systems of linear equationsfacstaff.cbu.edu/~wschrein/media/m105 notes/m105c2.pdf1. systems of...

CHAPTER 2

Systems of Linear Equations

1. Systems of Linear Equations

Solution of an equation with one variable:

5 is a solution of 2x + 6 = 16 since 2(5) + 6 = 16 is a true statement.

Solution of an equation with two variables:

(2, 7) is a solution of 2x� y = �3 since 2(2)� 7 = �3 is a true statment. Wemight also say that x = 2 and y = 7 is a solution. Also, (5, 3) is a solution ofx = 5 (thought of as x + 0y = 5) since 5 = 5 is a true statement.

Definition. A system of equations in two variables is a collection of twoor more linear equations with two variables. The solution to the system ofequations, if it exists, is the point(s) of intersection of the graphs of the equa-tions. In other terms, a solution is an ordered pair that satisfies each of theequations in the system.

Note. This definition is easily generalized to more variables.

29

30 2. SYSTEMS OF LINEAR EQUATIONS

Graphical Method. It is hard to be accurate here.

Example.

(1) x� 2y = 2

(2) x + y = 5

The solution appears to be x = 4, y = 1 or (4, 1), but we need to check to besure, because maybe it is (3.98, 1.02). There is no visual way to tell.

4� 2(1) = 2

4 + 1 = 5

The algebra convinces us that we have the correct solution. These equationsare consistent (and independent), i.e., they have a single solution. ⇤

1. SYSTEMS OF LINEAR EQUATIONS 31

Example.

(1) x + 2y = �4

(2) 2x + 4y = 8

The graphs do not intersect since the lines are parallel. Thus there is no solution.These equations are inconsistent. ⇤


Example.

(1) 2x + 4y = 8

(2) x + 2y = 4

Both equations simplify to y = �1

2x + 2, so they have the same graph. These

equations are consistent and dependent, i.e., there are infinitely many solutions.You can say the solution is all points on the line 2x + 4y = 8 or all points onthe line x + 2y = 4. ⇤


Example. Solve using the calculator:

(1) 6x� 2y = 10

(2)2

3x + y = 4

We need to solve these equations for y:

(1) 2y = 6x� 10 =) y = 3x� 5

(2) y = �2

3x + 4

Find the approximate solution by using the Technology Tip on page 55. Weget

x = 2.4545455, y = 2.3636364.

The exact solution is

x =27

11, y =

26

11.

Substitution Method. This method is appropriate for cases where a variablehas a coe�cient of ±1. Solve for that variable in that equation, and thensubstitute into the other equation.

Example.

3x� y = 7

2x + 3 y = 1

��3x� 7 = y

2x + 3(3x� 7) = 1

2x + 9x� 21 = 1

11x = 22

x = 2

y = 3(2)� 7

y = �1


Elimination (by Addition).

Example.

4x + 3y = 26 (⇥11) To give opposite

3x� 11y = �7 (⇥3) coe�cients for y.

��44x + 33y = 286

9x� 33y = �21

��53x = 265

x = 5

Substitute this value into any previous equation with two variables to get y:

4(5) + 3y = 26

20 + 3y = 26

3y = 6

y = 2

Example.

2x + 5y = �3 (⇥(�2))

4x + 10y = 2

�� 4x� 10y = 6

4x + 10y = 2

��0 = 8

When a problem reduces to a false numerical equation, there is no solution, likethe second graphical example. When a problem reduces to a true numericalequation, the two lines coincide, like the third graphical example. ⇤


Example.

(1) y = .25x + 1

(2) y = �4x + 18

(3) y = �x + 6

We need to find points common to all three lines. We first pick any two linesto see if there is a solution. If not, the problem has no solution. We will chooseequations (1) and (2) and use substitution.

.25x + 1 = �4x + 18

x + 4 = �16x + 72

17x = 68

x = 4

Now substitute this value for x into one of the two equations we chose. Wesubstitute into equation (2).

y = �4(4) + 18

y = �16 + 18

y = 2

We have now found a solution for (1) and (2). We now substitute this into (3)to see if we get a true statement.

2 = �4 + 6 (TRUE)

Since we get a true statement, (4, 2) is the solution of this system of threeequations.

If we had gotten a false statement from (3), such as if (3) was y = �x + 10,then there is no solution.


Problem (page 75 #28).

t = # of years since the 1990-91 school year.

F (t) = # of Female athletes in year t.

F (t) = 0.104t + 1.83 million

M(t) = # of Male athletes in year t.

M(t) = 0.0606t + 3.33 million

According to the model, when will the number of female athletes surpass thenumber of male athletes? Solve

y = 0.104t + 1.83

y = 0.0606t + 3.33

Use substitution:

0.104t + 1.83 = 0.0606t + 3.33

.0434t = 1.5

t =1.5

.0434= 34.56

Thus the number of female athletes would surpass the number of male athletesin the 2025-26 school year if the model continued to hold. ⇤

2. USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION 37

2. Using Matrices to Solve Linear Systems of Equation

Definition. A =

242 3 1 3

3 2 0 54 2 3 1

35 is a matrix, an array of numbers in rows

and columns. The plural of matrix is matrices. The 12 numbers are the entriesof A. A is a 3⇥ 4 matrix since it has 3 rows and 4 columns.

An n⇥ n matrix (same number of rows as columns) is a square matrix.

B =

24 2 1 6�3 4 01 0 2

35 is a 3⇥ 3 square matrix.

The 4⇥ 1 matrix

2664

1234

3775is a column matrix.

The 1⇥ 3 matrix⇥5 �3 2

⇤is a row matrix.

The entries of a 3⇥ 4 matrix A may be represented as

A =

24a11 a12 a13 a14

a21 a22 a23 a24

a31 a32 a33 a34

35

a23 means the entry in row 2, column 3. ⇤


Particular System General System

2x + 8y = 13 a11x + a12y = k1

5x� 3y = �6 a21x + a22y = k2

a21 means 2nd equation, coe�cient of the 1st variable.

k1 means constant or RHS (right-hand side) of 1st equation.

Coe�cient matrices:2 85 �3

� a11 a12

a21 a22

�

Constant or RHS matrices:13�6

� k1

k2

�

Augmented matrices:2 8 | 135 �3 | �6

� a11 a12 | k1

a21 a22 | k2

�

We will often skip the vertical bar.

Since an augmented matrix clearly represents a system of linear equations, wecan work with these matrices to solve the equations.

Equivalent Systems of Equations

The following operations on a system of equations result in a system of equationswith the same solution as the original system:

(1) Interchange (change the position) of two equations.

(2) Multiply an equation by a nonzero number.

(3) Add a nonzero multiple of one equation to a nonzero multiple of anotherequation and replace either equation with the result.


Related to this, we have the following.

Row Operations

For any augmented matrix of a system of equations, the following row operationsyield an augmented matrix of an equivalent system of equations:

(1) Interchange (change the position) of two rows.

(2) Multiply a row by a nonzero number.

(3) Add a nonzero multiple of one row to a nonzero multiple of another rowand replace either row with the result.

Our goal :

A reduced matrix, a matrix in reduced row echelon form (RREF).

Definition. An augmented matrix A is in RREF if:

(1) The leading entry (first nonzero entry) in each row is a 1.

(2) The leading entry in each row is the only nonzero entry in its correspondingcolumn.

(3) The leading entry of each row is to the right of the leading entry in the rowabove it.

(4) All rows of zeros are at the bottom of the matrix. ⇤

Problem (page 89 #12–20 even).

(12) RREF

(14) not RREF — The leading entry in row 3 does not have 0’s in the rest ofthe column–the 2 is the problem.

(16) RREF

(18) RREF

(20) not RREF — the leading entry in row 1 is not 1


In general:

With two equations, our goal is 1 0 �0 1 �

�

where “�” can be any number of columns;

With three equations, our goal is241 0 0 �

0 1 0 �0 0 1 �

35


With four equations, our goal is2664

1 0 0 0 �0 1 0 0 �0 0 1 0 �0 0 0 1 �

3775


and so on.

We need to make 1’s and 0’s.

To make 1’s, we swap rows (row operation 1) or multiply a row by a nonzeronumber (row operation 2).

To make 0’s, we use (high school) elimination (row operation 3).


Example.

3x� 4y = �8

2x + 3y = 63 �4 �82 3 6

�Our goal is

1 0 �0 1 �

�

��

3 �4 �80 17 34

�Making a 0

�2R1 + 3R2 ! R2

�6 8 166 9 18� � �0 17 34

��3 �4 �80 1 2

�Making a 1

(1/17)R2 ! R2

��

3 0 00 1 2

�4R2 + R1 ! R1

Making a 0

3 �4 �80 4 8� � �3 0 0

��1 0 00 1 2

�(1/3)R1 ! R1

Making a 1

��x = 0y = 2


Example.

2x� 2y + z = 3

3x + y � z = 7

x� 3y + 2z = 0242 �2 1 3

3 1 �1 71 �3 2 0

35 Our goal is

241 0 0 �

0 1 0 �0 0 1 �

35

��241 �3 2 0

3 1 �1 72 �2 1 3

35 R3 $ R1

Making a 1R1 $ R3

��

241 �3 2 0

0 10 �7 70 4 �3 3

35 Making 0’s�3R1 + R2 ! R2

�2R1 + R3 ! R3

�3 9 �6 03 1 �1 7� � � �0 10 �7 7

�2 6 �4 02 �2 1 3� � � �0 4 �3 3

��


2410 0 �1 21

0 10 �7 70 0 �1 1

35 10R1 + 3R2 ! R1

Making 0’s�2R2 + 5R3 ! R3

10 �30 20 00 30 �21 21� � � �10 0 �1 21

0 �20 14 �140 20 �15 15� � � �0 0 �1 1

��2410 0 �1 21

0 10 �7 70 0 1 �1

35 Making 1’s�R3 ! R3

��

2410 0 0 20

0 10 0 00 0 1 �1

35 R3 + R1 ! R1

7R3 + R2 ! R2

Making 0’s

0 0 1 �110 0 �1 21� � � �10 0 0 20

0 0 7 �70 10 �7 7� � � �0 10 0 0

��241 0 0 2

0 1 0 00 0 1 �1

35 (1/10)R1 ! R1

(1/10)R2 ! R2

Making 1’s

x = 2y = 0

z = �1


Example.

2x + 3y + 5z = 21

x� y � 5z = �2

2x + y � z = 11

��242 3 5 21

1 �1 �5 �22 1 �1 11

35 Our goal is

241 0 0 �

0 1 0 �0 0 1 �

35

��241 �1 �5 �2

2 3 5 212 1 �1 11

35 R2 $ R1

R1 $ R2

making 1’s

��

241 �1 �5 �2

0 5 15 250 3 9 15

35 Making 0’s�2R1 + R2 ! R2

�2R1 + R3 ! R3

�2 2 10 42 3 5 21� � � �0 5 15 25

�2 2 10 42 1 �1 11� � � �0 3 9 15

��241 �1 �5 �2

0 1 3 50 3 9 15

35 Making 1’s

(1/5)R2 ! R2

��


241 0 �2 3

0 1 3 50 0 0 0

35 R2 + R1 ! R1

Making 0’s�3R2 + R3 ! R3

0 1 3 51 �1 �5 �2� � � �1 0 �2 3

0 �3 �9 �150 3 9 15� � � �0 0 0 0

��How do we find a solution in a case like this where we have lost a row?

Finding Solutions from Reduced Matrices:

Example. (1) 241 0 0 3

0 1 0 00 0 1 5

35

x y z

Every column except the last is associated with a variable.

x = 3

y = 0

z = 5

This is the kind of case with a single solution.

(2) Look to the previous example:

x� 2z = 3

y + 3z = 5

Solve each equation for the leftmost variable, assigning parameters (such as t ors) to variables that are not associated with leading entries, i.e. are not leftmostvariables in any equation.


x = 2z + 3

y = 5� 3z

z = t

Then, for the general solution, replace each variable to the right of the = signby its parameter.

x = 2t + 3

y = 5� 3t

z = t

Cases with parameters have infinitely many solutions. To get some specific solutions,replace the parameters by various real numbers.

t = 0 t = 1 t = 2 t = �3—— —— —— ——x = 3 x = 5 x = 7 x = �3y = 5 y = 2 y = �1 y = 14z = 0 z = 1 z = 2 z = �3

(3) 241 3 0 6

0 0 1 40 0 0 0

35

��

Our goal was

241 0 0 �

0 1 0 �0 0 1 �

35

��x + 3y = 6

z = 4


x = 6� 3y

y = t

z = 4

��x = 6� 3t

y = t

z = 4

(4) 1 4 2 00 0 0 1

�

��

Our goal was

1 0 �0 1 �

�

The second row represents

0x + 0y + 0z = 1,

which is impossible. As soon as you encounter a row where the only nonzeronumber is the last number, you conclude there is no solution.

(5) 1 �2 0 3 40 0 1 2 �1

�

w x y z

��

Our goal was

1 0 �0 1 �

�


w � 2x + 3z = 4

y + 2z = �1

��w = 4 + 2x� 3z

x = s

y = �1� 2z

z = t

The general solution is

w = 4 + 2s� 3t

x = s

y = �1� 2t

z = t

��Two particular solutions are

s = 0, t = 1 s = 1, t = 0————— —————

w = 1 w = 6x = 0 x = 1

y = �3 y = �1z = 1 z = 0


9x� 6y = 0

4x + 5y = 23

x � z = 6

��

2. USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION 49249 �6 0 0

4 5 0 231 0 �1 6

35

��241 0 �1 6

4 5 0 239 �6 0 0

35 R3 $ R1

Making 1’sR1 $ R3

��

241 0 �1 6

0 5 4 �10 �6 9 �54

35 Making 0’s�4R1 + R2 ! R2

�9R1 + R3 ! R3

�4 0 4 �244 5 0 23� � � �0 5 4 �1

�9 0 9 �549 �6 0 0� � � �0 �6 9 �54

��241 0 �1 6

0 1 �13 550 �6 9 �54

35 Making 1’s�R3 �R2 ! R2

a di↵erent way

��241 0 �1 6

0 1 �13 550 0 �69 276

35 Making 0’s

6R2 + R3 ! R3

0 6 �78 3300 �6 9 �54� � � �0 0 �69 276

��

50 2. SYSTEMS OF LINEAR EQUATIONS241 0 �1 6

0 1 �13 550 0 1 �4

35 Making 1’s

(�1/69)R3 ! R3

��

241 0 0 2

0 1 0 30 0 1 �4

35 R1 + R3 ! R1

13R3 + R2 ! R2

making 0’s

1 0 �1 60 0 1 �4� � � �1 0 0 2

0 0 13 �520 1 �13 55� � � �0 1 0 3

��

x = 2

y = 3

z = �4

Using the TI:

(1) Enter the matrix above using the Technology Tip on page 87.

(2) Find the reduced version of the matrix by using the Technology Tip on page88. ⇤

Example.

4x� 2y + 2z = 5

� 6x + 3y � 3z = �2

10x� 5y + 9z = 4

��

2. USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION 5124 4 �2 2 5�6 3 �3 �210 �5 9 4

35

��24 1 �1

212

54

�6 3 �3 �210 �5 9 4

35 (1/4)R1 ! R1

Making 1’s

��24 1 �1

212

54

0 0 0 112

· · · · · · · · · · · ·

35 Making 0’s

6R1 + R2 ! R2

6 �3 3 152

�6 3 �3 �2� � � �0 0 0 11

2

��

We can stop here and say no solution since

0x + 0y + 0z =11

2

is impossible. ⇤


2x� 4y + 2z = 6

5x� 5y + z = 12 �4 2 65 �5 1 1

�

��

Our goal is

1 0 �0 1 �

�

��

52 2. SYSTEMS OF LINEAR EQUATIONS1 �2 1 35 �5 1 1

�(1/2)R1 ! R1

Making 1’s

��

1 �2 1 30 5 �4 �14

�Making 0’s

�5R1 + R2 ! R2

�5 10 �5 �155 �5 1 1� � � �0 5 �4 �14

��

5 0 �3 �130 5 �4 �14

�5R1 + 2R2 ! R1

Making 0’s

5 �10 5 150 10 �8 �28� � � �5 0 �3 �13

��1 0 �3

5 �135

0 1 �45 �

145

�(1/5)R1 ! R1

(1/5)R2 ! R2

��Making 1’s

x� 3

5z = �13

5

y � 4

5z = �14

5��

x =3

5z � 13

5

y =4

5z � 14

5z = t

��

3. LINEAR SYSTEM APPLICATIONS 53

x =3

5t� 13

5

y =4

5t� 14

5z = t

3. Linear System ApplicationsProblem (page 99 #2).

(a) Go to STAT/EDIT and enter t in L1, M in L2, and C in L3.

Turn on Plot1 as usual.

Turn on Plot2 with L3 for Ylist and + for Mark.

Set a WINDOW as [�1, 20]⇥ [�1, 15].

Go to Y= and clear any functions..

Hit GRAPH to see t on the x-axis and M and C on the y-axis, M with ⇤ andC with +.

Hit STAT/CALC/4:LinReg(ax+b)/VARS/Y-VARS/1:Function/

1:Y1/ENTER.

Hit STAT/CALC/4:LinReg(ax+b)/2nd 1/,/2nd 3/,/VARS/Y-VARS/

1:Function/2:Y2/ENTER.

Now hit GRAPH to see the two regression lines.

(b) By rounding, M(t) = �.18t + 11.72 and C(t) = .32t + 3.15. We solve:

M(t) = C(t)

�.18t + 11.72 = .32t + 3.15

�.5t = �8.57

t = 17.14

Or, if using the TI, assuming the recent graph window is showing, hit

2nd Calc/5:intersect/ENTER/ENTER/ENTER.


The output of the two models will be equal about two months into 1998.

(c) As per capita margarine consumption decreased, mozzarella cheese con-sumption has increased at almost twice the rate and has surpassed margarineconsumption in 1998.


x = # of Corollas bought.

y = # of Camry’s bought.

z = # of Prius bought.

x + y + z = 30

x = 5y

��x + y + z = 30

x� 5y = 0

��1 1 1 301 �5 0 0

�! (RREF )

��1 0 5

6 250 1 1

6 5

�

��

x +5

6z = 25

y +1

6z = 5

��

x = 25� 5

6z


y = 5� 1

6z

z = t

��

x = 25� 5

6t

y = 5� 1

6t

z = t

��We need x, y, and z to be whole numbers greater than or equal to 0 (and lessthan or equal to 30). Thus t is limited to multiples of 6.

t x y z—— —— —— ——

0 25 5 06 20 4 612 15 3 1218 10 2 1824 5 1 2430 0 0 30

These are all possible values.Problem (page 104 #28).

x = # of chicken bags.

y = # of lamb bags.

z = # of turkey bags.

8x + 8y + 8z = 120 (amount)

.25(8x) + .28(8y) + .26(8z) = .27(120) (protein)

.03(8x) + .03(8y) + .03(8z) = .03(120) (fiber)


Both the first and third equations reduce to

x + y + z = 15,

so this is really a two equation system. The second equation simplifies to

2x + 2.24y + 2.08z = 32.4.

1 1 1 152 2.24 2.08 32.4

�! (RREF )

1 0 2

3 50 1 1

3 10

�

x +2

3z = 5

y +1

3z = 10

��

x = 5� 2

3z

y = 10� 1

3z

z = t

��

x = 5� 2

3t

y = 10� 1

3t

z = t

Now x, y, and z must be integers greater than or equal to 0, so z must be amultiple of 3. Also,

cost = 7.99(x + y) + 8.49z.

Possible solutions are:


x y z cost—— —— —— ——

5 10 0 119.853 9 3 121.351 8 6 122.85

Problem (page 106 #40). Variables are given in the diagram. Note that

cars entering each intersection = cars leaving each intersection (hopefully).

(N + E) x4 + 400 = x1

(E + S) x1 = x2 + 150

(S + W ) x2 + 320 = x3 + 450

(W + N) x3 + 180 = x4 + 300

Rewriting,

x1 � x4 = 400

x1 � x2 = 150

x2 � x3 = 130

x3 � x4 = 120

2664

1 0 0 �1 4001 �1 0 0 1500 1 �1 0 1300 0 1 �1 120

3775! (RREF )

2664

1 0 0 �1 4000 1 0 �1 2500 0 1 �1 1200 0 0 0 0

3775

8><>:

x1 � x4 = 400

x2 � x4 = 250

x3 � x4 = 120

!

8>>><>>>:

x1 = x4 + 400

x2 = x4 + 250

x3 = x4 + 120

x4 = t

!

8>>><>>>:

x1 = t + 400

x2 = t + 250

x3 = t + 120

x4 = t


t = 100 t = 500———– ———–x1 = 500 x1 = 900x2 = 350 x2 = 750x3 = 220 x3 = 620x4 = 100 x4 = 500

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