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Section 1.1 Systems of Linear Equations

DEFINITION: A linear equation in the variables x1, . . . , xn is an equation that can be written in theform

a1x1 + . . .+ anxn = b,

where a1, . . . , an and b are constants, x1, . . . , xn are variables.

EXAMPLE: The equation2x1 + x2 − 7x3 =

√5

is linear. The equation3x1x2 + 2x2

3= 1

is NOT linear.

DEFINITION: A system of linear equations (or a linear system) is a collection of one or more linearequations.

EXAMPLE:

3x1 + 2x2 + 7x3 − x4 = 6x1 + x2 − x3 + x4 = 14x1 + 3x2 + 6x3 = 8

Solving a Linear System in Two Variables

EXAMPLE: Solve the following system of linear equations:

{

x1 −2x2 = −1−x1 +3x2 = 3

Solution: We have{

x1 −2x2 = −1−x1 +3x2 = 3

⇓{

x1 −2x2 = −1x2 = 2

⇓{

x1 = 3x2 = 2

1

EXAMPLE: Solve the system of equations.

Solution: Begin by solving for in Equation 1.

Next, substitute this expression for y into Equation 2 and solve the resulting single-variable equation forx.

Finally, you can solve for by back-substituting x = 3 into the equation y = 4− x, to obtain

The solution is the ordered pair (3, 1). You can check this solution as follows.

2

EXAMPLE: Solve the system of equations.

Solution: You can eliminate the y-terms by adding the two equations.

So, x =12

8=

3

2. By back-substituting into Equation 1, you can solve for y.

The solution is

(

3

2,−1

4

)

. You can check the solution algebraically by substituting into the original system.

3

EXAMPLE: Find all solutions of the system

2x+ y = 1

3x+ 4y = 14

Solution 1(Substitution Method): We solve for y in the first equation.

2x+ y = 1 ⇐⇒ y = 1− 2x

Now we substitute for y in the second equation and solve for x:

3x+ 4y = 14

3x+ 4(1− 2x) = 14

3x+ 4− 8x = 14

−5x = 10

x = −2

Finally, we back-substitute x = −2 into the equation y = 1− 2x:

y = 1− 2(−2) = 5

Solution 2(Elimination Method): We have

2x+ y = 1

3x+ 4y = 14⇐⇒

8x+ 4y = 4

3x+ 4y = 14⇐⇒

5x = −10

3x+ 4y = 14⇐⇒

x = −2

3x+ 4y = 14

Next we substitute x = −2 into the equation 3x+ 4y = 14:

3(−2) + 4y = 14 ⇐⇒ −6 + 4y = 14 ⇐⇒ 4y = 20 ⇐⇒ y = 5

4


2x+ 3y = −4

5x− 7y = 1

Solution 1(Substitution Method): We solve for x in the first equation.

2x+ 3y = −4 ⇐⇒ 2x = −4− 3y ⇐⇒ x = −4 + 3y

2

Now we substitute for x in the second equation and solve for y:

5x− 7y = 1

−54 + 3y

2− 7y = 1

−5(4 + 3y)− 14y = 2

−20− 15y − 14y = 2

−29y = 22

y = −22

29

Finally, we back-substitute y = −22

29into the equation x = −4 + 3y

2:

x = −4 + 3

(

−22

29

)

2= −25

29

Solution 2(Elimination Method): On the one hand, we have

2x+ 3y = −4

5x− 7y = 1⇐⇒

10x+ 15y = −20

10x− 14y = 2=⇒ 29y = −22 ⇐⇒ y = −22

29

On the other hand, we have

2x+ 3y = −4

5x− 7y = 1⇐⇒

14x+ 21y = −28

15x− 21y = 3=⇒ 29x = −25 ⇐⇒ x = −25

29


−3

5x+

1

2y =

7

2

1

3x+

4

5y =

3

2

5


−3

5x+

1

2y =

7

2

1

3x+

4

5y =

3

2

Solution (Elimination Method): We have

−3

5x+

1

2y =

7

2

1

3x+

4

5y =

3

2

⇐⇒

−6x+ 5y = 35

10x+ 24y = 45

On the one hand, we have

−6x+ 5y = 35

10x+ 24y = 45⇐⇒

−30x+ 25y = 175

30x+ 72y = 135=⇒ 97y = 310 ⇐⇒ y =

310

97

On the other hand, we have

−6x+ 5y = 35

10x+ 24y = 45⇐⇒

−144x+ 120y = 840

50x+ 120y = 225=⇒ −194x = 615 ⇐⇒ x = −615

194

EXAMPLE: Solve the system of linear equations

Solution: Because the coefficients in this system have two decimal places, you can begin by multiplyingeach equation by 100 to produce a system with integer coefficients.

Now, to obtain coefficients that differ only by sign, multiply revised Equation 1 by 3 and multiply revisedequation 2 by -2.

So, you can conclude that y =−322

−23= 14. Back-substitution this value into revised Equation 2 produces

the following.

6


x+ y = 1

x+ y = 2

Answer: No solution.


3x− 2y = 4

−6x+ 4y = 7

Solution: We have

3x− 2y = 4

−6x+ 4y = 7⇐⇒

3x− 2y = 4

3x− 2y = −7

2

It follows that the system has no solution (inconsistent).


x+ y = 1

x+ y = 1

Answer: The system has infinitely many solutions (dependent).


8x− 2y = −4

−4x+ y = 2

Solution: We have

8x− 2y = −4

−4x+ y = 2⇐⇒

8x− 2y = −4

8x− 2y = −4

It follows that the system has infinitely many solutions (dependent).

7

Larger Systems of Linear Equations

EXAMPLE: Solve the system of linear equations.

Solution: From Equation 3, you know the value z. To solve for y, substitute z = 2 into Equation 2 toobtain

Finally, substitute y = −1 and z = 2 into Equation 1 to obtain

The solution is x = 1, y = −1 and z = 2.


Solution: Because the leading coefficient of the first equation is 1, you can begin by saving the x at theupper left and eliminating the other x-terms from the first column.

Now that all but the first x have been eliminated from the first column, go to work on the second column.(You need to eliminate y from the third equation.)

Finally, you need a coefficient of 1 for z in the third equation.

This is the same system that was solved in Example 1. As in that example, you can conclude that thesolution is x = 1, y = −1, and z = 2.

8


2x− y + 3z = 1

x− 2y + z = 1

2x− 3y − z = 2

Solution (Elimination Method): We have

2x− y + 3z = 1

x− 2y + z = 1

2x− 3y − z = 2

⇐⇒

2x− y + 3z = 1

2x− 4y + 2z = 2

2x− 3y − z = 2

⇐⇒

2x− y + 3z = 1

−3y − z = 1

−2y − 4z = 1

⇐⇒

2x− y + 3z = 1

−12y − 4z = 4

−2y − 4z = 1

therefore

2x− y + 3z = 1

−12y − 4z = 4

10y = −3

⇐⇒

2x− y + 3z = 1

−12y − 4z = 4

y = − 3

10

⇐⇒

2x− y + 3z = 1

−12

(

− 3

10

)

− 4z = 4

y = − 3

10

⇐⇒

2x− y + 3z = 1

z = − 1

10

y = − 3

10

so

2x−(

− 3

10

)

+ 3

(

− 1

10

)

= 1

z = − 1

10

y = − 3

10

⇐⇒

x =1

2

z = − 1

10

y = − 3

10


Solution:

Because 0 = −2 is false statement, you can conclude that this system is inconsistent and so has nosolution. Moreover, because this system is equivalent to the original system, you can conclude that theoriginal system also has no solution.

9


Solution:

This result means that Equation 3 depends on Equations 1 and 2 in the sense that it gives us no additionalinformation about the variables. So, the original system is equivalent to the system

{

x+ y − 3z = −1

y − z = 0

In the last equation, solve for y in terms of z to obtain y = z. Back-substituting y = z in the first equationproduces x = 2z − 1. Finally, letting z = a, where a is a real number, the solutions to the given systemare all of the form

x = 2a− 1, y = a, and z = a

So, every ordered triple of the form (2a− 1, a, a) is a solution of the system.

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The Augmented Matrix of a Linear System

We can write a system of linear equations as a matrix, called the augmented matrix of the system, bywriting only the coefficients and constants that appear in the equations. Here is an example.

The next Example demonstrates the elementary row operations described above.

EXAMPLE:

(a) Add −2 times the first row of the original matrix to the third row.

(b) Multiply the first row of the original matrix by1

2.

(c) Interchange the first and second rows of the original matrix.

11


x− y + 3z = 4

x+ 2y − 2z = 10

3x− y + 5z = 14

Solution: Our goal is to eliminate the x-term from the second equation and the x- and y-terms from thethird equation. For comparison, we write both the system of equations and its augmented matrix.

Now we use back-substitution to find that

z = 3 y − 2z = 1 x− y + 3z = 4

y − 2(3) = 1 x− 7 + 3(3) = 4

y − 6 = 1 x+ 2 = 4

y = 7 x = 2

The solution is (2, 7, 3).


x− 2y + 3z = 9

−x+ 3y + z = −2

2x− 5y + 5z = 17

12


x− 2y + 3z = 9

−x+ 3y + z = −2

2x− 5y + 5z = 17

Solution:

Now we use back-substitution to find that

z = 2 y + 4z = 7 x− 2y + 3z = 9

y + 4(2) = 7 x− 2(−1) + 3(2) = 9

y + 8 = 7 x+ 8 = 9

y = −1 x = 1

The solution is (1,−1, 2).

13

EXAMPLE: Show that the following system has no solutions.

x− 3y + 2z = 12

2x− 5y + 5z = 14

x− 2y + 3z = 20

Solution: We have

Now if we translate the last row back into equation form, we get 0x + 0y + 0z = 1, or 0 = 1, which isfalse. No matter what values we pick for x, y, and z, the last equation will never be a true statement.This means the system has no solution.

EXAMPLE: Show that the following system has infinitely many solutions.

−3x− 5y + 36z = 10

−x+ 7z = 5

x+ y − 10z = −4

Solution: We have

The third row corresponds to the equation 0 = 0. This equation is always true, no matter what values areused for x, y, and z. The complete solution of this system will be discussed in the next section.

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