Lecture 3
Systems of equations and
matricies
Systems of Two Equations
in Two Variables

We are given the linear system
ax + by = c
dx + ey = f
A solution is an ordered pair (x0, y0) that will satisfy
each equation (make a true equation when
substituted into that equation). The solution set is
the set of all ordered pairs that satisfy both
equations. In this section, we wish to find the
solution set of a system of linear equations.
Solve by Graphing? Substitution?


This is practical for two-variable systems, but very
impractical once problem gets larger:
Solve the following system:
3x + 5y = -9
x+ 4y = -10
Solve by Substitution
3x + 5y = -9
x+ 4y = -10
Substitution: Solve equation 1 for x, replace for x in second
equation.
3x =- 5y -9 , x=(-5/3)y-3
(-5/3)y-3+ 4y = -10
(-5/3)y+(12/3)y=-7
(7/3)y=-7
Y=-3
Plug back into either equation to solve for x.
3x+5(-3)=-93x=6x=2… OR x+4(-3)=-10x=2
Solution? (2,-3)
Method of Substitution

The method of substitution is an algebraic one. It
works well when the coefficients of x or y are either
1 or -1. For example, let’s solve the system
2x + 3 = y
x + 2y = -4
using the method of substitution.
Method of Substitution
(continued)
The steps for this method are as follows:
1) Solve one of the equations for either x or y.
2) Substitute that result into the other equation to
obtain an equation in a single variable (either x or
y).
3) Solve the equation for that variable.
4) Substitute this value into any convenient equation
to obtain the value of the remaining variable.
Example
(continued)
2x+3 = y
x +2y = -4

Substitute y from first
equation into second
equation
x + 2(2x + 3) = -4
x + 4x + 6 = -4
5x + 6 = -4
5x = -10
x = -2

Solve the resulting equation

After we find x = -2, then
from the first equation, we
have 2(-2)+3 = y or y = -1.
Our solution is (-2, -1)
Another Example

Solve the system using
substitution:
3x - 2y = 7
y = 2x - 3
Another Example
Solution

Solve the system using
substitution:
3x - 2y = 7
y = 2x - 3

Solution:
3x  2 y  7
y  2x  3
3x  2(2 x  3)  7
3x  4 x  6  7
x  1
x  1
y  2  1  3  y  5
Solve by Substitution
6 x1  3x2  x3  22
x1  4 x2  2 x3  12
4 x1  x2  5 x3  10
Notice how quickly this becomes very cumbersome. We will first
learn a method called Jordan elimination to learn how to manipulate
matricies, but this method too is also very cumbersome. Learning to
use some computer program is essential for larger problems. Solve
still works in Mathematica, but it too becomes very cumbersome.
X1=2, X2=3, X3=1, as before… but what did I just do and how?
Terminology

A consistent linear system is one that has one or more
solutions.
 If a consistent system has exactly one solution, it is said to
be independent. An independent system will occur when
the two lines have different slopes.
 If a consistent system has more than one solution, it is
said to be dependent. A dependent system will occur
when the two lines have the same slope and the same y
intercept. In other words, the graphs of the lines will
coincide. There will be an infinite number of points of
intersection.
Terminology
(continued)


An inconsistent linear system is one that has no
solutions. This will occur when two lines have the
same slope but different y intercepts. In this case,
the lines will be parallel and will never intersect.
Example: Determine if the system is consistent,
independent, dependent or inconsistent:
2x – 5y = 6
-4x + 10y = -1
Solution of Example



Solve each equation for y to
obtain the slope intercept
form of the equation:
Since each equation has the
same slope but different y
intercepts, they will not
intersect. This is an inconsistent
system.
Notice that Mathematica has
problems as well. You must
know what is going on…
2x  5 y  6
2x  6  5 y
2x  6
 y
5
2x
6

 y
5
5
4 x  10 y  1
10 y  4 x  1
4x 1 2
1
y
  x
10 10 5
10
Note, when I use the methods as I showed in the last 3 equation
3 variable example using Mathematica, I get an error that means
something… You will come back to this slide after we learn
about matricies
Elimination by Addition


The method of substitution is not preferable if none of the
coefficients of x and y are 1 or -1. For example, substitution is
not the preferred method for the system below:
2x – 7y = 3
-5x + 3y = 7
A better method is elimination by addition. The following
operations can be used to produce equivalent systems:
 1. Two equations can be interchanged.
 2. An equation can be multiplied by a non-zero constant.
 3. An equation can be multiplied by a non-zero constant
and then added to another equation.
Elimination by Addition
Example






For our system, we will seek to
eliminate the x variable. The
coefficients are 2 and -5. Our goal is
to obtain coefficients of x that are
additive inverses of each other.
We can accomplish this by
multiplying the first equation by 5,
and the second equation by 2.
Next, we can add the two equations
to eliminate the x-variable.
Solve for y
Substitute y value into original
equation and solve for x
Write solution as an ordered pair
2x-7y=3 

-5x+3y=7 
5(2 x  7 y )  5(3) 
{

2( 5 x  3 y )  2(7) 
10 x  35 y  15 

10 x  6 y  14 
0 x  29 y  29 
y  1
2 x  7( 1) { 3 
2x  7  3 
2 x  4
x  2
(2, 1)
Elimination by Addition
Another Example
Solve 2x - 5y = 6
-4x + 10y = -1
Elimination by Addition
Another Example
Solve 2x - 5y = 6
-4x + 10y = -1
Solution:
1. Eliminate x by multiplying
equation 1 by 2 .
2. Add two equations
3. Upon adding the equations,
both variables are eliminated
producing the false equation
0 = 11
2x  5 y  6


4 x  10 y  1
4 x  10 y  12 

4 x  10 y  1
0  11
4. Conclusion: If a false
equation arises, the system
is inconsistent and there is
no solution. See how
Mathematica responds…
answer is empty set
Application

A man walks at a rate of 3 miles per hour and jogs at
a rate of 5 miles per hour. He walks and jogs a total
distance of
3.5 miles in 0.9 hours. How long does the man jog?
Application


A man walks at a rate of 3 miles per hour and jogs at a rate of
5 miles per hour. He walks and jogs a total distance of
3.5 miles in 0.9 hours. How long does the man jog?
Solution: Let x represent the amount of time spent walking
and y represent the amount of time spent jogging. Since the
total time spent walking and jogging is 0.9 hours, we have the
equation x + y = 0.9.
We are given the total distance traveled as 3.5 miles. Since
Distance = Rate x Time, distance walking = 3x and distance
jogging = 5y. Then total distance is 3x+5y= 3.5.
Application
(continued)
We can solve the system using
substitution.
1. Solve the first equation for y
2. Substitute this expression into
the second equation.
3. Solve second equation for x
4. Find the y value by substituting
this x value back into the first
equation.
5. Answer the question: Time
spent jogging is 0.4 hours.

Solution:
x  y  0.9
y  0.9  x
3 x  5 y  3.5
3 x  5(0.9  x)  3.5
3 x  4.5  5 x  3.5
2 x   1
x  0.5
0.5  y  0.9
y  0.4
Supply and Demand
The quantity of a product that people are willing to buy during
some period of time depends on its price. Generally, the higher
the price, the less the quantity demanded; the lower the price,
the greater the quantity demanded.
Similarly, the quantity of a product that a supplier is willing to
sell during some period of time also depends on the price.
Generally, a supplier will be willing to supply more of a
product at higher prices and less of a product at lower prices.
The simplest supply and demand model is a linear model
where the graphs of a demand equation and a supply equation
are straight lines. You will see these quite a bit.
Supply and Demand
(continued)
In supply and demand problems we are usually interested in
finding the price at which supply will equal demand. This is
called the equilibrium price, and the quantity sold at that
price is called the equilibrium quantity.
If we graph the the supply equation and the demand equation
on the same axis, the point where the two lines intersect is
called the equilibrium point. Its horizontal coordinate is the
value of the equilibrium quantity, and its vertical coordinate is
the value of the equilibrium price.
Supply and Demand
Example
Example: Suppose that the supply equation for long-life
light bulbs is given by
p = 1.04 q - 7.03,
and that the demand equation for the bulbs is
p = -0.81q + 7.5
where q is in thousands of cases. Find the equilibrium price
and quantity, and graph the two equations in the same
coordinate system.
Supply and Demand
(Example continued)
If we graph the two equations and find the intersection point,
we see the graph below.
Demand Curve
Supply Curve
Thus the equilibrium point is (7.854, 1.14), the
equilibrium price is $1.14 per bulb, and the equilibrium
quantity is 7,854 cases.
Now, Solve the Opening Example

A restaurant serves two types of fish dinners- small
for $5.99 each and large for $8.99. One day, there
were 134 total orders of fish, and the total receipts
for these 134 orders was $1024.66. How many small
dinners and how many large dinners were ordered?
Solution

A restaurant serves two types of fish dinners- small
for $5.99 each and large for $8.99. One day, there
were 134 total orders of fish, and the total receipts
for these 134 orders was $1024.66. How many small
dinners and how many large dinners were ordered?

Answer: 60 small orders and 74 large orders
Matrix Methods

It is impractical to solve more complicated linear
systems by hand. Computers and calculators now
have built in routines to solve larger and more
complex systems. Matrices, in conjunction with
graphing utilities and or computers are used for
solving more complex systems. In this section, we
will develop certain matrix methods for solving two
by two systems.
Matrices
A matrix is a rectangular array
of numbers written within
brackets. Here is an example
of a matrix which has three
rows and three columns: The
subscripts give the “address”
of each entry of the matrix.
For example the entry a23 is
found in the second row and
third column
Since this matrix has 3 rows and 3
columns, the dimensions of the
matrix are 3 x 3. Remember,
columns hold up things…
 a11 a12

a
a
21
22

a
 31 a32
a13 

a23 
a33 
Each number in the matrix is
called an element.
Matrix Solution of Linear Systems
When solving systems of
linear equations, we can
represent a linear system of
equations by an augmented
matrix, a matrix which stores
the coefficients and constants
of the linear system and then
manipulate the augmented
matrix to obtain the solution
of the system. This gets
cumbersome quick, but for
now, it’s a useful way to get
used to seeing matricies.
Example:
x + 3y = 5
2x – y = 3
The augmented matrix
associated with the above
system is
 1 3 5


 2 1 3
Generalization

Linear system:
a11 x1  a12 x2  k1
a21 x1  a22 x2  k 2

Associated
augmented matrix:
 a11

a
 21
a12 k1 

a22 k2 
Operations that Produce
Row-Equivalent Matrices
Ri  R j

1. Two rows are interchanged:

2. A row is multiplied by a nonzero constant:

3. A constant multiple of one row is added to another row:
kR j  Ri  Ri
kRi  Ri
Augmented Matrix Method
Example 1
Solve
x + 3y = 5
2x – y = 3
1. Augmented system
2. Eliminate 2 in 2nd row by
row operation
3. Divide row two by -7 to
obtain a coefficient of 1.
4. Eliminate the 3 in first row,
second position.
5. Read solution from matrix
1 3 5


2

1
3


2 R1  R2  R2
1 3 5 


0

7

7


R2 /  7  R2 
1 3 5


0
1
1


3R2  R1  R1 
1 0 2 

  x  2, y  1;(2,1)
01 1 
Augmented Matrix Method
Example 2
x  2y  4
Solve
x + 2y = 4
x + (1/2)y = 4
1.
2.
3.
4.
5.
6.
7.
Eliminate fraction in second equation
by multiplying by 2
Write system as augmented matrix.
Multiply row 1 by -2 and add to row 2
Divide row 2 by -3
Multiply row 2 by -2 and add to row 1.
Read solution : x = 4, y = 0
(4,0)
1
x
y  4  2x  y  8
2
1
2 4


2
1
8


1

0
2 4

3 0 
1

0
1

0
2 4

1 0
0 4

1 0
Augmented Matrix Method
Example 3
Solve
10x - 2y = 6
-5x + y = -3
1. Represent as augmented matrix.
2. Divide row 1 by 2
3. Add row 1 to row 2 and replace row 2 by
sum
4. Since 0 = 0 is always true, we have a
dependent system. The two equations
are identical, and there are infinitely
many solutions. See how Mathematica
tells you this…
10

 5
5

 5
5

0
2 6 

1 3 
1 3 

1 3 
1 3 

0 0
Augmented Matrix Method
Example 4





Solve
5 x  2 y  7
5
y  x 1
2
Rewrite second equation
Add first row to second row
The last row is the equivalent of 0x +
0y = -5
Since we have an impossible
equation, there is no solution. The
two lines are parallel and do not
intersect. Here Mathematica gives the
empty set.
5 x  2 y  7 

5 x  2 y  2 
 5 2 7 


 5 2 2 
5 2 7 


0 0 5 
Possible Final Matrix Forms for a
Linear System in Two Variables
Form 1: Unique Solution
(Consistent and Independent)
Form 2: Infinitely Many Solutions
(Consistent and Dependent)
Form 3: No Solution (Inconsistent)
1 0 m 
0 1 n 


1 m n 
0 0 0 


1 m
0 0

n
p 
Gauss-Jordan Elimination


Any linear system must have exactly one solution,
no solution, or an infinite number of solutions.
Previously we considered the 2x2 case, in which the
term consistent is used to describe a system with a
unique solution, inconsistent is used to describe a
system with no solution, and dependent is used for
a system with an infinite number of solutions. In this
section we will consider larger systems with more
variables and more equations, but the same three
terms are used to describe them.
Matrix Representations of Consistent,
Inconsistent and Dependent Systems
 The following matrix representations of three linear
equations in three unknowns illustrate the three different
cases:

Case I: consistent
1 0 0

0 1 0
0 0 1

3

4
5 

From this matrix
representation, you can
determine that
x=3
y=4
z=5
Matrix Representations
(continued)

Case 2: inconsistent
1 2 3

0 0 0
0 0 0

4

6
0 

From the second row of the
matrix, we find that
0x + 0y +0z =6
or
0 = 6,
an impossible equation.
From this, we conclude that
there are no solutions to
the linear system.
Matrix Representations
(continued)

Case 3: dependent
1 2 3

0 0 0
0 0 0

4

0
0 

When there are fewer nonzero rows of a system than
there are variables, there
will be infinitely many
solutions, and therefore the
system is called dependent.
Reduced Row Echelon Form

A matrix is said to be in reduced row echelon form
or, more simply, in reduced form, if
 Each row consisting entirely of zeros is below any
row having at least one non-zero element.
 The leftmost nonzero element in each row is 1.
 All other elements in the column containing the
leftmost 1 of a given row are zeros.
 The leftmost 1 in any row is to the right of the
leftmost 1 in the row above.
Examples of Reduced
Row Echelon Form
1 2 3

0 0 0
0 0 0

4

6

0
1 3 0 0

0 0 1 0
0 0 0 1

1 0 0

0 1 0
0 0 1

 2

7 
8 
3

4
5 
Solving a System
Using Gauss-Jordan Elimination

Example: Solve
x + y – z = -2
2x – y + z = 5
-x + 2y + 2z = 1
Solving a System
Using Gauss-Jordan Elimination

Example: Solve
x + y – z = -2
2x – y + z = 5
-x + 2y + 2z = 1

Solution:
We begin by writing the system as an augmented
matrix
Example (continued)
We already have a 1 in the
diagonal position of first column.
Now we want 0’s below the 1.
The first 0 can be obtained by
multiplying row 1 by -2 and
adding the results to row 2:
 Row 1 is unchanged
 (-2) times Row 1 is added to
Row 2
 Row 3 is unchanged
 1 1 1 2 
 2 1 1 5  


 1 2 2 1 
Example (continued)
The second 0 can be
obtained by adding
row 1 to row 3:
 Row 1 is unchanged
 Row 2 is unchanged
 Row 1 is added to
Row 3
Example (continued)
Moving to the second
column, we want a 1 in
the diagonal position
(where there was a –3).
We get this by dividing
every element in row 2 by
-3:
 Row 1 is unchanged
 Row 2 is divided by –3
 Row 3 is unchanged
Example (continued)
To obtain a 0 below
the 1 , we multiply row
2 by -3 and add it to
the third row:
 Row 1 is unchanged
 Row 2 is unchanged
 (-3) times row 2 is
added to row 3
Example (continued)
To obtain a 1 in the third
position of the third row, we
divide that row by 4. Rows 1
and 2 do not change.
Example (continued)
We can now work upwards to get
zeros in the third column, above
the 1 in the third row.
 Add R3 to R2 and replace R2
with that sum
 Add R3 to R1 and replace R1
with the sum.
Row 3 will not be changed.
All that remains to obtain
reduced row echelon form is to
eliminate the 1 in the first row,
2nd position.
1 1 0

0 1 0
0 0 1

0

 1

2
Example (continued)
To get a zero in the
first row and second
position, we multiply
row 2 by -1 and add
the result to row 1 and
replace row 1 by that
result. Rows 2 and 3
remain unaffected.
1 1 0

0 1 0
0 0 1

1 0 0

0 1 0
0 0 1

0

 1
2 
1

 1
2 
Final Result
We can now “read” our
solution from this last matrix.
We have
x = 1,
y = -1
z = 2.
Written as an ordered triple,
we have (1, -1, 2). This is a
consistent system with a
unique solution.
1 0 0

0 1 0
0 0 1

1

 1

2
Example 2
Example: Solve the system
3x – 4y + 4z = 7
x – y – 2z = 2
2x – 3y + 6z = 5
Example 2
Example: Solve the system
3x – 4y + 4z = 7
x – y – 2z = 2
2x – 3y + 6z = 5
Solution:
Begin by representing the
system as an augmented
matrix:
 3 4 4 7 


 1  1 2 2 
 2 3 6 5 


Example 2
(continued)

Since the first number
in the second row is a
1, we interchange rows
1 and 2 and leave row 3
unchanged:
 3 4 4 7 


 1  1 2 2 
 2 3 6 5 


1

3

2
1
4
2
3
6
4
2

7
5 
Example 2
(continued)


In this step, we will get
zeros in the entries
beneath the 1 in the first
column:
Multiply row 1 by -3 , add
to row 2 and replace row 2:
-3*R1+R2 → R2.
Multiply row 1 by -2, add
to row 3 and replace row 3:
-2*R1+R3 → R3.
 1 1

 3 4

 2 3
 1 1

 0 1

 0 1
2
4
6
2
2

7
5 
2

10 1 

1
10

Final Result
To get a zero in the third row,
second entry we multiply row
2 by -1 and add the result to
R3 and replace R3 by that
sum: Notice this operations
“wipes out” row 3 so row 3
consists entirely of zeros.
Any time you have fewer nonzero rows than variables you
will have a dependent system.
 1 1

 0 1

 0 1
2
2

10 1 

10 1 
 1 1 2

 0 1 10
0 0 0

2

1
0 
Representation of a Solution of a
Dependent System


 1 1 2

 0 1 10
0 0 0

2

1
0 
We can interpret the second row
of this matrix as
–y + 10z = 1, or
10z – 1 = y
So, if we let z = t (arbitrary real
number,) then in terms of t,
y = 10t - 1.
Next we can express the variable x in
terms of t as follows: From the first row
of the matrix, we have x – y -2z = 2.
If z = t and y = 10t – 1, we have
x – (10t-1) - 2t = 2 or x = 12t+1.
Our general solution can now be
expressed in terms of t:
(12t+1,10t-1,t),
where t is an arbitrary real number.
Procedure for
Gauss-Jordan Elimination
Step 1. Choose the leftmost nonzero column and use
appropriate row operations to get a 1 at the top.
Step 2. Use multiples of the row containing the 1 from step 1 to
get zeros in all remaining places in the column containing this 1.
Step 3. Repeat step 1 with the submatrix formed by (mentally)
deleting the row used in step 2 and all rows above this row.
Step 4. Repeat step 2 with the entire matrix, including the rows
deleted mentally. Continue this process until the entire matrix
is in reduced form.
Note: If at any point in this process we obtain a row with all
zeros to the left of the vertical line and a nonzero number to the
right, we can stop because we will have a contradiction.
Applications
Example: Purchasing. A company that rents small moving
trucks wants to purchase 25 trucks with a combined capacity
of 28,000 cubic feet. Three different types of trucks are
available: a 10-foot truck with a capacity of 350 cubic feet, a
14-foot truck with a capacity of 700 cubic feet, and a 24-foot
truck with a capacity of 1,400 cubic feet. How many of each
type of truck should the company purchase?
Solution
Step 1. The question in this example indicates that the relevant
variables are the number of each type of truck.
x = number of 10-foot trucks
y = number of 14-foot trucks
z = number of 24-foot trucks
We form the mathematical model:
x + y + z = 25
(Total number of trucks)
350x + 700y + 1,400z = 28,000 (Total capacity)
With two equations and three unknowns… we won’t get a solution
to our problem, but we can find a general solution
Solution
(continued)
Step 2. Now we form the augmented coefficient matrix of
the system and solve by using Gauss-Jordan elimination:
1
1
25 
 1
350 700 1, 400 28, 000 


1 1 1 25
1 2 4 80 


1 1 1 25
0 1 3 55 


1 0 2 30
0 1 3 55 


(1/350)R2
R2
-R1 + R2
R2
-R2 + R1
R1
Matrix is in reduced form.
x - 2z = -30 or x = 2z -30,
y + 3z = 55 or y = -3z + 55.
Solution
(continued)
Let z = t. Then for t any real number
x = 2t – 30
y = -3t + 55
z=t
is a solution to our mathematical model.
Step 3. We must interpret this solution in terms of the original
problem. Since the variables x, y, and z represent numbers of trucks,
they must be nonnegative. And since we can’t purchase a fractional
number of trucks, each must be a nonnegative whole number.
Solution (continued)
Since t = z, it follows that t must also
be a nonnegative whole number. The
first and second equations in the
model place additional restrictions on
the values t can assume:
x = 2t - 30 > 0 implies that t > 15
y = -3t + 55 > 0 implies that t < 55/3
Thus the only possible values of t that
will produce meaningful solutions to
the original problem are 15, 16, 17,
and 18. A table is a convenient way to
display these solutions.
t
10-ft 14-ft. 24-ft
truck truck truck
x
y
z
15
0
10
15
16
2
7
16
17
4
4
17
18
6
1
18