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142 Chapter 2. Matrices and Linear Equations

2.3 Systems of Linear Equations

A linear equation in variables x1, x2, . . . is an equation such as 2x1 −3x2 + x3 −

√2x4 = 17, where each variable appears by itself (and to the

first power) and the coefficients are scalars. A set of one or more linear

equations is called a linear system. To solve a linear system means to find

values of the variables that make each equation true.

In this section, we describe a procedure for solving systems of linear

equations. Many people could find numbers x and y so that

x + y = 7

x − 2y = −2(1)

via some ad hoc process and, in fact, we have assumed you have this abil-

ity several times already in previous sections of this book. Suppose you

were faced with a system of eight equations in 17 unknowns, however.

How would you go about finding a solution in this case?

Two linear systems are said to equivalent if they have the same solutions.

For example, system (1) is equivalent to

x + y = 7

3y = 9(2)

because each system has the same solution, x = 4, y = 3.

Our procedure for solving a linear system involves transforming the given

system to another system that is equivalent but easier to solve. For ex-

ample, system (2) makes it immediately clear that y = 3, so it is easier to

solve than (1).

There are three very useful ways to change a system of linear equations

to an equivalent system. First, we can change the order of the equations.

For example, the numbers x, y , and z that satisfy

−2x + y + z = −9

x + y + z = 0

x + 3y = −3(3)

are the same as the numbers that satisfy

x + y + z = 0

−2x + y + z = −9

x + 3y = −3,

(interchanging the first two equations of (3)), so the two systems are

equivalent.

Second, in any system of equations, we can multiply any equation by a

number (different from 0) without changing the solution. For example,

x + 3y = −3 if and only if (multiplying by13)

13x +y = −1.

2.3. Systems of Linear Equations 143

Third, and this is less obvious, in any system of equations, we may always

add or subtract a multiple of one equation from another without chang-

ing the solution. Suppose, for instance, that we replace the first equation

in system (3) by the first equation plus twice the third, giving

−2x + y + z + 2(x + 3y) = −9+ 2(−3)

x + y + z = 0

x + 3y = −3.(4)

Any values of x, y , and z that satisfy this system must also satisfy sys-

tem (3) since the first equation in that system is a consequence of the

equations in (4):

−2x +y + z = [(−2x +y + z)+ 2(x + 3y)]− 2(x + 3y)

= [−9+ 2(−3)]− 2(−3) = −9.

Conversely, any values of x, y , and z that satisfy the original system (3)

also satisfy (4) since

−2x +y + z + 2(x + 3y) = −9+ 2(−3).

Thus systems (4) and (3) are equivalent.

In summary, we have described three ways that a system of equations

can be changed to an equivalent system:

1. interchange two equations;

2. multiply an equation by any number other than 0;

3. replace an equation by that equation minus any multiple of another.

By the way, it makes no difference whether we say “minus” or “plus” in

Statement 3 since, for example, adding 2E is the same as subtracting−2E,

but we prefer the language of subtraction for reasons that will become

clear later.

We illustrate how these principles can be applied to solve system (3).

First, interchange the first two equations to get

x + y + z = 0

−2x + y + z = −9

x + 3y = −3.

Now replace the second equation by that equation plus twice the first—

we denote this as E2→ E2+ 2(E1)—to obtain

x + y + z = 0

3y + 3z = −9

x + 3y = −3.(5)


We describe the transformation from system (3) to system (5) like this:

−2x + y + z = −9

x + y + z = 0

x + 3y = −3

E1↔E2−→x + y + z = 0

−2x + y + z = −9

x + 3y = −3

E2→E2+2(E1)−→x + y + z = 0

3y + 3z = −9

x + 3y = −3.

We continue with our solution, replacing equation 3 with equation 3 mi-

nus equation 1:

x + y + z = 0

3y + 3z = −9

x + 3y = −3

E3→E3−E1−→x + y + z = 0

3y + 3z = −9

2y − z = −3.

Now we multiply the equation 2 by13 and replace equation 3 with equa-

tion 3 minus twice equation 2:

x + y + z = 0

3y + 3z = −9

2y − z = −3

E2→ 13E2−→

x + y + z = 0

y + z = −3

2y − z = −3

E3↔E3−2(E2)−→x + y + z = 0

y + z = −3

− 3z = 3.

We have reached our goal, which was to transform the original system

to one which is upper triangular, like this: @@ . At this point, we can

obtain a solution in a straightforward manner by using a technique called

back substitution, which means solving the equations from the bottom

up. The third equation gives z = −1. The second says y + z = −3, so

y = −3− z = −2. The first says x +y + z = 0, so x = −y − z = 3.

It is easy to check that x = 3, y = −2, z = −1 satisfies the original system

(3), so we have our solution. We strongly advise you always to check

solutions!

Did you get tired of writing the x, the y , the z, and the = in the solution

just illustrated? Only the coefficients change from one stage to another,

so it is only the coefficients we really have to write down. Let’s solve

system (3) again, by different means.

First we write the system in matrix form, like this:

−2 1 1

1 1 1

1 3 0

x

y

z

=

−9

0

3

.


This is Ax = b with A =

−2 1 1

1 1 1

1 3 0

, the matrix of coefficients, x =

x

y

z

and b =

−9

0

−3

. Next, we write down the augmented matrix

[A|b] =

−2 1 1 −9

1 1 1 0

1 3 0 −3

,

which consists of A followed by a vertical line and then b.

Just keeping track of coefficients, with augmented matrices, the solution

to system (3) now looks like this:

−2 1 1 −9

1 1 1 0

1 3 0 −3

R1↔R2−→

1 1 1 0

−2 1 1 −9

1 3 0 −3

R2→ R2+ 2(R1)R3→ R3− R1−→

1 1 1 0

0 3 3 −9

0 2 −1 −3

R2→ 13R2−→

1 1 1 0

0 1 1 −3

0 2 −1 −3

(6)

R3→R3−2(R2)−→

1 1 1 0

0 1 1 −3

0 0 −3 3

R3→− 13R3−→

1 1 1 0

0 1 1 −3

0 0 1 −1

.

Here we have used R for row rather than E for equation to describe the

changes to each system, writing, for example, R3 → R3 − 2(R2) at the

last step, rather than E3 → E3 − 2(E2) as we did before. The three ba-

sic operations that we have been performing on equations now become

operations on rows, called the elementary row operations.

2.3.1 The Elementary Row Operations.

1. Interchange two rows.

2. Multiply a row by any scalar except 0.

3. Replace a row by that row minus a multiple of another.

We could equally well have stated the third elementary row operation as

3. replace a row by that row plus a multiple of another,

since R1 + c(R2) = R1 − (−c)(R2), but, as we said before, there are ad-

vantages to using the language of subtraction. In particular, when doing

so, the multiple has a special name. It’s called a multiplier, −c in our

example, a concept that will be important to us later.


2.3.2 Definitions. The main diagonal of an m × n matrix A is the list of ele-

ments a11, a22, . . . . A matrix is diagonal if its only nonzero entries lie on

the main diagonal, upper triangular if all entries below (and to the left)

of the main diagonal are 0, and lower triangular if all entries above (and

to the right of) the main diagonal are 0.

[

@@@

] [

@@@0⋆] [

@@@0

⋆

]

Diagonal Upper triangular Lower triangular

Figure 2.2: Schematic representations of a diagonal, lower triangu-

lar, and upper triangular matrix.

2.3.3 Examples. The main diagonal of

1 2 3

4 5 6

7 8 9

is 1,5,9, and the main diag-

onal of

[

7 9 3

6 5 4

]

is 7,5. The matrix

[

−3 4 1 1

0 8 2 −3

]

is upper triangular

while the matrix

[

1 0

−2 1

]

is lower triangular. The matrix

1 0 0

0 −7 0

0 0 2

is diagonal. ℄

The procedure illustrated in (6), whereby a matrix is transformed to anThe author is a ninthgeneration student ofGauss, throughGudermann,Weierstrass,Frobenius, Schur,Brauer, Bruck,Kleinfeld, andAnderson.

upper triangular matrix via elementary row operations, is called Gaussian

elimination after the German mathematician Karl Friedrich Gauss (1777–

1855), who first used this process to solve systems of 17 linear equations

in order to determine the orbit of a new planet. (See 2.3.2 for a precise

definition of upper triangular matrix.)

2.3.4 Gaussian Elimination. To move a matrix to one that is upper triangular,

1. If the matrix is all 0, do nothing. Otherwise, change the order of

the rows, if necessary, to ensure that the first non-zero entry in the

first row is a leftmost nonzero entry in the matrix. This number is

called a pivot and the column containing this pivot is the first pivot

column.

2. Use the third elementary row operation to get 0s in the pivot column

below the pivot.

3. Repeat steps 1 and 2 on the submatrix immediately to the right of

and below the pivot.


4. Repeat step 3 until you reach the last nonzero row. The first nonzero

entry in this row (assuming this row has never been multiplied by a

scalar) is the last pivot.

When doing calculations by hand, it is often helpful to make the first

nonzero entry in a row a 1 (using the second elementary row operation)

since a 1 makes it especially easy to get 0s below it. (Note, however, that

by so doing, some pivots will disappear.)

2.3.5 Remark . We emphasize that the elementary row operations change a

matrix A to a matrix U with the property that the systems of equations

Ax = b and Ux = b are equivalent, that is, they have the same solutions.

In particular, this is true we change A to an upper triangular matrix U .

Since it is easy to solve Ux = b (by back substitution), we have a proce-

dure for solving Ax = b.

2.3.6 To Solve a System of Linear Equations

1. Write the system in the form Ax = b, where A is the matrix of co-

efficients, x is the vector of unknowns, and b is the vector whose

components are the constants to the right of the equals signs.

2. Add a vertical line and b to to the right of A to form the augmented

matrix [A|b].

3. Reduce the augmented matrix to an upper triangular matrix by Gaus-

sian elimination.

4. Write down the equations that correspond to the triangular matrix

and solve by back substitution.

To illustrate this last point, look again at the final matrix in the Gaussian

Elimination process demonstrated in (6). This matrix is

1 1 1 0

0 1 1 −3

0 0 1 −1

.

The corresponding equations are

x + y + z = 0

y + z = −3

z = −1.

Back substitution means solving these equations from the bottom up:

We get z = −1, then y = −2, and then x = 3. Notice that this solution


is unique because it is the only one; 3, −2, and −1 are the only values of

x, y , and z that satisfy the original system (3). Equivalently,

x

y

z

is the

only vector that satisfies

−2 1 1

1 1 1

1 3 0

x

y

z

=

−9

0

−3

.

As we shall soon see, solutions are not always unique. In fact, solutions

may not even exist.

2.3.7 Problem . Solve the systemx + 2y = 1

4x + 8y = 3.

Solution. This is Ax = b with A =[

1 2

4 8

]

, x =[

x

y

]

, and b =[

1

3

]

. Since

the (1,1) entry of A is not zero, no row interchanges are necessary to

start. The first pivot is that 1 in the (1,1) position. Gaussian elimination

on the augmented matrix requires only one step.

[A|b] =[

1 2 1

4 8 3

]

R2→R2−4(R1)−→[

1 2 1

0 0 −1

]

.

The equations that correspond to this triangular matrix are

x + 2y = 1

0 = −1.

The last equation is not true, so there is no solution, a fact we might have

noted earlier. If x + 2y = 1, then certainly 4x + 8y = 4(x + 2y) = 4, not

3. So if the first equation is satisfied, the second one is not. A system like

this, with no solution, is called inconsistent. �

2.3.8 Problem . Solve the system2x − 3y + 6z = 14

x + y − 2z = −3.

Solution. This is Ax = b, with

A =[

2 −3 6

1 1 −2

]

, x =

x

y

z

, and b =[

14

−3

]

. (7)

We form the augmented matrix

[A|b] =[

2 −3 6 14

1 1 −2 −3

]

.


While as before, there is no need for initial row interchanges, it is easier

to work with a 1 rather than a 2 in the (1,1) position, so we interchange

rows one and two (the 1 becomes the first pivot), and then apply the third

elementary row operation to get a 0 below the pivot.

[

2 −3 6 14

1 1 −2 −3

]

R1↔R2−→[

1 1 −2 −3

2 −3 6 14

]

R2→R2−2(R1)−→[

1 1 −2 −3

0 −5 10 20

]

Now we focus on the submatrix immediately to the right and below the

first pivot. This is [−5 10 20]. The second pivot is the −5. The matrix

is upper triangular. We can stop here, or divide the last row by the −5,

obtaining[

1 1 −2 −3

0 1 −2 −4

]

.

The equations corresponding to this matrix are

x + y − 2z = −3

y − 2z = −4.

Applying back substitution, the last equation says y = 2z − 4 and the

first says x +y − 2z = −3. So x = −y + 2z− 3 = −(2z− 4)+ 2z− 3 = 1.

The solution isx = 1

y = 2z − 4

z = z.

Our solution shows that x = 1 and that z can be chosen arbitrarily, but

that y is then determined in terms of z. To emphasize the idea that z can

be any number, we introduce the parameter t and let z = t. The solution

then takes the formx = 1

y = 2t − 4

z = t.

The variable z is called free because, as the equation z = t tries to

show, z can be any number. The system of equations in this problem has

infinitely many solutions, one for each value of t. We write the solution

as a vector, like this:

x

y

z

=

1

2t − 4

t

=

1

−4

0

+ t

0

2

1

.

Such an equation should look familiar. It is the equation of a line, the one

through (1,−4,0) with direction d =

0

2

1

. Geometrically, this is what we


expect. Each of the two given equations describes a plane in 3-space.

When we solve the system, we are finding the points that lie on both

planes. Since the planes are not parallel, the points that lie on both form

a line. �

Reading Challenge 1. How do you know that the planes described by the

equations in Problem 2.3.8 are not parallel?

Each solution to the linear system in Problem 2.3.8 is the sum of the

vector

1

−4

0

and a scalar multiple of

0

2

1

. The solution

1

−4

0

is called a

particular solution. The vector

0

2

1

is a solution to the system Ax = 0.

It is characteristic of linear systems that whenever there is more than

one solution, each solution is the sum of a particular solution and the

solution to Ax = 0. We shall have more to say about this in Section 2.4.

Reading Challenge 2. Verify that

1

−4

0

is a solution to the system Ax = b

in Problem 2.3.8, and that

0

2

1

is a solution to Ax =[

0

0

]

.

2.3.9 Problem . Solve the system

3y − 12z − 7w = −15

y − 4z = 2

−2x + 4y + 5w = −1.

Solution. This is Ax = b with

A =

0 3 −12 −7

0 1 −4 0

−2 4 0 5

, x =

x

y

z

w

, b =

−15

2

−1

.

We form the augmented matrix and immediately interchange rows one

and three in order to get a nonzero entry in the (1,1) position.

0 3 −12 −7 −15

0 1 −4 0 2

−2 4 0 5 −1

R1↔R3−→

−2 4 0 5 −1

0 1 −4 0 2

0 3 −12 −7 −15

.

The first pivot is −2. Now we continue with the submatrix immediately

to the right of and below this pivot. The second pivot is the 1 in the (2,2)


position. We use the third elementary row operation to put a 0 below it

R3→R3−3(R2)−→

−2 4 0 5 −1

0 1 −4 0 2

0 0 0 −7 −21

,

obtaining an upper triangular matrix. Back substitution proceeds like

this:−7w = −21 so w = 3,

y − 4z = 2, which we write as y = 2+ 4z.

This suggests the z is free, so we set z = t and obtain y = 2+4t. Finally, Free variables arethose associated withthose columns whichdo not contain pivots.

−2x + 4y + 5w = −1,

so −2x = −1−4y−5w = −1−4(2+4t)−15 = −24−16t and x = 12+8t.

Our solution vector is

x =

12+ 8t

2+ 4t

t

3

=

12

2

0

3

+ t

8

4

1

0

.

As before, the vector

12

2

0

3

is a particular solution to Ax = b and

8

4

1

0

is

a solution to Ax = 0. �

In general, there are three possible outcomes when we attempt to solve a

system of linear equations.

2.3.10 A system of linear equations may have

1. a unique solution,

2. no solution, or

3. infinitely many solutions.

The solution to system (3) was unique. In Problem 2.3.7, you met an in-

consistent system (no solution) and, in Problems 2.3.8 and 2.3.9, systems

with infinitely many solutions. Interestingly, a linear system can’t have

just three solutions or five solutions or ten solutions, because if a solu-

tion is not unique, there is a free variable that can be chosen in infinitely

many ways. But we are getting a bit ahead of ourselves!

Reading Challenge 3. How many solutions does the systemx2 +y2 = 8

x −y = 0

have? (Note that this system is not linear!)


⋆⋆

⋆⋆

At this point, we confess that the goal of

Gaussian elimination is to transform a ma-

trix not simply into any upper triangular

matrix, but into a special kind of upper

triangular matrix called a row echelon ma-

trix. A row echelon matrix takes its name

from the French word “echelon” meaning “step.” When a matrix is in row

echelon form, the path formed by the pivots resembles a staircase.

2.3.11 Definition . A row echelon matrix is an upper triangular matrix with the

following properties:

1. All rows consisting entirely of 0s are at the bottom;

2. The pivots step from left to right as you read down the matrix (a

pivot is the first nonzero entry in a nonzero row);

3. All the entries in a column below a pivot are 0.

A row echelon form of a matrix A is a row echelon matrix U to which A

can be moved via elementary row operations.

Any (nonzero) matrix, in row echelon form or not, has pivots.

2.3.12 Definition . A pivot of a row echelon matrix U is the first nonzero el-

ement in a nonzero row. A column containing a pivot is called a pivot

column of U . If U is a row echelon form of a matrix A, then a pivot

column of A is a column that becomes a pivot column of U . If U was

obtained without any multiplication of rows by nonzero scalars, then the

pivots of A are the pivots of U .

2.3.13 Examples. • The matrix U =

0 0 2k1 5

0 0 0 0 1k0 0 0 0 0

is a row echelon matrix.

The pivots are the circled entries. The pivot columns are columns three

and five.

• U =

0 0 2 1 5

0 0 0 0 1

0 0 0 0 3

is an upper triangular matrix that is not row

echelon. If row echelon, the 1 in the (2,5) position would be a pivot,

but then the number below it should be 0, not 3.

• The matrix U =

7k1 3 4 5

0 0−2l3 1

0 0 0 0 4k

is a row echelon matrix. The pivots


are circled. Since these are in columns one, three, and five, the pivot

columns are columns one, three, and five. ℄

2.3.14 Examples. Here are some matrices that are not in row echelon form:

2 0 7

0 0 1

0 0 1

,

6 0 3 2

0 0 5 4

0 1 0 2

,

3 0 −1 5

0 0 0 0

0 −4 2 −3

. ℄

Reading Challenge 4. Explain why each of the matrices in Example 2.3.14

is not in row echelon form.

2.3.15 Examples. • Let A =

3 −2 1 −4

6 −4 8 −7

9 −6 9 −12

. Then

A

R2→ R2− 2(R1)R3→ R3− 3(R1)−→

3 −2 1 −4

0 0 6 1

0 0 6 0

R3→R3−6(R2)−→

3 −2 1 −4

0 0 6 1

0 0 0 −1

= U,

so the pivots of A are 3, 6, and −1, and the pivot columns of A are one,

three, and four.

• Let A =

−2 3 4

1 0 1

4 −5 6

. Then

AR1↔R2−→

1 0 1

−2 3 4

4 −5 6

R2→ R2+ 2(R1)R3→ R3− 4(R1)−→

1 0 1

0 3 6

0 −5 2

R3→R3+ 53 (R2)−→

1 0 1

0 3 6

0 0 12

= U,

so the pivots of A are 1, 3, and 12, and the pivot columns of A are

columns one, two, and three.

Perhaps a more natural way to have reduced to row echelon form is


this:

A→

1 0 1

−2 3 4

4 −5 6

→

1 0 1

0 3 6

0 −5 2

→

1 0 1

0 1 2

0 −5 2

→

1 0 1

0 1 2

0 0 12

.

This is another row echelon form of A, but we have lost the pivots

(but not the pivot columns). We also see here that row echelon form

is not unique. Moreover, the pivots are not unique. Had we pro-

ceeded without interchanging rows one and two at the beginning,

like this,

A→

−2 3 4

032 3

0 1 14

→

−2 3 4

032

3

0 0 12

,

the pivots would have been −2,32 , and 12. ℄

Reading Challenge 5. Find the pivots and identify the pivot columns of

A =[

−3 6 4

2 −4 0

]

.

2.3.16 Remark . As we have observed, row echelon form is not unique, since

different sequences of elementary row operations can transform a matrix

A into different row echelon matrices. Thus we say a row echelon form,

not the row echelon form. On the other hand, the columns in which the

pivots eventually appear—that is, the pivot columns—are unique. This

follows from the uniqueness of the reduced row echelon matrix. See

Definition 2.7.15 and Appendix A of Linear Algebra and its Applications

by David C. Lay, Addison-Wesley (2000).

2.3.17 Remark . The definition of row echelon form varies from author to au-

thor. It was long the custom, and some authors still require, that the

leading nonzero entry in a nonzero row of a row echelon matrix be a 1.

This makes a lot of sense when doing calculations by hand, but is much

less important in today’s highly technological world. Having said this,

in the rest of this section, we will regularly divide each row containing a

pivot by that pivot, thus turning the first nonzero entry in that row into

a 1. Putting 0s below a 1 by hand is much easier (and less susceptible to

errors) than putting 0s below23 , for instance.


2.3.18 Example. Suppose we wish to solve the system

2x − y + z = −7

x + y + z = −2

3x + y + z = 0

without the use of a computer. The steps of Gaussian elimination we

would probably employ are these:

[A|b] =

2 −1 1 −7

1 1 1 −2

3 1 1 0

R1↔R2−→

1 1 1 −2

2 −1 1 −7

3 1 1 0

R2←− R2− 2(R1)R3←− R3− 3(R1)−→

1 1 1 −2

0 −3 −1 −3

0 −2 −2 6

R2↔R3−→

1 1 1 −2

0 −2 −2 6

0 −3 −1 −3

R2←−(− 12 )R2−→

1 1 1 −2

0 1 1 −3

0 −3 −1 −3

R3→R3+3(R2)−→

1 1 1 −2

0 1 1 −3

0 0 2 −12

R3→ 12 (R3)−→

1 1 1 −2

0 1 1 −3

0 0 1 −6

.

The final matrix is in row echelon form and, by back substitution, we

quickly read off the solution: z = −6, y = −3−z = 3, x = −2−y−z = 1.

We went to a lot of effort to make all leading nonzero entries 1. Look,

however, at how much more efficiently the Gaussian elimination process

would proceed if we didn’t so insist:

[A|b] =

2 −1 1 −7

1 1 1 −2

3 1 1 0

R2→ R2− 12 (R1)

R3→ R3− 32 (R1)−→

2 −1 1 −7

032

12

32

052−1

2212

R3→R3− 53 (R2)−→

2 −1 1 −7

032

12

32

0 0 −43 8

(two steps instead of six). The first elementary row operation—multiply

a row by a nonzero scalar—is by far the least important. It is useful when

doing calculations by hand, but not otherwise. ℄


Free Variables Correspond to Nonpivot Columns

The pivot columns of a matrix are extremely important, for many rea-

sons. The variable that corresponds to a pivot column is never free. For

example, look at the pivot in column three of

1 1 3 4 5

0 0 1 3 1

0 0 0 1 −3

. (8)

The second equation reads x3 + 3x4 = 1. Thus x3 = 1 − x4 is not free;When a system ofequations involvesmore than two orthree variables, weusually label thevariablesx1, x2, x3, . . . (so thatwe don’t run out ofletters).

it is determined by x4. On the other hand, a variable that corresponds to

a column that is not a pivot column is always free. The equations that

correspond to the rows of the matrix in (8) are

x1 + x2 + 3x3 + 4x4 = 5

x3 + 3x4 = 1

x4 = −3.

The variables x1, x3 and x4 are not free since

x4 = −3

x3 = 1− 3x4

x1 = 5− x2 − 3x3 − 4x4,

but there is no equation of the form x2 = ∗, so x2 is free. To summarize,

2.3.19Free variables are those that correspond to columns that are

not pivot columns.

Reading Challenge 6. Suppose a system of equations in the variables x1, x2, x3, x4

leads to the row echelon matrix

0 1 2 7 5

0 0 0 1 1

0 0 0 0 0

. What are the free

variables?

2.3.20 Problem . Solve2x1 − 2x2 − x3 + 4x4 = 9

−x1 + x2 + 2x3 + x4 = −3.

Solution. The given system is Ax = b where

A =[

2 −2 −1 4

−1 1 2 1

]

, x =

x1

x2

x3

x4

, and b =[

9

−3

]

.

The augmented matrix is

[A|b] =[

2 −2 −1 4 9

−1 1 2 1 −3

]

.


Gaussian elimination begins[

2 −2 −1 4 9

−1 1 2 1 −3

]

E1↔E2−→[

−1 1 2 1 −3

2 −2 −1 4 9

]

(the first pivot is the −1 in the (1,1) position)

E2→E2+2(E1)−→[

−1 1 2 1 −3

0 0 3 6 3

]

(using the third elementary row operation to put a 0 below the first pivot).

We have row echelon form. The second pivot is the 3, the first nonzero

entry of the last row. The pivot columns are columns one and three. The

columns (of A) which are not pivot columns are columns two and four.

The corresponding variables are free and we so signify by setting x2 = tand x4 = s. The equations corresponding to the rows of the row echelon

matrix are−x1 + x2 + 2x3 + x4 = −3

3x3 + 6x4 = 3.

We solve these by back substitution (that is, from the bottom up).

The second equation says 3x3 = 3− 6x4 = 3− 6s, so x3 = 1− 2s.

The first equation says −x1 = −x2 − 2x3 − x4 − 3

= −t − 2(1− 2s)− s − 3

= −t + 3s − 5, so x1 = t − 3s + 5.

.

The solution isx1 = t − 3s + 5

x2 = t

x3 = 1− 2s

x4 = s

which, in vector form, is

x =

x1

x2

x3

x4

=

t − 3s + 5

t

1− 2s

s

=

5

0

1

0

+ t

1

1

0

0

+ s

−3

0

−2

1

.

Note that

5

0

1

0

is a particular solution to Ax = b while t

1

1

0

0

+ s

−3

0

−2

1

is

the solution to Ax = 0. �

2.3.21 Problem . Express the vector

[

9

−3

]

as a linear combination of the columns

of A =[

2 −2 −1 4

−1 1 2 1

]

.


Solution. We seek a, b, c, and d so that[

9

−3

]

= a[

2

−1

]

+ b[

−2

1

]

+ c[

−1

2

]

+ d[

4

1

]

.

By 2.1.33, this is just A

a

b

c

d

=[

9

−3

]

, the system considered in Prob-

lem 2.3.20. There are many solutions, one of which is

a

b

c

d

=

5

0

1

0

. So,

for instance,

[

9

−3

]

= 5

[

2

−1

]

+ 0

[

−2

1

]

+ 1

[

−1

2

]

+ 0

[

4

1

]

, the coefficients

being the components of the vector

5

0

1

0

. �

2.3.22 Problem . Is

[

1

−6

]

a linear combination of

[

1

−4

]

,

[

2

0

]

, and

[

3

−1

]

?

Solution. The question asks if there exist scalars a, b, c so that

[

1

−6

]

= a[

1

−4

]

+ b[

2

0

]

+ c[

3

−1

]

=[

1 2 3

−4 0 −1

]

a

b

c

.

So we want to know whether there exists a solution to Ax = b, with

x =

a

b

c

and b =[

1

−6

]

. We apply Gaussian elimination to the augmented

matrix as follows:

[A|b] =[

1 2 3 1

−4 0 −1 −6

]

→[

1 2 3 1

0 8 11 −2

]

.

This is row echelon form. There is one column that is not a pivot column,

column three. Thus the third variable, c, is free. There are infinitely

many solutions. Yes, the vector

[

1

−6

]

is a linear combination of the other

three. �

Reading Challenge 7. Find specific numbers a, b, and c so that

[

1

−6

]

=

a

[

1

−4

]

+ b[

2

0

]

+ c[

3

−1

]

.


Do you, dear reader, feel more comfortable with numbers than letters?

Why? Here are a couple of worked problems we hope will alleviate fears.

2.3.23 Problem . Show that the system

x + y + z = a

x + 2y − z = b

2x − 2y + z = chas a solution for

any values of a, b, and c.

Solution. We apply Gaussian elimination to the augmented matrix.

1 1 1 a

1 2 −1 b

2 −2 1 c

→

1 1 1 a

0 1 −2 b − a0 −4 −1 c − 2a

→

1 1 1 a

0 1 −2 b − a0 0 −9 (c − 2a)+ 4(b − a)

.

The equation which corresponds to the last row is −9z = −6a + 4b + c.

For any values of a, b, c, we can solve for z. The equation corresponding

to the second row says y = 2z+b−a hence knowing z gives us y . Finally

we determine x from the first row. �

2.3.24 Problem . Find conditions on a and b under which the system

x + y + z = −2

x + 2y − z = 1

2x + ay + bz = 2

has

1. no solution;

2. infinitely many solutions.

3. a unique solution;

Solution. We apply Gaussian elimination to the augmented matrix (ex-

actly as if the letters were numbers).

1 1 1 −2

1 2 −1 1

2 a b 2

→

1 1 1 −2

0 1 −2 3

0 a− 2 b − 2 6

→

1 1 1 −2

0 1 −2 3

0 0 (b − 2)+ 2(a− 2) 6− 3(a− 2)

.

The equation corresponding to the last row reads 2a+b− 6 = −3a+ 12.


1. If 2a+b−6 = 0 while −3a+12 6= 0, there is no solution. Thus there

is no solution if a 6= 4 and 2a+ b − 6 = 0.

2. If 2a+b−6 = 0 and −3a+12 = 0, the equation corresponding to the

last row reads 0 = 0. The variable z is free and there are infinitely

many solutions. So there are infinitely many solutions in the case

a = 4, b = −2.

3. If 2a+ b − 6 6= 0, the equation corresponding to the last row allows

us to determine z and then, continuing with back substitution, we

can determine y and x. There is a unique solution if 2a+b−6 6= 0.

�

Answers to Reading Challenges

1. The two normals,

2

−3

6

and

1

1

−2

, are not parallel.

2. A

1

−4

0

=[

2 −3 6

1 1 −2

]

1

−4

0

=[

14

−3

]

= b;

A

0

2

1

=[

2 −3 6

1 1 −2

]

0

2

1

=[

0

0

]

.

3. Setting x = y in the first equation, we obtain 2x2 = 8, so x2 = 4. The

system has two solutions, x = 2, y = 2 and x = −2, y = −2.

4. Since A →[

−3 6 4

0 083

]

= U , the pivots are −3 and83, and the pivot

columns are columns one and three.

5. The 1 in the (2,3) position of

2 0 7

0 0 1

0 0 1

is a pivot, so the number below

it should be a 0.

The leading nonzero entries in

6 0 3 2

0 0 5 4

0 1 0 2

are 6, 5, and 1, but these do

not step from left to right as you read down the matrix.

The zero row of

3 0 −1 5

0 0 0 0

0 −4 2 −3

is not at the bottom of the matrix.

6. There are pivots in columns two and four. There are no pivots in columns

one and three, so the corresponding variables x1 and x3 are free.


7. We must solve the system whose augmented matrix has row echelon form[

1 2 3 1

0 8 11 −2

]

. Thus c = t is free, 8b = −2 − 11c, so b = − 14− 11

8t

and a+2b+3c = 1, so that a = 1−2b−3c = 1−2(− 14− 11

8t)−3t = 3

2− 1

4t.

One particular solution is obtained with t = 0, giving a = 32, b = − 1

4, c = 0.

Checking our answer, we have

[

1

−6

]

= 32

[

1

−4

]

− 14

[

2

0

]

+ 0

[

3

−1

]

.

True/False Questions

Decide, with as little calculation as possible, whether each of the following state-

ments is true or false and explain your answer whenever you say “false.” (An-

swers can be found in the back of the book.)

1. πx1 +√

3x2 + 5√x3 = ln 7 is a linear equation.

2. The systems

x − y − z = 0

−4x + y + 2z = −2

2x + y − z = 3and

x − y − z = 0

−4x + y + 2z = −2

3y + z = 3

have the same solutions.

3. It is possible for a system of linear equations to have exactly two (different)

solutions.

4. The matrix

0 4 3 −4 1

0 0 0 5 1

0 0 0 0 −1

is in row echelon form.

5. If U is a row echelon form of A, the systems Ax = b and Ux = b have the

same solutions.

6. The matrix

[

0 1 1 0 0

2 0 1 5 1

]

is in row echelon form.

7. If a matrix A has row echelon form U , the pivot columns of A are the pivot

columns of U .

8. If Gaussian elimination on the augmented matrix of a system of linear

equations leads to the matrix

0 1 3 −4 1

0 0 0 1 1

0 0 0 0 1

, the system has a

unique solution.

9. In the solution to a system of linear equations, free variables are those that

correspond to columns of row echelon form that contain the pivots.

10. If a linear system Ax = b has a solution, then b is a linear combination of

the columns of A.


11. The pivots of the matrix

4 0 2

0 7 3

0 0 1

are 4, 7, and 1.


4 0 2

0 7 1

0 0 0

are 4, 7, and 0.


4 0 2

0 7 1

0 1 1

are 4, 7, and67.

14. If every column of a matrix A is a pivot column, then the system Ax = b

has a solution for every vector b.

EXERCISES

Solutions to exercises marked [BB] can be found in the Back of the Book.

1. Which of the following matrices are not in row echelon form? Give a reason

in each case.

(a) [BB]

[

0 1

1 0

]

(b)

1 2 3

0 1 0

0 0 1

(c)

1 5

0 0

0 1

0 0

0 0

(d)

1 2 3 4

0 0 −1 10

0 0 0 0

2. Reduce each of the following matrices to row echelon form. In each case,

identify the pivots and the pivot columns.

(a) [BB]

1 −1 −2

2 −3 −5

−1 4 5

(b)

2 1 4 2

3 0 2 1

5 2 3 5

(c) [BB]

1 4 5 2

3 13 20 8

−2 −10 −16 −4

1 10 38 28

(d)

1 2 1 3 1

−1 −1 2 −1 4

2 8 15 11 26

0 2 8 −2 21

2.3 Systems of Linear Equations - University of South Alabama · 2.3 Systems of Linear Equations A linear equation in variables x1 ... for solving systems of linear equations. ...

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