Algebra 2 Interactive Chalkboard
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Lesson 4-1 Introduction to Matrices
Lesson 4-2 Operations with Matrices
Lesson 4-3 Multiplying Matrices
Lesson 4-4 Transformations with Matrices
Lesson 4-5 Determinants
Lesson 4-6 Cramer’s Rule
Lesson 4-7 Identity and Inverse Matrices
Lesson 4-8 Using Matrices to Solve Systems
of Equations
Example 1 Organize Data in a Matrix
Example 2 Dimensions of a Matrix
Example 3 Solve an Equation Involving Matrices
College Kaitlin wants to attend one of three Iowa
universities next year. She has gathered information
about tuition (T), room and board (R/B), and
enrollment (E) for the universities. Use a matrix to
organize the information. Which university’s total
cost is lowest?
Iowa State University:
T - $3132
R/B - $4432
E - 26,845
University of Iowa:
T - $3204
R/B - $4597
E - 28,311
University of Northern Iowa:
T - $3130
R/B - $4149
E - 14,106
Organize the data into labeled columns and rows.
T
R/B
E
ISU
UI
UNI
Answer: The University of Northern Iowa has the
lowest total cost.
Dining Out Justin is going out for lunch. The
information he has gathered from the two fast-food
restaurants is listed below. Use a matrix to organize
the information. When is each restaurant’s total cost
less expensive?
Burger Complex
Lunch Express
Hamburger Meal
$3.39 Hamburger Meal
$3.49
Cheeseburger Meal
$3.59 Cheeseburger Meal
$3.79
Chicken Sandwich Meal
$4.99 Chicken Sandwich Meal $4.89
Answer:
Hamburger
Meal
Cheeseburger
Meal
Chicken
Sandwich
Meal
Burger Complex
Lunch Express
The Burger Complex has the best price for hamburgers
and cheeseburgers. Lunch Express has the best price for
chicken sandwiches.
State the dimensions of matrix G if
2 rows
4 columns
Answer: Since matrix G has 2 rows and 4 columns, the
dimensions of matrix G are 2  4.
State the dimensions of matrix G if
Answer: 3  2
Solve
for x and y.
Since the matrices are equal, the corresponding
elements are equal. When you write the sentences to
solve this equation, two linear equations are formed.
This system can be solved using substitution.
Second equation
Substitute 3x – 2 for y.
Distributive Property
Add 4 to each side.
Divide each side by 7.
To find the value for y, substitute 1 for x in
either equation.
First equation
Substitute 1 for x.
Simplify.
Answer: The solution is (1, 1).
Solve
Answer: (2, 5)
for x and y.
Example 1 Add Matrices
Example 2 Subtract Matrices
Example 3 Use Matrices to Model Real-World Data
Example 4 Multiply a Matrix by a Scalar
Example 5 Combination of Matrix Operations
Find
Definition of
matrix addition
Add corresponding
elements.
Answer:
Simplify.
Find
Answer: Since the dimensions of A are 2  3 and the
dimensions of B are 2  2, these matrices
cannot be added.
a. Find
Answer: Since their dimensions are different, these
matrices cannot be added.
b. Find
Answer:
Find
Definition of
matrix subtraction
Subtract
corresponding
elements.
Answer:
Simplify.
Find
Answer:
The table below shows the total number of
households in the United States and the number of
U.S. households that had a computer (PC) in given
years. Use matrices to find the number of U.S.
households without a PC.
U.S. Households
Year
Total households
(millions)
Households with PCs
(millions)
1995
97.7
33.2
1996
98.9
38.7
1997
100.0
44.0
1998
101.0
47.8
1999
101.7
51.9
Source: The New York Times Almanac
The data can be organized into two matrices. Find the
difference of the matrix that represents the total U.S.
households and the matrix that represents the number
of households with PCs.
total
households
# households
with a PC
Subtract
corresponding
elements.
# households
without a PC
1995
1996
1997
1998
1999
Answer: There were 64.5 million households without
a PC in 1995, 60.2 million in 1996, 56.0
million in 1997, 53.2 million in 1998, and
49.8 million in 1999.
Tests The table below shows the percentage of
students at Clark High School who passed the 9th
and 10th grade proficiency tests in 2001 and 2002.
Use matrices to find how the percentage of passing
students changed from 2001 to 2002.
Proficiency Tests Passing Percentages
Year
Grade
2001
9
2002
10
9
10
Math
86.2
87.3
88.4
89.6
Reading
83.5
90.1
81.9
91.2
Science
79.6
89.7
85.0
89.9
Citizenship
86.1
87.4
86.4
85.7
Answer:
9th grade 10th grade
Math
Reading
Science
Citizenship
Substitution
Multiply each
element by 2.
Answer:
Simplify.
Answer:
Perform the scalar multiplication first. Then subtract
the matrices.
Substitution
Multiply each element in
the first matrix by 4 and
each element in the
second matrix
by 3.
Simplify.
Subtract corresponding
elements.
Answer:
Simplify.
Answer:
Example 1 Dimensions of Matrix Products
Example 2 Multiply Square Matrices
Example 3 Multiply Matrices with Different Dimensions
Example 4 Commutative Property
Example 5 Distributive Property
Determine whether the product of
is defined. If so, state the dimensions of the product.
A
34
•
B
42
=
AB
32
The inner dimensions are equal so the matrix product
is defined. The dimensions of the product are 3  2.
Determine whether the product of
is
defined. If so, state the dimensions of the product.
A
32
•
B
43
The inner dimensions are not equal, so the matrix
product is not defined.
Determine whether each matrix product is defined.
If so, state the dimensions of the product.
a.
Answer: The matrix product is defined.
The dimensions are 3  3.
b.
Answer: The matrix product is not defined.
Find RS if
Step 1 Multiply the numbers in the first row of R by
the numbers in the first column of S, add the
products, and put the result in the first row,
first column of RS.
Step 2 Multiply the numbers in the first row of R by
the numbers in the second column of S, add
the products, and put the result in the first row,
second column of RS.
Step 3 Multiply the numbers in the second row of R by
the numbers in the first column of S, add the
products, and put the result in the second row,
first column of RS.
Step 4 Multiply the numbers in the second row of R by
the numbers in the second column of S, add
the products, and put the result in the second
row, second column of RS.
Step 5 Simplify the product matrix.
Answer: So,
Find RS if
Answer:
Chess Three teams competed in the final round of
the Chess Club’s championships. For each win, a
team was awarded 3 points and for each draw a team
received 1 point. Find the total number of points for
each team. Which team won the tournament?
Team
Wins
Draws
Blue
5
6
4
4
3
5
Red
Green
Explore
The final scores can be found by
multiplying the wins and draws by
the points for each.
Plan
Write the results from the championship and
the points in matrix form. Set up the matrices
so that the number of rows in the points
matrix equals the number of columns in
the results matrix.
Results
Points
Solve
Multiply the matrices.
Write an equation.
Multiply columns
by rows.
Simplify.
The labels for the product matrix are
shown below.
Total Points
Blue
Red
Green
Answer: The blue team had 19 points, the red team had
21 points, and the green team had 17 points. The red
team won the championship.
Examine R is a 3  2 matrix and P is a 2  1 matrix.
Their product should be a 3  1 matrix.
Basketball In Thursday night’s basketball game,
three of the players made the points listed below in
the chart. They scored 1 point for the free-throws,
2 points for the 2-point shots, and 3 points for the
3-points shots. How many points did each player
score and who scored the most points?
Player
Free-throws
2-point
3-point
Warton
2
3
2
Bryant
5
1
0
Chris
2
4
5
Answer: Warton scored 14 points, Bryant scored
7 points, and Chris scored 25 points. Chris scored
the most points in the game.
Find KL if
Substitution
Multiply columns
by rows.
Answer:
Simplify.
Find LK if
Substitution
Multiply columns
by rows.
Answer:
Simplify.
Find each product if
a. AB
Answer:
b. BA
Answer:
Find A(B + C) if
Substitution
Add
corresponding
elements.
Answer:
or
Multiply
columns by
rows.
Find AB + AC if
Substitution
Multiply
columns by
rows.
Simplify.
Answer:
Add
corresponding
elements.
Find each product if
a. A(B + C)
Answer:
b. AB + AC
Answer:
Example 1 Translate a Figure
Example 2 Find a Translation Matrix
Example 3 Dilation
Example 4 Reflection
Example 5 Rotation
Find the coordinates of the vertices of the image
of quadrilateral ABCD with A(–5, –1), B(–2, –1),
C(–1, –4), and D(–3, –5) if it is moved 3 units to
the right and 4 units up. Then graph ABCD and its
image ABCD.
Write the vertex matrix for quadrilateral ABCD.
To translate the quadrilateral 3 units to the right,
add 3 to each x-coordinate.
To translate the figure 4 units up, add 4 to
each y-coordinate.
This can be done by adding the translation matrix
to the vertex matrix of ABCD.
Vertex Matrix
of ABCD
Translation
Matrix
Vertex Matrix
of A'B'C'D'
The coordinates of ABCD are A(–2, 3), B(1, 3),
C(2, 0), D(0, –1).
Graph the preimage and the image. The two graphs have
the same size and shape.
Answer: A(–2, 3), B(1, 3), C(2, 0), D(0, –1)
Find the coordinates of the vertices of the image of
quadrilateral HIJK with H(2, 3), I(3, –1), J(–1, –3), and
K(–2, 5) if it is moved 2 units
to the left and 2 units up.
Then graph HIJK and its
image HIJK.
Answer: H(0, 5), I(1, 1),
J(–3, –1), K(–4, 7)
Short-Response Test Item Rectangle EFGH is
the result of a translation of the rectangle EFGH.
A table of the vertices of each rectangle is shown.
Find the coordinates of F and G.
Rectangle EFGH
Rectangle EFGH
E(–2, 2)
F
G(4, –2)
H(–2, –2)
E(–5, 0)
F(1, 0)
G
H(–5, –4)
Read the Test Item
You are given the coordinates of the preimage and
image of points E and H. Use this information to find
the translation matrix. Then you can use the
translation matrix to find the coordinates of F and G.
Solve the Test Item
Write a matrix equation. Let (a, b) represent the
coordinates of F and let (c, d) represent the
coordinates of G.
Since these two matrices are equal, corresponding
elements are equal.
Solve an equation for x.
Solve an equation for y.
Use the values for x and y to find the variables for
F(a, b) and G(c, d).
Answer: So, the coordinates of F are (4, 2) and the
coordinates of G are (1, –4).
Short-Response Test Item Rectangle ABCD is
the result of a translation of the rectangle ABCD. A
table of the vertices of each rectangle is shown.
Find the coordinates of D and A.
Rectangle ABCD
Rectangle ABCD
A(–4, 5)
B(–1, 5)
A
B(8, 0)
C(–1, 0)
D
C(8, –5)
D(5, –5)
Answer: The coordinates of D are (–4, 0) and the
coordinates of A are (5, 0).
XYZ has vertices X(1, 2), Y(3, –1), and Z(–1, –2).
Dilate XYZ so that its perimeter is twice the
original perimeter. What are the coordinates of the
vertices of XYZ?
If the perimeter of a figure is twice the original figure,
then the lengths of the sides of the figure will be twice
the measure of the original lengths. Multiply the vertex
matrix by the scale factor of 2.
The coordinates of the vertices of XYZ are
X(2, 4), Y(6, –2), and Z(–2, –4).
Graph XYZ and XYZ.
The triangles are not
congruent. The image has
sides that are half the length
of those of the original figure.
Answer: X(2, 4), Y(6, –2),
Z(–2, –4)
ABC has vertices A(2, 1), B(–3, –2), and C(1, 4).
Dilate ABC so that its perimeter is four times the
original perimeter. What are the coordinates of the
vertices of ABC?
Answer: A(8, 4), B(–12, –8), C(4, 16)
Find the coordinates of the vertices of the image of
pentagon PENTA with P(–3, 1), E(0, –1), N(–1, –3),
T(–3, –4), and A(–4, –1) after a reflection across
the x-axis.
Write the ordered pairs as a vertex matrix. Then multiply
the vertex matrix by the reflection matrix for the x-axis.
Answer: The coordinates of
the vertices of PENTA are
P(–3, –1), E(0, 1), N(–1, 3),
T(–3, 4), and A(–4, 1). Notice
that the preimage and image are
congruent. Both figures have the
same size and shape.
Find the coordinates of the vertices of the image of
pentagon PENTA with P(–5, 0), E(–3, 3), N(1, 2),
T(1, –1), and A(–4, –2) after a reflection across the
line y = x.
Answer: P(0, –5), E(3, –3), N(2, 1),
T(–1, 1), A(–2, –4)
Find the coordinates of the vertices of the image of
DEF with D(4, 3), E(1, 1), and F(2, 5) after it is
rotated 90 counterclockwise about the origin.
Write the ordered pairs in a vertex matrix. Then multiply
the vertex matrix by the rotation matrix.
Answer: The coordinates of
the vertices of triangle DEF
are D(–3, 4), E(–1, 1), and
F(–5, 2). The image is
congruent to the preimage.
Find the coordinates of the vertices of the image of
TRI with T(–1, 2), R(–3, 0), and I(–2, –2) after it is
rotated 180 counterclockwise about the origin.
Answer: T(1, –2), R(3, 0), I(2, 2)
Example 1 Second-Order Determinant
Example 2 Expansion by Minors
Example 3 Use Diagonals
Example 4 Area of a Triangle
Find the value of
Definition of determinant
Multiply.
Answer:
Simplify.
Find the value of
Definition of determinant
Multiply.
Answer:
Simplify.
Find the value of each determinant.
a.
Answer: –2
b.
Answer: 0
Evaluate
using expansion by minors.
Expansion
by minors
Evaluate
22
matrices.
Multiply.
Answer:
Simplify.
Evaluate
Answer: –71
using expansion by minors.
Evaluate
using diagonals.
Step 1 Rewrite the first 2 columns to the right of
the determinant.
Step 2 Find the product of the elements of
the diagonals.
9
0
–4
Step 2 Find the product of the elements of
the diagonals.
1 0
12
9
–4
0
Step 3 Add the bottom products and subtract the
top products.
Answer: The value of the determinant is –8.
Evaluate
Answer: 79
using diagonals.
Find the area of a triangle whose vertices are located
at (0, –1), (–2, –6), and (3, –2).
Assign values to a, b, c, d, e, and f and substitute them
into the Area Formula. Then evaluate.
Area Formula
(a, b) = (0, –1),
(c, d) = (–2, –6),
(e, f ) = (3, –2)
Expansion by
minors
Evaluate the
22
determinants.
Multiply.
Simplify.
Answer: The area is 8.5 square units.
Find the area of a triangle whose vertices are located
at (2, 3), (–2, 2), and (0, 0).
Answer: 5 units2
Example 1 System of Two Equations
Example 2 Use Cramer’s Rule
Example 3 System of Three Equations
Use Cramer’s Rule to solve the system of equations.
Cramer’s Rule
a = 5 , b = 4, c = 3 ,
d = –2, e = 28,
and f = 8
Evaluate each
determinant.
Simplify.
Answer: The solution is (4, 2).
Use Cramer’s Rule to solve the system of equations.
Answer: (–1, 2)
Voting In a vote for the school colors of a new high
school, blue-and-gold received 440 votes from 10th
and 11th graders and red-and-black received 210
votes from 10th and 11th graders. In the 10th grade,
blue-and-gold received 72% of the votes, and red-andblack received 28% of the votes. In the 11th grade,
blue-and-gold received 64% of the votes and red-andblack received 36%.
Write a system of equations that represents the total
number of votes cast for each set of colors in these
two grades.
Let t represent the total number of 10th grade votes.
Let e represent the total number of 11th grade votes.
Answer:
Votes for blue-and-gold
Votes for red-and-black
Voting In a vote for the school colors of a new high
school, blue-and-gold received 440 votes from 10th
and 11th graders and red-and-black received 210
votes from 10th and 11th graders. In the 10th grade,
blue-and-gold received 72% of the votes, and red-andblack received 28% of the votes. In the 11th grade,
blue-and-gold received 64% of the votes and red-andblack received 36%.
Find the total number of votes cast in the 10th and in
the 11th grade.
Cramer’s Rule
Answer: The solution to the system is (300, 350).
So, the 10th graders cast 300 votes and the 11th
graders cast 350 votes.
Voting In a vote for the student council president,
Greg received 550 votes from 11th and 12th graders
and Amber received 300 votes from 11th and 12th
graders. In the 11th grade, Greg received 70% of the
votes, and Amber received 30% of the votes. In the
12th grade, Greg received 60% of the votes and Amber
received 40%.
a. Write a system of equations that represents the total
number of votes cast for each student in these two
grades. Let t represent the total number of 12th grade
votes and let e represent the total number of 11th
grade votes.
Answer:
b. Find the total number of votes cast in the 11th and in
the 12th grade.
Answer: The 11th graders cast 400 votes and the 12th
graders cast 450 votes.
Use Cramer’s Rule to solve the system of equations.
Use a calculator to evaluate each determinant.
Answer: The solution is
Use Cramer’s Rule to solve the system of equations.
Answer: (–2, 0, 1)
Example 1 Verify Inverse Matrices
Example 2 Find the Inverse of a Matrix
Example 3 Use Inverses to Solve a Problem
Determine whether X and Y are inverses.
Check to see if X • Y = I.
Write an
equation.
Matrix
multiplication
Now find Y • X.
Write an
equation.
Matrix
multiplication
Answer: Since X • Y = Y • X = I, X and Y are inverses.
Determine whether P and Q are inverses.
Check to see if P • Q = I.
Write an
equation.
Matrix
multiplication
Answer: Since P • Q  I, they are not inverses.
Determine whether each pair of matrices are inverses.
a.
Answer: no
b.
Answer: yes
Find the inverse of the matrix, if it exists.
Find the value of the determinant.
Since the determinant is not equal to 0, S –1 exists.
Definition of
inverse
a = –1, b = 0,
c = 8, d = –2
Answer:
Simplify.
Check:
Find the inverse of the matrix, if it exists.
Find the value of the determinant.
Answer: Since the determinant equals 0, T –1 does
not exist.
Find the inverse of each matrix, if it exists.
a.
Answer: No inverse exists.
b.
Answer:
Use the table to assign a
number to each letter in the
message ALWAYS_SMILE.
Then code the message with
Code
the matrix
Convert the message to
numbers using the table.
_ 0
I
9
R 18
A 1
J 10
S 19
B 2
K 11
T 20
C 3
L 12
U 21
D 4
M 13
V 22
E 5
N 14
W23
F 6
O 15
X 24
G 7
P 16
Y 25
H 8
Q 17
Z 26
A
L
W
A
Y
S
_
S
M
I
L
E
1
12
23
1
25
19
0
19
13
9
12
5
Write the message in matrix form. Then multiply the
message matrix B by the coding matrix A.
Write an equation.
Matrix multiplication
Simplify.
Answer: The coded message is 13 | 38 | 24 | 49 | 44 |
107 | 19 | 57 | 22 | 53 | 17 | 39.
Use the inverse matrix A–1 to decode 13 | 38 | 24 | 49 |
44 | 107 |19 | 57 | 22 | 53 | 17 | 39.
Definition of inverse
a = 1, b = 2,
c = 1, d = 3
Simplify.
Next, decode the message by multiplying the coded
matrix C by A–1.
Code
Use the table again to convert
the numbers to letters. You can
now read the message.
Answer:
_ 0
I
9
R 18
A 1
J 10
S 19
B 2
K 11
T 20
C 3
L 12
U 21
D 4
M 13
V 22
E 5
N 14
W23
F 6
O 15
X 24
G 7
P 16
Y 25
H 8
Q 17
Z 26
1
12
23
1
25
19
0
19
13
9
12
5
A
L
W
A
Y
S
_
S
M
I
L
E
a. Use the table to assign a
number to each letter in the
message FUN_MATH.
Then code the message
Code
_ 0
I
9
R 18
A 1
J 10
S 19
B 2
K 11
T 20
C 3
L 12
U 21
D 4
M 13
V 22
E 5
N 14
W23
Answer: 12 | 63 | 28 | 14 | 26 |
F 6
O 15
X 24
16 | 40 | 44
G 7
P 16
Y 25
H 8
Q 17
Z 26
with the matrix A =
b. Use the inverse matrix of
Code
to decode
the message 12 | 63 | 28 |
14 | 26 | 16 | 40 | 44.
Answer:
_ 0
I
9
R 18
A 1
J 10
S 19
B 2
K 11
T 20
C 3
L 12
U 21
D 4
M 13
V 22
E 5
N 14
W23
F 6
O 15
X 24
6
21
14
0
13
1
20
8
G 7
P 16
Y 25
F
U
N
_
M
A
T
H
H 8
Q 17
Z 26
Example 1 Two-Variable Matrix Equation
Example 2 Solve a Problem Using a Matrix Equation
Example 3 Solve a System of Equations
Example 4 System of Equations with No Solution
Write a matrix equation for the system of equations.
Determine the coefficient, variable, and constant matrices.
Write the matrix equation.
A
Answer:
• X = B
Write a matrix equation for the system of equations.
Answer:
Fabrics The table below shows the composition of
three types of fabrics and cost per yard of each type.
Type
Wool
Silk
Cotton
Cost
R
10%
20%
70%
$7
S
20%
30%
50%
$8
T
20%
50%
30%
$10
Write a system of equations that represents the total
cost for each of the three fabric components.
Let w represent the percent of wool.
Let s represent the percent of silk.
Let c represent the percent of cotton.
Write an equation for the cost of each
type of fabric.
Answer:
Type R:
Type S:
Type T:
Write a matrix equation for the system of equations.
Determine the coefficient, variable, and constant matrices.
Write the matrix equation.
A
Answer:
• X = B
Snacks The table below shows the composition of
two types of snack mixes and the cost per pound of
each type.
Mix
Nuts
Candy
Cost Per
Pound
A
50%
50%
$4
B
75%
25%
$5
a. Write a system of equations that represents the cost
per pound for each of the two snack mix components.
Answer:
b. Write a matrix equation for the system of equations.
Answer:
Use a matrix equation to solve the system
of equations.
The matrix equation
when
Step 1
Find the inverse of the coefficient matrix.
Step 2
Multiply each side of the matrix equation by
the inverse matrix.
Multiply
each side
by A–1.
Multiply
matrices.
Answer: The solution is (5, –4). Check this solution in
the original equations.
Use a matrix equation to solve the system
of equations.
Answer: (2, –4)
Use a matrix equation to solve the system
of equations.
The matrix equation is
when
Find the inverse of the coefficient matrix.
The determinant of the coefficient matrix
so A–1 does not exist.
is 0,
Graph the system of equations.
Since the lines are parallel, this
system has no solution.
The system is inconsistent.
Answer: There is no solution of
this system.
Use a matrix equation to solve the system
of equations.
Answer: no solution
Explore online information about the
information introduced in this chapter.
Click on the Connect button to launch your browser
and go to the Algebra 2 Web site. At this site, you
will find extra examples for each lesson in the
Student Edition of your textbook. When you finish
exploring, exit the browser program to return to this
presentation. If you experience difficulty connecting
to the Web site, manually launch your Web browser
and go to www.algebra2.com/extra_examples.
Click the mouse button or press the
Space Bar to display the answers.
Click the mouse button or press the
Space Bar to display the answers.
Click the mouse button or press the
Space Bar to display the answers.
Click the mouse button or press the
Space Bar to display the answers.
Click the mouse button or press the
Space Bar to display the answers.
Click the mouse button or press the
Space Bar to display the answers.
Click the mouse button or press the
Space Bar to display the answers.
Click the mouse button or press the
Space Bar to display the answers.
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