Download 1. Systems of Linear Equations Solution of an equation with one

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. We
might also say that x = 2 and y = 7 is a solution. Also, (5, 3) is a solution of
x = 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 two
or more linear equations with two variables. The solution to the system of
equations, if it exists, is the point(s) of intersection of the graphs of the equations. In other terms, a solution is an ordered pair that satisfies each of the
equations 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 be
sure, 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 equations
are 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. ⇤
32
2. SYSTEMS OF LINEAR EQUATIONS
Example.
(1) 2x + 4y = 8
(2) x + 2y = 4
1
Both equations simplify to y =
x + 2, so they have the same graph. These
2
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 on
the line x + 2y = 4. ⇤
1. SYSTEMS OF LINEAR EQUATIONS
33
Example. Solve using the calculator:
(1) 6x 2y = 10
2
(2)
x+y =4
3
We need to solve these equations for y:
(1) 2y = 6x 10 =) y = 3x 5
2
(2) y =
x+4
3
Find the approximate solution by using the Technology Tip on page 55. We
get
x = 2.4545455, y = 2.3636364.
The exact solution is
27
26
x= , y= .
11
11
Substitution Method. This method is appropriate for cases where a variable
has a coefficient of ±1. Solve for that variable in that equation, and then
substitute 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)
y=
1
7
34
2. SYSTEMS OF LINEAR EQUATIONS
Elimination (by Addition).
Example.
4x + 3y = 26
3x 11y = 7
(⇥11)
(⇥3)
To give opposite
coefficients 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
4x + 10y = 2
(⇥( 2))
4x 10y = 6
4x + 10y = 2
0=8
When a problem reduces to a false numerical equation, there is no solution, like
the second graphical example. When a problem reduces to a true numerical
equation, the two lines coincide, like the third graphical example. ⇤
1. SYSTEMS OF LINEAR EQUATIONS
35
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 lines
to see if there is a solution. If not, the problem has no solution. We will choose
equations (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. We
substitute into equation (2).
y=
y=
4(4) + 18
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 three
equations.
If we had gotten a false statement from (3), such as if (3) was y =
then there is no solution.
x + 10,
36
2. SYSTEMS OF LINEAR EQUATIONS
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 the
number 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
1.5
t=
= 34.56
.0434
Thus the number of female athletes would surpass the number of male athletes
in 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
2
3
2 3 1 3
Definition. A = 43 2 0 55 is a matrix, an array of numbers in rows
4 2 3 1
and columns. The plural of matrix is matrices. The 12 numbers are the entries
of 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.
2
3
2 1 6
B = 4 3 4 05 is a 3 ⇥ 3 square matrix.
1 0 2
2 3
1
627
7
The 4 ⇥ 1 matrix 6
435is a column matrix.
4
⇥
⇤
The 1 ⇥ 3 matrix 5 3 2 is a row matrix.
The entries of a 3 ⇥ 4 matrix A may be represented as
2
3
a11 a12 a13 a14
A = 4a21 a22 a23 a245
a31 a32 a33 a34
a23 means the entry in row 2, column 3. ⇤
38
2. SYSTEMS OF LINEAR EQUATIONS
Particular System
2x + 8y = 13
5x 3y = 6
General System
a11x + a12y = k1
a21x + a22y = k2
a21 means 2nd equation, coefficient of the 1st variable.
k1 means constant or RHS (right-hand side) of 1st equation.
Coefficient matrices:

2 8
5 3

a11 a12
a21 a22
Constant or RHS matrices:

13
6

k1
k2
Augmented matrices:

2 8 | 13
5 3 | 6
We will often skip the vertical bar.

a11 a12 | k1
a21 a22 | k2
Since an augmented matrix clearly represents a system of linear equations, we
can 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 equations
with 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 another
equation and replace either equation with the result.
2. USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION
39
Related to this, we have the following.
Row Operations
For any augmented matrix of a system of equations, the following row operations
yield 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 row
and 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 corresponding
column.
(3) The leading entry of each row is to the right of the leading entry in the row
above 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 of
the column–the 2 is the problem.
(16) RREF
(18) RREF
(20) not RREF — the leading entry in row 1 is not 1
40
2. SYSTEMS OF LINEAR EQUATIONS
In general:
With two equations, our goal is

1 0
0 1
where “ ” can be any number of columns;
With three equations, our goal is
2
3
1 0 0
40 1 0 5
0 0 1
where “ ” can be any number of columns;
With four equations, our goal is
2
1 0 0 0
60 1 0 0
6
40 0 1 0
0 0 0 1
where “ ” can be any number of columns;
3
7
7
5
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 nonzero
number (row operation 2).
To make 0’s, we use (high school) elimination (row operation 3).
2. USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION
Example.
3x 4y = 8
2x + 3y = 6


3 4 8
1 0
Our goal is
2 3 6
0 1

3 4 8
0 17 34
6 8 16
6 9 18
Making a 0
2R1 + 3R2 ! R2
0 17 34

3 4 8
Making a 1
0 1 2 (1/17)R2 ! R2

3 0 0 4R2 + R1 ! R1
0 1 2
Making a 0

1 0 0 (1/3)R1 ! R1
0 1 2 Making a 1
x=0
y=2
3
0
4 8
4 8
3
0
0
41
42
2. SYSTEMS OF LINEAR EQUATIONS
Example.
2x 2y + z = 3
3x + y
z=7
x 3y + 2z = 0
2
3
2
2 2 1 3
1 0 0
43 1
1 75 Our goal is 40 1 0
1 3 2 0
0 0 1
2
3
1 3 2 0 R3 $ R1
43 1
1 75 Making a 1
2 2 1 3 R1 $ R3
2
3
1 3 2 0
Making 0’s
40 10
7 75 3R1 + R2 ! R2
0 4
3 3
2R1 + R3 ! R3
3
5
3 9
3 1
6 0
1 7
0
7 7
10
2 6
4 0
2
2 1 3
0
4
3 3
2. USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION
43
10 30 20
0
0 30
21 21
2
10 0
4 0 10
0 0
3
1 21 10R1 + 3R2 ! R1
7 75
Making 0’s
1 1
2R2 + 5R3 ! R3
10
0
1
0
0
20 14
14
20
15 15
0
0
1
2
3
10 0
1 21
4 0 10 7 7 5 Making 1’s
0 0 1
1
R3 ! R3
2
3
10 0 0 20 R3 + R1 ! R1
4 0 10 0 0 5 7R3 + R2 ! R2
0 0 1 1
Making 0’s
0 0
10 0
1
1
1 21
10 0
0
0 0 7
7
0 10 7 7
0 10 0
2
3
1 0 0 2 (1/10)R1 ! R1
40 1 0 0 5 (1/10)R2 ! R2
0 0 1 1
Making 1’s
x=2
y=0
z= 1
20
0
21
1
44
2. SYSTEMS OF LINEAR EQUATIONS
Example.
2x + 3y + 5z = 21
x
y 5z = 2
2x + y
z = 11
2
3
2
2 3 5 21
1 0 0
41 1 5 25 Our goal is 40 1 0
2 1
1 11
0 0 1
2
3
1 1 5 2 R2 $ R1
42 3 5 21 5 R1 $ R2
2 1
1 11 making 1’s
3
5
2 2 10 4
2 3 5 21
2
3
1 1 5 2
Making 0’s
40 5 15 25 5 2R1 + R2 ! R2
0 3 9 15
2R1 + R3 ! R3
0
2 2 10 4
2 1
1 11
0
2
3
1 1 5 2
Making 1’s
40 1 3 5 5 (1/5)R2 ! R2
0 3 9 15
5 15 25
3
9 15
2. USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION
2
3
1 0 2 3 R2 + R1 ! R1
40 1 3 55
Making 0’s
0 0 0 0
3R2 + R3 ! R3
45
0
1
1 3
1 5
5
2
1
0
3
0
0
3 9 15
3 9 15
0
0
2
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)
2
3
1 0 0 3
40 1 0 05
0 0 1 5
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 or
s) to variables that are not associated with leading entries, i.e. are not leftmost
variables in any equation.
46
2. SYSTEMS OF LINEAR EQUATIONS
x = 2z + 3
y = 5 3z
z=t
Then, for the general solution, replace each variable to the right of the = sign
by 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
——
x=3
y=5
z=0
(3)
t=1
——
x=5
y=2
z=1
t=2
——
x=7
y= 1
z=2
t= 3
——
x= 3
y = 14
z= 3
2
3
1 3 0 6
40 0 1 45
0 0 0 0
2
1 0 0
Our goal was 40 1 0
0 0 1
x + 3y = 6
z=4
3
5
2. USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION
47
x = 6 3y
y=t
z=4
x = 6 3t
y=t
z=4
(4)

1 4 2 0
0 0 0 1
The second row represents

1 0
Our goal was
0 1
0x + 0y + 0z = 1,
which is impossible. As soon as you encounter a row where the only nonzero
number is the last number, you conclude there is no solution.
(5)

1 2 0 3 4
0 0 1 2 1
w x y z

1 0
Our goal was
0 1
48
2. SYSTEMS OF LINEAR EQUATIONS
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
—————
w=1
x=0
y= 3
z=1
Problem (page 90 #26).
9x 6y
4x + 5y
x
= 0
= 23
z= 6
s = 1, t = 0
—————
w=6
x=1
y= 1
z=0
2. USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION
2
3
9 6 0 0
44 5 0 235
1 0
1 6
49
2
3
1 0
1 6 R3 $ R1
44 5 0 235 Making 1’s
9 6 0 0 R1 $ R3
2
1 0
1
40 5 4
0 6 9
2
1 0
40 1
0 6
2
1 0
40 1
0 0
4 0
4 5
4
0
24
23
0
4
1
9 0 9
9
6 0
54
0
0
6 9
54
0
0
6
6
78 330
9
54
0
0
69 276
3
6
Making 0’s
1 5 4R1 + R2 ! R2
54
9R1 + R3 ! R3
5
3
1 6
Making 1’s
13 55 5 R3 R2 ! R2
9
54 a di↵erent way
3
1 6
Making 0’s
13 55 5
69 276 6R2 + R3 ! R3
50
2
1 0
40 1
0 0
32. SYSTEMS OF LINEAR EQUATIONS
1 6
Making 1’s
13 55 5
1
4 ( 1/69)R3 ! R3
2
3
1 0 0 2
R1 + R3 ! R1
40 1 0 3 5 13R3 + R2 ! R2
0 0 1 4
making 0’s
1 0
0 0
1
1
6
4
1 0
0
2
0 0
0 1
13
52
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 page
88. ⇤
Example.
4x 2y + 2z =
6x + 3y 3z =
10x 5y + 9z =
5
2
4
2
2. USING MATRICES TO SOLVE LINEAR SYSTEMS OF EQUATION
51
3
4
2 2 5
4 6 3
3 25
10
5 9 4
2
1
2
1
2
5
4
3
(1/4)R1 ! R1
1
4 6 3
3 25
Making 1’s
10
5 9 4
2
1
2
1
2
5
4
11
2
3
1
Making 0’s
4 0 0 0
5 6R1 + R2 ! R2
··· ··· ··· ···
We can stop here and say no solution since
0x + 0y + 0z =
is impossible. ⇤
Problem (page 91 #34).
2x
5x

2
5

1 0
Our goal is
0 1
4y + 2z = 6
5y + z = 1
4 2 6
5 1 1
11
2
6
3 3
6 3
3
15
2
0
11
2
0
0
2
52
2. SYSTEMS OF LINEAR EQUATIONS

1
5
2 1 3 (1/2)R1 ! R1
5 1 1 Making 1’s

1 2 1
0 5
4

5 0
0 5
3
4

1 0
0 1
3
5
4
5
x
y
3
z=
5
4
z=
5
3
x= z
5
4
y= z
5
z=t
3
14
Making 0’s
5R1 + R2 ! R2
13 5R1 + 2R2 ! R1
14
Making 0’s
13
5
14
5
13
5
14
5
13
5
14
5
(1/5)R1 ! R1
(1/5)R2 ! R2
Making 1’s
5 10
5
5
5 1
15
1
0
14
5
4
5
0
10 5 15
10
8 28
5
0
3
13
3. LINEAR SYSTEM APPLICATIONS
3
x= t
5
4
y= t
5
z=t
53
13
5
14
5
3. Linear System Applications
Problem (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 ⇤ and
C 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.
54
2. SYSTEMS OF LINEAR EQUATIONS
The output of the two models will be equal about two months into 1998.
(c) As per capita margarine consumption decreased, mozzarella cheese consumption has increased at almost twice the rate and has surpassed margarine
consumption in 1998.
Problem (page 103 #22).
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 30
! (RREF )
1 5 0 0

1 0
0 1
5
6
1
6
25
5
5
x + z = 25
6
1
y+ z=5
6
x = 25
5
z
6
3. LINEAR SYSTEM APPLICATIONS
y=5
55
1
z
6
z=t
x = 25
y=5
5
t
6
1
t
6
z=t
We need x, y, and z to be whole numbers greater than or equal to 0 (and less
than or equal to 30). Thus t is limited to multiples of 6.
t
x
y
z
—— —— —— ——
0
25
5
0
6
20
4
6
12
15
3
12
18
10
2
18
24
5
1
24
30
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)
56
2. SYSTEMS OF LINEAR EQUATIONS
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
15
1 0
! (RREF )
2 2.24 2.08 32.4
0 1
2
3
1
3
5
10
2
x+ z =5
3
1
y + z = 10
3
2
z
3
1
y = 10
z
3
z=t
x=5
2
t
3
1
y = 10
t
3
z=t
x=5
Now x, y, and z must be integers greater than or equal to 0, so z must be a
multiple of 3. Also,
cost = 7.99(x + y) + 8.49z.
Possible solutions are:
3. LINEAR SYSTEM APPLICATIONS
x
——
5
3
1
y
——
10
9
8
z
——
0
3
6
57
cost
——
119.85
121.35
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)
(E + S)
(S + W )
(W + N )
x4 + 400 = x1
x1 = x2 + 150
x2 + 320 = x3 + 450
x3 + 180 = x4 + 300
Rewriting,
x1
x1
x2
x2
x3
x3
x4 = 400
= 150
= 130
x4 = 120
2
3
2
1 0 0
1 400
1
61 1 0 0 1507
6
6
7 ! (RREF ) 60
40 1
40
1 0 1305
0 0 1
1 120
0
8
>
<x1
x2
>
:x
3
x4 = 400
x4 = 250
x4 = 120
8
x1
>
>
>
<x
2
!
>
x3
>
>
:
x4
= x4 + 400
= x4 + 250
= x4 + 120
=t
0
1
0
0
0 1
0 1
1 1
0 0
8
x1
>
>
>
<x
2
!
>
x3
>
>
:
x4
3
400
2507
7
1205
0
= t + 400
= t + 250
= t + 120
=t
58
2. SYSTEMS OF LINEAR EQUATIONS
t = 100
———–
x1 = 500
x2 = 350
x3 = 220
x4 = 100
t = 500
———–
x1 = 900
x2 = 750
x3 = 620
x4 = 500