row. The vertical line replaces the equal signs.

Try It 4.73

ⓐ { 3 x + 8 y = −3 2 x = −5 y − 3 { 3 x + 8 y = −3 2 x = −5 y − 3 ⓑ { 2 x − 5 y + 3 z = 8 3 x − y + 4 z = 7 x + 3 y + 2 z = −3 { 2 x − 5 y + 3 z = 8 3 x − y + 4 z = 7 x + 3 y + 2 z = −3

Try It 4.74

ⓐ { 11 x = −9 y − 5 7 x + 5 y = −1 { 11 x = −9 y − 5 7 x + 5 y = −1 ⓑ { 5 x − 3 y + 2 z = −5 2 x − y − z = 4 3 x − 2 y + 2 z = −7 { 5 x − 3 y + 2 z = −5 2 x − y − z = 4 3 x − 2 y + 2 z = −7

It is important as we solve systems of equations using matrices to be able to go back and forth between the system and the matrix. The next example asks us to take the information in the matrix and write the system of equations.

Example 4.38

Write the system of equations that corresponds to the augmented matrix:

[ 4 −3 3 1 2 −1 −2 −1 3 | −1 2 −4 ] . [ 4 −3 3 1 2 −1 −2 −1 3 | −1 2 −4 ] .

We remember that each row corresponds to an equation and that each entry is a coefficient of a variable or the constant. The vertical line replaces the equal sign. Since this matrix is a 4 × 3 4 × 3 , we know it will translate into a system of three equations with three variables.

Try It 4.75

Write the system of equations that corresponds to the augmented matrix: [ 1 −1 2 3 2 1 −2 1 4 −1 2 0 ] . [ 1 −1 2 3 2 1 −2 1 4 −1 2 0 ] .

Try It 4.76

Write the system of equations that corresponds to the augmented matrix: [ 1 1 1 4 2 3 −1 8 1 1 −1 3 ] . [ 1 1 1 4 2 3 −1 8 1 1 −1 3 ] .

Use Row Operations on a Matrix

Once a system of equations is in its augmented matrix form, we will perform operations on the rows that will lead us to the solution.

To solve by elimination, it doesn’t matter which order we place the equations in the system. Similarly, in the matrix we can interchange the rows.

When we solve by elimination, we often multiply one of the equations by a constant. Since each row represents an equation, and we can multiply each side of an equation by a constant, similarly we can multiply each entry in a row by any real number except 0.

In elimination, we often add a multiple of one row to another row. In the matrix we can replace a row with its sum with a multiple of another row.

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

Row Operations

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.

Performing these operations is easy to do but all the arithmetic can result in a mistake. If we use a system to record the row operation in each step, it is much easier to go back and check our work.

We use capital letters with subscripts to represent each row. We then show the operation to the left of the new matrix. To show interchanging a row:

To multiply row 2 by −3 −3 :

To multiply row 2 by −3 −3 and add it to row 1:

Example 4.39

Perform the indicated operations on the augmented matrix:

ⓐ Interchange rows 2 and 3.

ⓑ Multiply row 2 by 5.

ⓒ Multiply row 3 by −2 −2 and add to row 1.

ⓐ We interchange rows 2 and 3.

ⓑ We multiply row 2 by 5.

ⓒ We multiply row 3 by −2 −2 and add to row 1.

Try It 4.77

Perform the indicated operations sequentially on the augmented matrix:

ⓐ Interchange rows 1 and 3.

ⓑ Multiply row 3 by 3.

ⓒ Multiply row 3 by 2 and add to row 2.

[ 5 −2 −2 4 −1 −4 −2 3 0 | −2 4 −1 ] [ 5 −2 −2 4 −1 −4 −2 3 0 | −2 4 −1 ]

Try It 4.78

ⓐ Interchange rows 1 and 2,

ⓑ Multiply row 1 by 2,

ⓒ Multiply row 2 by 3 and add to row 1.

[ 2 −3 −2 4 1 −3 5 0 4 | −4 2 −1 ] [ 2 −3 −2 4 1 −3 5 0 4 | −4 2 −1 ]

Now that we have practiced the row operations, we will look at an augmented matrix and figure out what operation we will use to reach a goal. This is exactly what we did when we did elimination. We decided what number to multiply a row by in order that a variable would be eliminated when we added the rows together.

Given this system, what would you do to eliminate x ?

This next example essentially does the same thing, but to the matrix.

Example 4.40

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: [ 1 −1 4 −8 | 2 0 ] . [ 1 −1 4 −8 | 2 0 ] .

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

Try It 4.79

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: [ 1 −1 3 −6 | 2 2 ] . [ 1 −1 3 −6 | 2 2 ] .

Try It 4.80

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: [ 1 −1 −2 −3 | 3 2 ] . [ 1 −1 −2 −3 | 3 2 ] .

Solve Systems of Equations Using Matrices

To solve a system of equations using matrices, we transform the augmented matrix into a matrix in row-echelon form using row operations. For a consistent and independent system of equations, its augmented matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a 1 and all entries below the diagonal are zeros.

Row-Echelon Form

For a consistent and independent system of equations, its augmented matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a 1 and all entries below the diagonal are zeros.

Once we get the augmented matrix into row-echelon form, we can write the equivalent system of equations and read the value of at least one variable. We then substitute this value in another equation to continue to solve for the other variables. This process is illustrated in the next example.

Example 4.41

How to solve a system of equations using a matrix.

Solve the system of equations using a matrix: { 3 x + 4 y = 5 x + 2 y = 1 . { 3 x + 4 y = 5 x + 2 y = 1 .

Try It 4.81

Solve the system of equations using a matrix: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Try It 4.82

Solve the system of equations using a matrix: { 2 x + y = −4 x − y = −2 . { 2 x + y = −4 x − y = −2 .

The steps are summarized here.

Solve a system of equations using matrices.

• Step 1. Write the augmented matrix for the system of equations.
• Step 2. Using row operations get the entry in row 1, column 1 to be 1.
• Step 3. Using row operations, get zeros in column 1 below the 1.
• Step 4. Using row operations, get the entry in row 2, column 2 to be 1.
• Step 5. Continue the process until the matrix is in row-echelon form.
• Step 6. Write the corresponding system of equations.
• Step 7. Use substitution to find the remaining variables.
• Step 8. Write the solution as an ordered pair or triple.
• Step 9. Check that the solution makes the original equations true.

Here is a visual to show the order for getting the 1’s and 0’s in the proper position for row-echelon form.

We use the same procedure when the system of equations has three equations.

Example 4.42

Solve the system of equations using a matrix: { 3 x + 8 y + 2 z = −5 2 x + 5 y − 3 z = 0 x + 2 y − 2 z = −1 . { 3 x + 8 y + 2 z = −5 2 x + 5 y − 3 z = 0 x + 2 y − 2 z = −1 .

Try It 4.83

Solve the system of equations using a matrix: { 2 x − 5 y + 3 z = 8 3 x − y + 4 z = 7 x + 3 y + 2 z = −3 . { 2 x − 5 y + 3 z = 8 3 x − y + 4 z = 7 x + 3 y + 2 z = −3 .

Try It 4.84

Solve the system of equations using a matrix: { −3 x + y + z = −4 − x + 2 y − 2 z = 1 2 x − y − z = −1 . { −3 x + y + z = −4 − x + 2 y − 2 z = 1 2 x − y − z = −1 .

So far our work with matrices has only been with systems that are consistent and independent, which means they have exactly one solution. Let’s now look at what happens when we use a matrix for a dependent or inconsistent system.

Example 4.43

Solve the system of equations using a matrix: { x + y + 3 z = 0 x + 3 y + 5 z = 0 2 x + 4 z = 1 . { x + y + 3 z = 0 x + 3 y + 5 z = 0 2 x + 4 z = 1 .

Try It 4.85

Solve the system of equations using a matrix: { x − 2 y + 2 z = 1 −2 x + y − z = 2 x − y + z = 5 . { x − 2 y + 2 z = 1 −2 x + y − z = 2 x − y + z = 5 .

Try It 4.86

Solve the system of equations using a matrix: { 3 x + 4 y − 3 z = −2 2 x + 3 y − z = −12 x + y − 2 z = 6 . { 3 x + 4 y − 3 z = −2 2 x + 3 y − z = −12 x + y − 2 z = 6 .

The last system was inconsistent and so had no solutions. The next example is dependent and has infinitely many solutions.

Example 4.44

Solve the system of equations using a matrix: { x − 2 y + 3 z = 1 x + y − 3 z = 7 3 x − 4 y + 5 z = 7 . { x − 2 y + 3 z = 1 x + y − 3 z = 7 3 x − 4 y + 5 z = 7 .

Try It 4.87

Solve the system of equations using a matrix: { x + y − z = 0 2 x + 4 y − 2 z = 6 3 x + 6 y − 3 z = 9 . { x + y − z = 0 2 x + 4 y − 2 z = 6 3 x + 6 y − 3 z = 9 .

Try It 4.88

Solve the system of equations using a matrix: { x − y − z = 1 − x + 2 y − 3 z = −4 3 x − 2 y − 7 z = 0 . { x − y − z = 1 − x + 2 y − 3 z = −4 3 x − 2 y − 7 z = 0 .

Access this online resource for additional instruction and practice with Gaussian Elimination.

• Gaussian Elimination

Section 4.5 Exercises

Practice makes perfect.

In the following exercises, write each system of linear equations as an augmented matrix.

ⓐ { 3 x − y = −1 2 y = 2 x + 5 { 3 x − y = −1 2 y = 2 x + 5 ⓑ { 4 x + 3 y = −2 x − 2 y − 3 z = 7 2 x − y + 2 z = −6 { 4 x + 3 y = −2 x − 2 y − 3 z = 7 2 x − y + 2 z = −6

ⓐ { 2 x + 4 y = −5 3 x − 2 y = 2 { 2 x + 4 y = −5 3 x − 2 y = 2 ⓑ { 3 x − 2 y − z = −2 −2 x + y = 5 5 x + 4 y + z = −1 { 3 x − 2 y − z = −2 −2 x + y = 5 5 x + 4 y + z = −1

ⓐ { 3 x − y = −4 2 x = y + 2 { 3 x − y = −4 2 x = y + 2 ⓑ { x − 3 y − 4 z = −2 4 x + 2 y + 2 z = 5 2 x − 5 y + 7 z = −8 { x − 3 y − 4 z = −2 4 x + 2 y + 2 z = 5 2 x − 5 y + 7 z = −8

ⓐ { 2 x − 5 y = −3 4 x = 3 y − 1 { 2 x − 5 y = −3 4 x = 3 y − 1 ⓑ { 4 x + 3 y − 2 z = −3 −2 x + y − 3 z = 4 − x − 4 y + 5 z = −2 { 4 x + 3 y − 2 z = −3 −2 x + y − 3 z = 4 − x − 4 y + 5 z = −2

Write the system of equations that corresponds to the augmented matrix.

[ 2 −1 1 −3 | 4 2 ] [ 2 −1 1 −3 | 4 2 ]

[ 2 −4 3 −3 | −2 −1 ] [ 2 −4 3 −3 | −2 −1 ]

[ 1 0 −3 1 −2 0 0 −1 2 | −1 −2 3 ] [ 1 0 −3 1 −2 0 0 −1 2 | −1 −2 3 ]

[ 2 −2 0 0 2 −1 3 0 −1 | −1 2 −2 ] [ 2 −2 0 0 2 −1 3 0 −1 | −1 2 −2 ]

In the following exercises, perform the indicated operations on the augmented matrices.

[ 6 −4 3 −2 | 3 1 ] [ 6 −4 3 −2 | 3 1 ]

ⓐ Interchange rows 1 and 2

ⓑ Multiply row 2 by 3

ⓒ Multiply row 2 by −2 −2 and add row 1 to it.

[ 4 −6 3 2 | −3 1 ] [ 4 −6 3 2 | −3 1 ]

ⓑ Multiply row 1 by 4

ⓒ Multiply row 2 by 3 and add row 1 to it.

[ 4 −12 −8 4 −2 −3 −6 2 −1 | 16 −1 −1 ] [ 4 −12 −8 4 −2 −3 −6 2 −1 | 16 −1 −1 ]

ⓐ Interchange rows 2 and 3

ⓒ Multiply row 2 by −2 −2 and add to row 3.

[ 6 −5 2 2 1 −4 3 −3 1 | 3 5 −1 ] [ 6 −5 2 2 1 −4 3 −3 1 | 3 5 −1 ]

ⓑ Multiply row 2 by 5

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: [ 1 2 −3 −4 | 5 −1 ] . [ 1 2 −3 −4 | 5 −1 ] .

Perform the needed row operations that will get the first entry in both row 2 and row 3 to be zero in the augmented matrix: [ 1 −2 3 3 −1 −2 2 −3 −4 | −4 5 −1 ] . [ 1 −2 3 3 −1 −2 2 −3 −4 | −4 5 −1 ] .

In the following exercises, solve each system of equations using a matrix.

{ 2 x + y = 2 x − y = −2 { 2 x + y = 2 x − y = −2

{ 3 x + y = 2 x − y = 2 { 3 x + y = 2 x − y = 2

{ − x + 2 y = −2 x + y = −4 { − x + 2 y = −2 x + y = −4

{ −2 x + 3 y = 3 x + 3 y = 12 { −2 x + 3 y = 3 x + 3 y = 12

{ 2 x − 3 y + z = 19 −3 x + y − 2 z = −15 x + y + z = 0 { 2 x − 3 y + z = 19 −3 x + y − 2 z = −15 x + y + z = 0

{ 2 x − y + 3 z = −3 − x + 2 y − z = 10 x + y + z = 5 { 2 x − y + 3 z = −3 − x + 2 y − z = 10 x + y + z = 5

{ 2 x − 6 y + z = 3 3 x + 2 y − 3 z = 2 2 x + 3 y − 2 z = 3 { 2 x − 6 y + z = 3 3 x + 2 y − 3 z = 2 2 x + 3 y − 2 z = 3

{ 4 x − 3 y + z = 7 2 x − 5 y − 4 z = 3 3 x − 2 y − 2 z = −7 { 4 x − 3 y + z = 7 2 x − 5 y − 4 z = 3 3 x − 2 y − 2 z = −7

{ x + 2 z = 0 4 y + 3 z = −2 2 x − 5 y = 3 { x + 2 z = 0 4 y + 3 z = −2 2 x − 5 y = 3

{ 2 x + 5 y = 4 3 y − z = 3 4 x + 3 z = −3 { 2 x + 5 y = 4 3 y − z = 3 4 x + 3 z = −3

{ 2 y + 3 z = −1 5 x + 3 y = −6 7 x + z = 1 { 2 y + 3 z = −1 5 x + 3 y = −6 7 x + z = 1

{ 3 x − z = −3 5 y + 2 z = −6 4 x + 3 y = −8 { 3 x − z = −3 5 y + 2 z = −6 4 x + 3 y = −8

{ 2 x + 3 y + z = 12 x + y + z = 9 3 x + 4 y + 2 z = 20 { 2 x + 3 y + z = 12 x + y + z = 9 3 x + 4 y + 2 z = 20

{ x + 2 y + 6 z = 5 − x + y − 2 z = 3 x − 4 y − 2 z = 1 { x + 2 y + 6 z = 5 − x + y − 2 z = 3 x − 4 y − 2 z = 1

{ x + 2 y − 3 z = −1 x − 3 y + z = 1 2 x − y − 2 z = 2 { x + 2 y − 3 z = −1 x − 3 y + z = 1 2 x − y − 2 z = 2

{ 4 x − 3 y + 2 z = 0 −2 x + 3 y − 7 z = 1 2 x − 2 y + 3 z = 6 { 4 x − 3 y + 2 z = 0 −2 x + 3 y − 7 z = 1 2 x − 2 y + 3 z = 6

{ x − y + 2 z = −4 2 x + y + 3 z = 2 −3 x + 3 y − 6 z = 12 { x − y + 2 z = −4 2 x + y + 3 z = 2 −3 x + 3 y − 6 z = 12

{ − x − 3 y + 2 z = 14 − x + 2 y − 3 z = −4 3 x + y − 2 z = 6 { − x − 3 y + 2 z = 14 − x + 2 y − 3 z = −4 3 x + y − 2 z = 6

{ x + y − 3 z = −1 y − z = 0 − x + 2 y = 1 { x + y − 3 z = −1 y − z = 0 − x + 2 y = 1

{ x + 2 y + z = 4 x + y − 2 z = 3 −2 x − 3 y + z = −7 { x + 2 y + z = 4 x + y − 2 z = 3 −2 x − 3 y + z = −7

Writing Exercises

Solve the system of equations { x + y = 10 x − y = 6 { x + y = 10 x − y = 6 ⓐ by graphing and ⓑ by substitution. ⓒ Which method do you prefer? Why?

Solve the system of equations { 3 x + y = 12 x = y − 8 { 3 x + y = 12 x = y − 8 by substitution and explain all your steps in words.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

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• Authors: Lynn Marecek, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Intermediate Algebra 2e
• Publication date: May 6, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/4-5-solve-systems-of-equations-using-matrices

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How to Solve Matrices

Last Updated: October 7, 2022 Fact Checked

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. There are 10 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 163,374 times.

A matrix is a very useful way of representing numbers in a block format, [1] X Research source which you can then use to solve a system of linear equations. If you only have two variables, you will probably use a different method. See Solve a System of Two Linear Equations and Solve Systems of Equations for examples of these other methods. But when you have three or more variables, a matrix is ideal. By using repeated combinations of multiplication and addition, you can systematically reach a solution.

Setting up the Matrix for Solving

• If you have fewer equations than the number of variables, you will be able to learn some limiting information about the variables (such as x = 3y and y = 2z), but you cannot get a precise solution. For this article, we will be working toward getting a unique solution only.

• If you have more variables, you will just continue the line as long as necessary. For example, if you are trying to solve a system with six variables, your standard form would look like Au+Bv+Cw+Dx+Ey+Fz =G. For this article, we will focus on systems with only three variables. Solving a larger system is exactly the same, but just takes more time and more steps.
• Note that in standard form, the operations between the terms is always addition. If your equation has subtraction instead of addition, you will need to work with this later my making your coefficient negative. If it helps you remember, you can rewrite the equation and make the operation addition and the coefficient negative. For example, you can rewrite the equation 3x-2y+4z=1 as 3x+(-2y)+4z=1.

• For example, suppose you have a system that consists of the three equations 3x+y-z=9, 2x-2y+z=-3, and x+y+z=7. The top row of your matrix will contain the numbers 3,1,-1,9, since these are the coefficients and solution of the first equation. Note that any variable that has no coefficient showing is assumed to have a coefficient of 1. The second row of the matrix will be 2,-2,1,-3, and the third row will be 1,1,1,7.
• Be sure to align the x-coefficients in the first column, the y-coefficients in the second, the z-coefficients in the third, and the solution terms in the fourth. When you finish working with the matrix, these columns will be important in writing your solution.

• You can indicate any specific position in a matrix by using a combination of R and C. For example, to pinpoint the term in the second row, third column, you could call it R2C3.

Learning the Operations for Solving a System with a Matrix

• You will be working with some basic operations to create the “solution matrix.” The solution matrix will look like this: [6] X Research source
• Notice that the matrix consists of 1’s in a diagonal line with 0’s in all other spaces, except the fourth column. The numbers in the fourth column will be your solution for the variables x, y and z.

• It is common to use fractions in scalar multiplication, because you often want to create that diagonal row of 1s. Get used to working with fractions. It will also be easier, for most steps in solving the matrix, to be able to write your fractions in improper form, and then convert them back to mixed numbers for the final solution. Therefore, the number 1 2/3 is easier to work with if you write it as 5/3.
• For example, the first row (R1) of our sample problem begins with the terms [3,1,-1,9]. The solution matrix should contain a 1 in the first position of the first row. In order to “change” our 3 into a 1, we can multiply the entire row by 1/3. Doing this will create the new R1 of [1,1/3,-1/3,3].
• Be careful to keep any negative signs where they belong.

• You can use some shorthand and indicate this operation as R2-R1=[0,-1,2,6].
• Recognize that adding and subtracting are merely opposite forms of the same operation. You can either think of adding two numbers or subtracting the opposite. For example, if you begin with the simple equation 3-3=0, you could consider this instead as an addition problem of 3+(-3)=0. The result is the same. This seems basic, but it is sometimes easier to think of a problem in one form or the other. Just keep track of your negative signs.

• Suppose you have a Row 1 of [1,1,2,6] and a Row 2 of [2,3,1,1]. You want to create a 0 term in the first column of R2. That is, you want to change the 2 into a 0. To do this, you need to subtract a 2. You can get a 2 by first multiplying Row 1 by the scalar multiplication 2, and then subtract the first row from the second row. In shorthand, you can think of this as R2-2*R1. First multiply R1 by 2 to get [2,2,4,12]. Then subtract this from R2 to get [(2-2),(3-2),(1-4),(1-12)]. Simplify this and your new R2 will be [0,1,-3,-11].

• A common mistake occurs when conducting a combined multiplication and addition step in one move. Suppose, for example, you need to subtract double R1 from R2. When you multiply R1 by 2 to do this step, remember that you are not changing R1 in the matrix. You are only doing the multiplication to change R2. Copy R1 first in its original form, then make the change to R2.

• 1. Create a 1 in the first row, first column (R1C1).
• 2. Create a 0 in the second row, first column (R2C1).
• 3. Create a 1 in the second row, second column (R2C2).
• 4. Create a 0 in the third row, first column (R3C1).
• 5. Create a 0 in the third row, second column (R3C2).
• 6. Create 1 in the third row, third column (R3C3).

• Create a 0 in the second row, third column (R2C3).
• Create a 0 in first row, third column (R1C3).
• Create a 0 in the first row, second column (R1C2).

Putting the Steps Together to Solve the System

• Notice that multiplication and division are merely inverse functions of each other. We can say we are multiplying by 1/3 or dividing by 3, and the result is the same.

• Copy down the unaffected row 3 as R3=[1,1,1,7].
• Be very careful with subtracting negative numbers, to make sure you keep the signs correct.
• For now, leave the fractions in their improper forms. This will make later steps of the solution easier. You can simplify fractions in the final step of the problem.

• Notice that as the left half of the row starts looking like the solution with the 0 and 1, the right half may start looking ugly, with improper fractions. Just carry them along for now.
• Remember to continue copying the unaffected rows, so R1=[1,1/3,-1/3,3] and R3=[1,1,1,7].

• Continue to copy along R1=[1,1/3,-1/3,3] and R2=[0,1,-5/8,27/8]. Remember that you only change one row at a time.

• Notice that the fractions, which appeared quite complicated in the previous step, have already begun to resolve themselves.
• Continue to carry along R1=[1,1/3,-1/3,3] and R2=[0,1,-5/8,27/8].
• Notice that at this point, you have the diagonal of 1’s for your solution matrix. You just need to transform three more items of the matrix into 0’s to find your solution.

• Copy along R1=[1,1/3,-1/3,3] and R3=[0,0,1,1].

• Copy the unchanged R2=[0,1,0,4] and R3=[0,0,1,1].

Verifying Your Solution

• Recall that the original equations for this problem were 3x+y-z=9, 2x-2y+z=-3, and x+y+z=7. When you replace the variables with their solved values, you get 3*2+4-1=9, 2*2-2*4+1=-3, and 2+4+1=7.

• Because each equation simplifies to a true mathematical statement, your solutions are correct. If any of them did not resolve correctly, you would have to go back through your work and look for any errors. Some common mistakes occur in dropping negative signs along the way or confusing the multiplication and addition of fractions.

Expert Q&A

• If your system of equations is very complicated, with many variables, you may be able to use a graphing calculator instead of doing the work by hand. For information on this, see Use a Graphing Calculator to Solve a System of Equations. Thanks Helpful 0 Not Helpful 0

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• ↑ https://www.mathsisfun.com/algebra/matrix-introduction.html
• ↑ https://www.mathsisfun.com/algebra/systems-linear-equations-matrices.html
• ↑ https://www.khanacademy.org/math/precalculus/precalc-matrices/representing-systems-with-matrices/a/representing-systems-with-matrices
• ↑ https://www.cuemath.com/algebra/solve-matrices/
• ↑ https://www.khanacademy.org/math/precalculus/precalc-matrices/multiplying-matrices-by-scalars/a/multiplying-matrices-by-scalars
• ↑ https://people.richland.edu/james/lecture/m116/matrices/operations.html
• ↑ https://www.math.tamu.edu/~dallen/m640_03c/lectures/chapter2.pdf
• ↑ https://www.varsitytutors.com/hotmath/hotmath_help/topics/solving-matrix-equations
• ↑ https://math.libretexts.org/Bookshelves/Algebra/Intermediate_Algebra_(OpenStax)/04%3A_Systems_of_Linear_Equations/4.06%3A_Solve_Systems_of_Equations_Using_Matrices
• ↑ https://www.cliffsnotes.com/study-guides/algebra/algebra-ii/linear-equations-in-three-variables/linear-equations-solutions-using-matrices-with-three-variables

About This Article

By properly setting up a matrix, you can use them to solve a system of linear equations. Start by writing out your equations and then transfer the numbers from them into a matrix by copying the coefficients and results into a single row. Stack the rows one on top of each other to form a block-looking format. Add a large square bracket around your full matrix and use the abbreviation “R” for the rows and “C” for the columns. This allows you to refer to a specific position in the matrix with a combination of R and C, such as R4C1. To solve the matrix, you can use different operations. For instance, you could use row-addition or row-subtraction, which allows you to add or subtract any two rows of the matrix. To learn about other ways to create a solution matrix, keep reading! Did this summary help you? Yes No

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Solving Matrix Equations

A matrix equation is an equation in which a variable stands for a matrix .

You can solve the simpler matrix equations using matrix addition and scalar multiplication .

Examples 1:

Solve for the matrix X : X + [ 3 2 1 0 ] = [ 6 3 7 − 1 ]

X + [ 3 2 1 0 ] − [ 3 2 1 0 ] = [ 6       3 7 − 1 ] − [ 3 2 1 0 ] X + [ 0 0 0 0 ] = [ 6 − 3       3 − 2 7 − 1 − 1 − 0 ] X = [ 3       1 6 − 1 ]

Examples 2:

Solve for the matrix X : X − [ − 9 − 3       6       0 ] = [ 4           0 12 − 10 ]

X − [ − 9 − 3       6       0 ] = [ 4           0 12 − 10 ] X − [ − 9 − 3       6       0 ] + [ − 9 − 3       6       0 ] = [ 4           0 12 − 10 ] + [ − 9 − 3       6       0 ] X − [ 0 0 0 0 ] = [     4 + ( − 9 )             0 + ( − 3 ) 12 +           6 − 10 +           0 ] X = [ − 5       − 3 18 − 10 ]

Solving systems of linear equations using matrices:

Matrix equations can be used to solve systems of linear equations by using the left and right sides of the equations.

Examples 3:

Solve the system of equations using matrices:  { 7 x + 5 y = 3 3 x − 2 y = 22

  7 x + 5 y = 3 3 x − 2 y = 22 → [ 7 x + 5 y 3 x − 2 y ] = [ 3 22 ]

Write the matrix on the left as the product of coefficients and variables.

[ 7 5 3 − 2 ]           [ x y ]     =   [ 3 22 ]

      ↑                                       ↑                                     ↑

coefficient         variable             constant matrix                               matrix                 matrix

First, find the inverse of the coefficient matrix.  The inverse of [ 7 5 3 − 2 ] is

1 7 ( − 2 ) − ( 3 ) ( 5 ) [ − 2 − 5 − 3 7 ] = − 1 29 [ − 2 − 5 − 3 7 ] = [ 2 29 5 29 3 29 − 7 29 ]

Next, multiply each side of the matrix equation by the inverse matrix .  Since matrix multiplication is not commutative, the inverse matrix should be at the left on each side of the matrix equation.

[ 2 29 5 29 3 29 − 7 29 ] [ 7 5 3 − 2 ] [ x y ] = [ 2 29 5 29 3 29 − 7 29 ] [ 3 22 ]

[ 1 0 0 1 ] [ x y ] = [ 4 − 5 ]

The identity matrix on the left verifies that the inverse matrix was calculated correctly.

[ x y ] = [ 4 − 5 ]

The solution is ( 4 , − 5 ) .

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• \begin{pmatrix}9&2&-4\\b+a&9&7\\0&c&8\end{pmatrix}=\begin{pmatrix}9&a&-4\\7&9&7\\0&16&8\end{pmatrix}
• \begin{pmatrix}4&0\\6&-2\\3&1\end{pmatrix}=\begin{pmatrix}x&0\\6&y+4\\\frac{z}{3}&1\end{pmatrix}
• x+\begin{pmatrix}3&2\\1&0\end{pmatrix}=\begin{pmatrix}6&3\\7&-1\end{pmatrix}
• 2\begin{pmatrix}1&2\\0&1\end{pmatrix}x+\begin{pmatrix}3&4\\2&1\end{pmatrix}=\begin{pmatrix}1&2\\3&4\end{pmatrix}

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Solving Linear Equations Using Matrix

Solving linear equations using matrix is done by two prominent methods, namely the matrix method and row reduction or the Gaussian elimination method. In this article, we will look at solving linear equations with matrix and related examples. With the study notes provided below, students will develop a clear idea about the topic.

Download Complete Chapter Notes of Matrices & Determinants Download Now

How to Solve Linear Equations Using Matrix Method?

Let the equations be:

The first method to find the solution to the system of equations is the matrix method. The steps to be followed are given below:

• All the variables in the equations should be written in the appropriate order.
• The variables, their coefficients and constants are to be written on the respective sides.

Solving a system of linear equations by the method of finding the inverse consists of two new matrices, namely

• Matrix A: which represents the variables
• Matrix B: which represents the constants

A system of equations can be solved using matrix multiplication.

We can write the above equations in the matrix form as follows:

Where, \(\begin{array}{l}A=\left[ \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right],\ X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right],\ B=\left[ \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \right]\end{array} \) .

A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

The second method to find the solution for the system of equations is row reduction or Gaussian elimination.

• The augmented matrix for the linear equations is written.
• Use elementary such that all the elements below the main diagonal are zero. If a zero is obtained on the diagonal, perform the row operation such that a nonzero element is obtained.
• Back substitution is used to find the solution.

Related Topics:

• Introduction to Matrices
• Types of Matrices
• Matrix Operations
• Adjoint and Inverse of a Matrix
• Rank of a Matrix and Special Matrices

Solution to a System of Equations

A set of values of x, y, and z which simultaneously satisfy all the equations, is called a solution to the system of equations.

Consider, x + y + z = 9

2x – y + z = 5

4x + y – z = 7

Here, the set of values – x = 2, y = 3, z = 4, is a solution to the system of linear equations.

Because, 2 + 3 + 4 = 9, 4 – 3 + 4 = 5, 8 + 3 – 4 = 7

Consistent Equations

If the system of equations has one or more solutions, then it is said to be a consistent system of equations; otherwise, it is an inconsistent system of equations. For example, the system of linear equations x + 3y = 5;  x – y = 1 is consistent because x = 2, y = 1 is a solution to it. However, the system of linear equations x + 3y = 5; 2x + 6y = 8 is inconsistent because there is no set of values of x and y, which may satisfy the two equations simultaneously.

Condition for consistency of a system of linear equation AX = B

(b) If |A| = 0, and (Adj A) B ≠ 0, then the system is inconsistent.

(c) If |A| = 0, and (Adj A) B = 0, then the system is consistent and has infinitely many solutions.

Note AX = 0 is known as the homogeneous system of linear equations, and here, B = 0. A system of homogeneous equations is always consistent.

The system has a non-trivial solution (non-zero solution), if | A | = 0

Theorem 1: Let AX = B be a system of linear equations, where A is the coefficient matrix. If A is invertible, then the system has a unique solution, given by X = A -1 B

Uniqueness: If AX = B has two sets of solutions X 1 and X 2 , then AX 1 = B and AX 2 = B (Each equal to B).

⇒  AX 1 = AX 2

By cancellation law, A is invertible.

⇒  X 1 = X 2

Hence, the given system AX = B has a unique solution.

Note: A homogeneous system of equations is always consistent.

Problems on Solving Linear Equations Using Matrix Method

Illustration: Let A = \(\begin{array}{l}\left[ \begin{matrix} x+y & y \\ 2x & x-y \\ \end{matrix} \right],\ B=\left[ \begin{matrix} 2 \\ -1 \\ \end{matrix} \right],\ C = \left[ \begin{matrix} 3 \\ 2 \\ \end{matrix} \right]\end{array} \) . If AB = C, then find the matrix A 2 .

By solving AB = C, we get the values of x and y. Then by substituting these values in A, we obtain A 2

Here \(\begin{array}{l}\left[ \begin{matrix} x+y & y \\ 2x & x-y \\ \end{matrix} \right]\left[ \begin{matrix} 2 \\ -1 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 \\ 2 \\ \end{matrix} \right]\\ \Rightarrow \left[ \begin{matrix} 2\left( x+y \right)-y \\ 2x.2-\left( x-y \right) \\ \end{matrix} \right]=\left[ \begin{matrix} 3 \\ 2 \\ \end{matrix} \right]\\ \Rightarrow 2\left( x+y \right)-y=3 \:\:and \:\:4x-\left( x-y \right)=2\end{array} \) \(\begin{array}{l}\Rightarrow \,\,2x+y=3\;\;\;\;and\;\;\;\;3x+y=2\end{array} \) Subtracting the two equations, we get, x = -1 So, y = 5 .

Illustration: Solve the following equations by matrix inversion

The given equation can be written in a matrix form as AX = D, and then by obtaining A -1 and multiplying it on both sides, we can solve the given problem.

then matrix A equals:

has no solution if a and b are

By applying row operation in the given matrices and comparing them, we can obtain the required result.

\(\begin{array}{l}\Rightarrow \,a=-3 \;\;and \;\;b\ne -1/3 \end{array} \) .

Applying \(\begin{array}{l}{{R}_{3}}\to {{R}_{3}}-2{{R}_{2}}+{{R}_{1}} \;\;and \;\;\;{{R}_{2}} to {{R}_{2}}-2{{R}_{1}}\end{array} \) we get;

Frequently Asked Questions

What do you mean by a linear equation.

A linear equation is an equation that has one or more variables having degree one.

Name two methods to solve linear equations using matrices.

The matrix multiplication method and the Gaussian elimination method are two methods to solve linear equations using matrices.

What do you mean by consistent equations?

A consistent system of equations is an equation having one or more solutions.

Give the formula used in the matrix multiplication method for solving linear equations.

We use the formula AX = B in the matrix multiplication method for solving linear equations. Here, A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

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Matrix Equation

Nonhomogeneous matrix equations of the form

can be solved by taking the matrix inverse to obtain

But from the original matrix , each of the entries in the first columns is zero since

Given a numerical solution to a matrix equation, the solution can be iteratively improved using the following technique. Assume that the numerically obtained solution to

Combining ( 11 ) and ( 12 ) then gives

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Cite this as:

Weisstein, Eric W. "Matrix Equation." From MathWorld --A Wolfram Web Resource. https://mathworld.wolfram.com/MatrixEquation.html

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What are systems of equations?

A system of equations is a set of one or more equations involving a number of variables..

The solutions to systems of equations are the variable mappings such that all component equations are satisfied—in other words, the locations at which all of these equations intersect. To solve a system is to find all such common solutions or points of intersection.

Systems of linear equations are a common and applicable subset of systems of equations. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. The system is said to be inconsistent otherwise, having no solutions. Systems of linear equations involving more than two variables work similarly, having either one solution, no solutions or infinite solutions (the latter in the case that all component equations are equivalent).

More general systems involving nonlinear functions are possible as well. These possess more complicated solution sets involving one, zero, infinite or any number of solutions, but work similarly to linear systems in that their solutions are the points satisfying all equations involved. Going further, more general systems of constraints are possible, such as ones that involve inequalities or have requirements that certain variables be integers.

Solving systems of equations is a very general and important idea, and one that is fundamental in many areas of mathematics, engineering and science.

Matrix Equation

A system of equations can be solved using matrices by writing it in the form of a matrix equation. We can also determine whether a system has a unique solution or infinite number of solutions or no solution using the matrix equation.

Let us learn how to solve matrix equations in different methods along with examples.

What is Matrix Equation?

A matrix equation is of the form AX = B where A represents the coefficient matrix, X represents the column matrix of variables, and B represents the column matrix of the constants that are on the right side of the equations in a system. Let us consider a system of n nonhomogenous equations in n variables.

a₁₁ x₁ + a₁₂ x₂ + ... + a₁ₙ xₙ = b₁

a₂₁ x₁ + a₂₂ x₂ + ... + a₂ₙ xₙ = b₂

aₙ₁ x₁ + aₙ₂ x₂ + ... + aₙₙ xₙ = bₙ

Then the matrix equation that corresponds to the above system is:

AX = B, where

• A = A matrix made of coefficients = \(\left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end{array}\right]\)
• X = Column matrix of variables = \(\left[\begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{array}\right]\)
• B = Column matrix of constants = \(\left[\begin{array}{c} b_{1} \\ b_{2} \\ \vdots \\ b_{n} \end{array}\right]\)

How to Write Matrix Equation?

To write a system of equations as a matrix equation:

• Make sure that the variables are in the same order in each of the equations.
• Make sure that only the variable terms are on the left side and the constant terms are on the right side.
• Then find the coefficient matrix A, variable matrix X, and the constant matrix B to write the given system as matrix equation AX = B.

Here is an example to understand these steps.

Example: Write the following system as a matrix equation.

3x + y + 2z - 10 = 0 z - 3y = -5 y + x - 2z = 7

Let us get all variable terms on one side (also in order) and constants on the right side.

3x + y + 2z = 10 0x - 3y + z = -5 x + y - 2z = 7

Now we can write this system as:

\(\left[\begin{array}{ccc} 3 & 1 & 2 \\ 0 & -3 & 1 \\ 1 & 1 & -2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 10 \\ -5 \\ 7 \end{array}\right]\)

This is the matrix equation of the given system which alternatively can be written as AX = B where, A = \(\left[\begin{array}{ccc} 3 & 1 & 2 \\ 0 & -3 & 1 \\ 1 & 1 & -2 \end{array}\right]\), X = \(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\), and B = \(\left[\begin{array}{c} 10 \\ -5 \\ 7 \end{array}\right]\).

How to Solve Matrix Equation?

Let us solve the matrix equation AX = B for X. For this, we left multiply both sides of the equation by the inverse of A (that can be written as A -1 ).

A -1 (AX) = A -1 B

We know that A -1 A = I, where I is the identity matrix of the same order as A.

IX = A -1 B

We also know that IX = X.

This gives the solution of the matrix equation. This is also known as inverse matrix equation and hence the process of using the above formula to solve a system of equations is known as the " inverse matrix method". Thus, here are the steps to solve a system of equations using matrices:

• Write the system as matrix equation AX = B.
• Find the inverse, A -1 .
• Multiply it by the constant matrix B to get the solution. i.e., X = A -1 B.

We can see the examples of solving a system using these steps in the " Matrix Equation Examples " section below.

Consistency of Solution of Matrix Equation

We know that we can find the inverse of a matrix only when it is nonsingular . i.e., A -1 exists only when det (A) ≠ 0. Thus, the solution (X = A -1 B) exists and it is unique only when det (A) ≠ 0.

In case, if det (A) = 0, how can we know how many solutions a system may have? In this case, we have to find (adj A) B, where adj A stands for " adjoint of A".

• If (adj A) B ≠ O, then the system has no solution and hence the system is said to be inconsistent.
• If (adj A) B = O, then the system may be "consistent with an infinite number of solutions" or "inconsistent (no solution)".

Here, 'O' is a null matrix . This is summarized in the flowchart below.

☛ Related Topics:

• Homogeneous System of Linear Equations
• Matrix Calculator
• Matrix Formula
• Matrix Multiplication Calculator

Matrix Equation Examples

Example 1: Write the following system of equations in terms of matrices: 2y - 3x = -2; 5x - 2y - 7 = 0.

Let us write the variables on the left and constants on the right side of the equation. Also, let us maintain the same order of variables in every equation.

-3x + 2y = -2 5x - 2y = 7

Now, the coefficient matrix is, A = \(\left[\begin{array}{ll} -3 & 2 \\ \\ 5 & -2 \end{array}\right]\).

The variable matrix is, X = \(\left[\begin{array}{ll} x \\ \\ y \end{array}\right]\).

The constant matrix is, B = \(\left[\begin{array}{ll} -2 \\ \\ 7 \end{array}\right]\).

Thus, the matrix equation is:

\(\left[\begin{array}{ll} -3 & 2 \\ \\ 5 & -2 \end{array}\right]\) \(\left[\begin{array}{ll} x \\ \\ y \end{array}\right]\) = \(\left[\begin{array}{ll} -2 \\ \\ 7 \end{array}\right]\).

Answer: The system of equations is written in matrix form.

Example 2: Using the matrix equation found in Example 1, find the solution of the given system.

From Example 1,

A = \(\left[\begin{array}{ll} -3 & 2 \\ \\ 5 & -2 \end{array}\right]\).

Using the formula of the inverse of 2x2 matrix ,

A -1 = 1 / (6 - 10) \(\left[\begin{array}{ll} -2 & -2 \\ \\ -5 & -3 \end{array}\right]\)

= -1/4 \(\left[\begin{array}{ll} -2 & -2 \\ \\ -5 & -3 \end{array}\right]\)

= \(\left[\begin{array}{ll} 1/2 & 1/2 \\ \\ 5/4 & 3/4 \end{array}\right]\)

The solution of the given system is,

=\(\left[\begin{array}{ll} 1/2 & 1/2 \\ \\ 5/4 & 3/4 \end{array}\right]\) \(\left[\begin{array}{ll} -2 \\ \\ 7 \end{array}\right]\)

= \(\left[\begin{array}{ll} (1/2) (-2) + (1/2) (7) \\ \\ (5/4)(-2) + (3/4) (7) \end{array}\right]\)

= \(\left[\begin{array}{ll} 5/2 \\ \\ 11/4 \end{array}\right]\)

= \(\left[\begin{array}{ll} x \\ \\ y \end{array}\right]\)

Answer: The solution is x = 5/4 and y = 11/4.

Example 3: Is the system that is solved in Example 2 consistent or inconsistent? How many solutions does it have? Justify your answer with another method.

In Example 2 , we have seen that the system has only one solution.

Since it has at least one solution, the system is consistent.

We can justify this using the determinant .

We have A = \(\left[\begin{array}{ll} -3 & 2 \\ \\ 5 & -2 \end{array}\right]\).

det (A) = (-3)(-2) - (5)(2) = 6 - 10 = -4 ≠ 0.

When det (A) ≠ 0, the system has a unique solution.

Answer: Consistent with a unique solution and the answer is justified using determinants.

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Practice Questions on Matrix Equation

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FAQs on Matrix Equation

What is the definition of matrix equation.

A matrix equation is of the form AX = B and it is writing the system of equations as a single equation in terms of matrices . Here,

• A = A matrix formed by the coefficients
• X = A column matrix formed by the variables
• B = A column matrix formed by the constants

How to Solve Matrix Equation AX = B?

To solve a matrix equation AX = B:

• Find A -1 (using the formula A -1 = (adj A) / (det A).
• Find the solution using X = A -1 B.

How to Solve System of Equations Using Matrix Equation?

To solve a system of equations using matrices:

• First, write all the variables on one side and the constants on the other side of the equations. Also, write the variables in the same order in every equation.
• Write the system in the form AX = B by writing all coefficients of variables in matrix A, all variables as a column in matrix X, and all constants in a column in matrix B.

What are the Methods Used to Solve a Matrix Equation?

Here are the methods to solve a matrix equation AX = B.

• Matrix inversion method (using the formula X = A -1 B)
• Cramer's rule. Click here to learn about it.
• Gauss Jordan method.

How to Write a System of Equations as a Matrix Equation?

We can write a system of equations as a matrix equation AX = B. Here are the steps for the same:

• Maintain the order of the variables to be the same in all the equations.
• Get all variables to the left side and send the constants to the right side of every equation.
• Write the matrix A with all the coefficients of variables, where each row of A represents the coefficients of variables in one equation.
• Write the matrix X with variables in order as a column.
• Write the matrix B with constants in order as a column.

How Many Solutions a Matrix Equation Has?

A matrix equation AX = B has:

• a unique solution (consistent) if det (A) ≠ 0.
• no solution (inconsistent) if det (A) = 0 and (adj A) B ≠ O.
• infinite solutions (consistent) or no solution (inconsistent) if det (A) = 0 and (adj A) B = O.

Here, O is the null matrix.

What is the Relation Between Matrix Equation and Rank of a Matrix?

For a matrix equation AX = B, [A B] represents the augmented matrix. Let r(A) and r(A, B) represent the ranks of matrices A and [A B].

• If r(A) = r(A, B) = number of variables, then the system is consistent and has unique solution.
• If r(A) = r(A, B) < number of variables, then the system is consistent and has infinitely many solutions.
• If r(A) ≠ r(A, B), then the system is inconsistent and has no solution.