Organizing Data Into Matrices

FeatureLesson

Algebra 2Algebra 2

LessonMain

Lesson 4-1

Use the table at the right.

(For help, go to Skills Handbook page 842.)

Organizing Data Into MatricesOrganizing Data Into Matrices

How many units were imported to theUnited States in 1996?

How many were imported in 2000?

U.S. Passenger Vehicles and Light Trucks Imports

And Exports (millions)

1996 1998 2000

Imports 4.678 5.185 6.964

Exports 1.295 1.331 1.402

Source: U.S. Department of Commerce.

Check Skills You’ll Need


4-1

4.678 million units

6.964 million units

FeatureLesson

Algebra 2Algebra 2

LessonMain

Write the dimensions of each matrix.

Lesson 4-1


a. 7 –412 9

The matrix has 2 rows and 2 columns and is

therefore a 2 2 matrix.

b. The matrix has 1 row and 3 columns and is

therefore a 1 3 matrix.0 6 15

Quick Check

Additional Examples

4-1

FeatureLesson

Algebra 2Algebra 2

LessonMain

Identify each matrix element.

K =

Lesson 4-1


3 –1 –8 51 8 4 98 –4 7 –5

a. k12 b. k32 c. k23 d. k34

Element k12 is –1. Element k32 is –4.

a. K =

k12 is the element in the first row and second column.

3 –1 –8 51 8 4 98 –4 7 –5

b. K =

k32 is the element in the third row and second column.

3 –1 –8 51 8 4 98 –4 7 –5

Additional Examples

4-1

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

K =

Lesson 4-1


3 –1 –8 51 8 4 98 –4 7 –5

a. k12 b. k32 c. k23 d. k34

Element k23 is 4. Element k34 is –5.

c. K =

k23 is the element in the second row and third column.

3 –1 –8 51 8 4 98 –4 7 –5

d. K =

k34 is the element in the third row and the fourth column.

3 –1 –8 51 8 4 98 –4 7 –5

Quick Check

Additional Examples

4-1

FeatureLesson

Algebra 2Algebra 2

LessonMain

Three students kept track of the games they won and lost in

a chess competition. They showed their results in a chart. Write a

2 3 matrix to show the data.

Let each row represent the number of wins and losses and each column represent a student.

Lesson 4-1


= Win X = Loss

Ed X X

Jo X

Lew X X X X

5 6 32 1 4

WinsLosses

Ed Jo Lew

Quick Check

Additional Examples

4-1

FeatureLesson

Algebra 2Algebra 2

LessonMain

Refer to the table.

a. Write a matrix N to represent the information.

Lesson 4-1


U.S. Passenger Car ImportsAnd Exports (millions)

1996 1998 2000

Imports 4.678 5.185 6.964

Exports 1.295 1.331 1.402

Source: U.S. Department of Commerce.Use 2 3 matrix.

4.678 5.185 6.9641.295 1.331 1.402

N = Import Exports

1996 1998 2000

Each column represents a different year.

Each row represents imports and exports.

Additional Examples

4-1

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

b. Which element represents exports for 2000?

Lesson 4-1


U.S. Passenger Car ImportsAnd Exports (millions)

1996 1998 2000

Imports 4.678 5.185 6.964

Exports 1.295 1.331 1.402Source: U.S. Department of Commerce.

Exports are in the second row.

The year 2000 is in the third column.

Element n23 represents the number of exports for 2000.

Quick Check

Additional Examples

4-1

FeatureLesson

Algebra 2Algebra 2

LessonMain

1. Write the dimensions of the matrix. M =

Lesson 4-1


8 4 0 1 9 3 –5 0–1 2 6 1

2. Identify the elements m24, m32, and m13 of the matrix M in

question 1.

3. The table shows the amounts of the deposits and withdrawals for the checking accounts of four bank customers. Show the data in a 2 4 matrix. Label the rows and columns.

Deposits Withdrawals

A $450 $370

B $475 $289

C $364 $118

D $420 $400

3 4

0, 2, 0

450 475 364 420370 289 118 400

A B C DDepositsWithdrawals

Lesson Quiz

4-1

FeatureLesson

Algebra 2Algebra 2

LessonMain

Simplify the elements of each matrix.

1. 2.

3. 4.

5. 6.

Lesson 4-2

Adding and Subtracting MatricesAdding and Subtracting Matrices


10 + 4 0 + 4–2 + 4 –5 + 4

5 – 2 3 – 2–1 – 2 0 – 2

–2 + 3 0 – 3 1 – 3 –5 + 3

3 + 1 4 + 9–2 + 0 5 + 7

8 – 4 –5 – 1 9 – 1 6 – 9

2 + 4 6 – 8 4 – 3 5 + 2



4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

1. =

2. =

3. =

4. =

5. =

6. =

Solutions

Lesson 4-2


10 + 4 0 + 4–2 + 4 –5 + 4

14 4 2 –1

5 – 2 3 – 2–1 – 2 0 – 2

3 1–3 –2

–2 + 3 0 – 3 1 – 3 –5 + 3

1 –3–2 –2

3 + 1 4 + 9–2 + 0 5 + 7

4 13–2 12

8 – 4 –5 – 1 9 – 1 6 – 9

4 –6 8 –3

2 + 4 6 – 8 4 – 3 5 + 2

6 –2 1 7


4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

The table shows information on ticket sales for a new

movie that is showing at two theaters. Sales are for children (C)

and adults (A).

Lesson 4-2


Theater C A C A

1 198 350 54 439 2 201 375 58 386

Matinee Evening

a. Write two 2 2 matrices to represent matinee and evening sales.

Theater 1 198 350Theater 2 201 375

MatineeC A

Theater 1 54 439Theater 2 58 386

EveningC A

Additional Examples

4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-2


b. Find the combined sales for the two showings.

198 350201 375

+ 54 43958 386

= 198 + 54 350 + 439201 + 58 375 + 386

= Theater 1 252 789Theater 2 259 761

C A

Quick Check

Additional Examples

4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

Find each sum.

Lesson 4-2


a. b. 9 0–4 6

+ 0 00 0

3 –8–5 1

+ –3 8 5 –1

= 9 + 0 0 + 0–4 + 0 6 + 0

= 3 + (–3) –8 + 8–5 + 5 1 + (–1)

= 9 0–4 6

= 0 00 0

Quick Check

Additional Examples

4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

A = and B = . Find A – B.

Method 1: Use additive inverses.

Lesson 4-2


4 8–2 0

7 –94 5

A – B = A + (–B) = + 4 8–2 0

–7 9–4 –5

Write the additive inverses of the elements of the second matrix.

4 + (–7) 8 + 9–2 + (–4) 0 + (–5)

Add corresponding elements

=

–3 17–6 –5

= Simplify.

Additional Examples

4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Method 2: Use subtraction.

Lesson 4-2


A – B = – 4 8–2 0

7 –94 5

4 – 7 8 – (–9)–2 – 4 0 – 5

Subtract corresponding elements

=

–3 17–6 –5

= Simplify.

Quick Check

Additional Examples

4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

Solve X – = for the matrix X.

Lesson 4-2


2 53 –18 0

10 –3–4 9 6 –9

X – + = + 2 53 –18 0

10 –3–4 9 6 –9

2 53 –18 0

2 53 –18 0

2 53 –18 0

Add

to each side of the equation.

X – =10 –3–4 9 6 –9

2 53 –18 0

12 2–1 814 –9

X = Simplify.

Quick Check

Additional Examples

4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

Determine whether the matrices in each pair are equal.

Lesson 4-2


M = ; N =8 + 9 5 –6 –1 0 0.7

17 54 – 10 –2 + 1

0 –79

Both M and N have three rows and two columns, but – 0.7. M and N are not equal matrices.

79

=/

a. M = ; N =8 + 9 5 –6 –1 0 0.7

17 54 – 10 –2 + 1

0 –79

Additional Examples

4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-2


b. P = ; Q =

Both P and Q have two rows and two columns, and their corresponding elements are equal. P and Q are equal matrices.

3 –440 –3

–

–

27 9

16 4

12 4

80.2

P = ; Q = 3 –440 –3

–

–

27 9

16 4

12 4

80.2

Quick Check

Additional Examples

4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

Solve the equation

Lesson 4-2


Since the two matrices are equal, their corresponding elements are equal.

2m – n –3 8 –4m + 2n

= for m and n.15 m + n 8 –30

2m – n = 15 –3 = m + n –4m + 2n = –30

2m – n –3 8 –4m + 2n

= 15 m + n 8 –30

Additional Examples

4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-2


The solutions are m = 4 and n = –7.

Solve for m and n.

2m – n = 15

m + n = –3

3m = 12 Add the equations.

m = 4 Solve for m.

4 + n = –3 Substitute 4 for m.

n = –7 Solve for n.

Quick Check

Additional Examples

4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

Find each sum or difference.

1. + 2. –

3. What is the additive identity for 2 4 matrices?

4. Solve the equation for x and y.

5. Are the following matrices equal?

6. Solve X – = for the matrix X.

Lesson 4-2


2 80 –12

–9 6 6 –5

–3 1 4 8–5 4

–3 –2 9 5 4 –6

–2x –1 5 x + y

18 –3x + 4yx – 2y –16

=

3 0.5

– ; 0.50

0.4 –0.6

622

523

4 31 5

2 70 6

–7 14 6 –17

0 3–5 3–9 10

0 0 0 00 0 0 0

x = –9, y = –7

6 101 11

no, – –0.623

=/

Lesson Quiz

4-2

FeatureLesson

Algebra 2Algebra 2

LessonMain

(For help, go to Lesson 4-2.)

Lesson 4-3

Matrix MultiplicationMatrix Multiplication

Find each sum.

1. + +3 52 8

3 52 8

3 52 8

2. + + + +–4 7

–4 7

–4 7

–4 7

–4 7

3. + + +–1 3 4 0 –2 –5

–1 3 4 0 –2 –5

–1 3 4 0 –2 –5

–1 3 4 0 –2 –5



4-3

FeatureLesson

Algebra 2Algebra 2

LessonMain

Solutions

Lesson 4-3


1. + + = =3 52 8

3 52 8

3 52 8

3 + 3 + 3 5 + 5 + 52 + 2 + 2 8 + 8 + 8

9 156 24

2. + + + + =–4 7

–4 7

–4 7

–4 7

–4 7

–4 + (–4) + (–4) + (–4) + (–4) 7 + 7 + 7 + 7 + 7

=–20 35

3. + + +–1 3 4 0 –2 –5

–1 3 4 0 –2 –5

–1 3 4 0 –2 –5

–1 3 4 0 –2 –5

= =4(–1) 4(3) 4(4) 4(0) 4(–2) 4(–5)

–4 12 16 0 –8 –20


4-3

FeatureLesson

Algebra 2Algebra 2

LessonMain

The table shows the salaries of the three managers (M1, M2, M3) in each of the two branches (A and B) of a retail clothing company. The president of the company has decided to give each manager an 8% raise. Show the new salaries in a matrix.

Lesson 4-3


Store M1 M2 M3

A $38,500 $40,000 $44,600 B $39,000 $37,800 $43,700

1.0838500 40000 4460039000 37800 43700

= Multiply each element by 1.08.

1.08(38500) 1.08(40000) 1.08(44600)1.08(39000) 1.08(37800) 1.08(43700)

Additional Examples

4-3

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-3


The new salaries at branch A are $41,580, $43,200, and $48,168.

= A B

41580 43200 4816842120 40824 47196

M1 M2 M3

The new salaries at branch B are $42,120, $40,824, and $47,196.

Quick Check

Additional Examples

4-3

FeatureLesson

Algebra 2Algebra 2

LessonMain

Find the sum of –3M + 7N for

M = and N = .

Lesson 4-3


2 –30 6

–5 –1 2 9

–3M + 7N = –3 + 7 2 –30 6

–5 –1 2 9

= + –6 9 0 –18

–35 –7 14 63

= –41 2 14 45

Quick Check

Additional Examples

4-3

FeatureLesson

Algebra 2Algebra 2

LessonMain

Solve the equation –3Y + 2 = .

Lesson 4-3


6 9–12 15

27 –1830 6

–3Y + 2 = 6 9–12 15

27 –1830 6

–3Y + = Scalar multiplication. 12 18–24 30

27 –1830 6

–3Y = – Subtract

from each side.

27 –1830 6

12 18–24 30

12 18–24 30

–3Y = Simplify.15 –3654 –24

Y = – =15 –3654 –24

13

–5 12–18 8

Multiply each side

by – and simplify.13

Additional Examples

4-3

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-3


–3Y + 2 = 6 9–12 15

27 –1830 6

–3 + 2 Substitute. 6 9–12 15

27 –1830 6

–5 12–18 8

+ Multiply. 12 18–24 30

27 –1830 6

15 –3654 –24

= Simplify.27 –1830 6

27 –1830 6

Check:

Quick Check

Additional Examples

4-3

FeatureLesson

Algebra 2Algebra 2

LessonMain

Find the product of and .

Multiply a11 and b11. Then multiply a12 and b21. Add the products.

Lesson 4-3


–2 5 3 –1

4 –42 6

–2 5 3 –1

4 –42 6

= (–2)(4) + (5)(2) = 2

The result is the element in the first row and first column. Repeat with the rest of the rows and columns.

–2 5 3 –1

4 –42 6

= (–2)(4) + (5)(6) = 382

–2 5 3 –1

4 –42 6

= (3)(4) + (–1)(2) = 102 38

Additional Examples

4-3

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-3


2 3810

–2 5 3 –1

4 –42 6

= (3)(–4) + (–1)(6) = –18

The product of and is .–2 5 3 –1

4 –42 6

2 3810 –18

Quick Check

Additional Examples

4-3

FeatureLesson

Algebra 2Algebra 2

LessonMain

Matrix A gives the prices of shirts and jeans on sale at a

discount store. Matrix B gives the number of items sold on one day.

Find the income for the day from the sales of the shirts and jeans.

A = $18 $22 B =

Lesson 4-3


Prices Number of Items SoldShirts Jeans

Shirts 109Jeans 76

Multiply each price by the number of items sold and add the products.

18 22 = (18)(109) + (22)(76) = 3634109 76

The store’s income for the day from the sales of shirts and jeans was $3634.

Quick Check

Additional Examples

4-3

FeatureLesson

Algebra 2Algebra 2

LessonMain

Use matrices P = and Q = .

Determine whether products PQ and QP are defined or undefined.

Lesson 4-3


3 –1 25 9 00 1 8

6 5 7 02 0 3 11 –1 5 2

Find the dimensions of each product matrix.

(3 3) (3 4) 3 4

PQ

productequal matrix

(3 4) (3 3)

QP

not equal

Product PQ is defined and is a 3 4 matrix.

Product PQ is undefined, because the number of columns of Q is not equal to the number of rows in P.Quick Check

Additional Examples

4-3

FeatureLesson

Algebra 2Algebra 2

LessonMain

Use matrices A, B, C, and D.

Lesson 4-3


A = B =

C = D =

2 3 –10 –5 4

–7 1 0 2 6 –6

294

–3 2 –1

1. Find 8A. 2. Find AC. 3. Find CD.

4. Is BD defined or undefined?

5. What are the dimensions of (BC)D?

16 24 –8 0 –40 32

27–29

–6 4 –2–27 18 –9–12 8 –4

undefined

2 3

Lesson Quiz

4-3

FeatureLesson

Algebra 2Algebra 2

LessonMain


Lesson 4-4

Geometric Transformations with MatricesGeometric Transformations with Matrices

1. y = x + 2; left 4 units 2. ƒ(x) = x + 2; up 5 units

3. g(x) = |x|; right 3 units 4. y = x; down 2 units

5. y = |x – 3|; down 2 units 6. ƒ(x) = –2|x|; right 2 units

Without using graphing technology, graph each function and its translation. Write the new function.

12

13



4-4

FeatureLesson

Algebra 2Algebra 2

LessonMain

1. y = x + 2;

left 4 units:

y = x + 6

2. ƒ(x) = x + 2;

up 5 units:

ƒ(x) = x + 7;

Solutions

Lesson 4-4


3. g(x) = |x|

right 3 units:

g(x) = |x – 3|

4. y = x;down 2 units:y = x – 2

12

12


4-4

FeatureLesson

Algebra 2Algebra 2

LessonMain

5. y = |x – 3|;

down 2 units:

y = |x – 3| – 2

6. ƒ(x) = –2|x|

right 2 units:

ƒ(x) = –2|x + 4|

Solutions (continued)

Lesson 4-4


13

13


4-4

FeatureLesson

Algebra 2Algebra 2

LessonMain

Triangle ABC has vertices A(1, –2), B(3, 1) and C(2, 3). Use

a matrix to find the vertices of the image translated 3 units left and 1

unit up. Graph ABC and its image ABC.

Lesson 4-4


The coordinates of the vertices of the image are A (–2, –1), B (0, 2), C (–1, 4).

Vertices of Translation Vertices ofthe Triangle Matrix the image

1 3 2–2 1 3

+ =–3 –3 –3 1 1 1

–2 0 –1–1 2 4

Subtract 3 from eachx-coordinate.

Add 1 to eachy-coordinate.

A B C A B C

Quick Check

Additional Examples

4-4

FeatureLesson

Algebra 2Algebra 2

LessonMain

The figure in the diagram is to be reduced by a factor of . Find

the coordinates of the vertices of the reduced figure.

Lesson 4-4


23

Quick Check

Write a matrix to represent the coordinates of the vertices.

A B C D E A B C D E

23

0 2 3 –1 –23 2 –2 –3 0

=0 2 – –

2 – –2 0

43

43

43

43

23 Multiply.

Additional Examples

4-4

The new coordinates are A (0, 2), B ( , ), C (2, – ),

D (– , –2), and E (– , 0).

43

43

43

23

43

FeatureLesson

Algebra 2Algebra 2

LessonMain

Lesson 4-4


Additional Examples

4-4

1 00 –1

2 3 4–1 0 –2

=2 3 41 0 2

Reflect the triangle with coordinates A(2, –1), B(3, 0), and C(4, –2) in each line. Graph triangle ABC and each image on the same coordinate plane.

a. x-axis

b. y-axis

c. y = x

–1 0 0 1

2 3 4–1 0 –2

=–2 –3 –4–1 0 –2

0 11 0

2 3 4–1 0 –2

=–1 0 –2 2 3 4

0 –1–1 0

2 3 4–1 0 –2

= 1 0 2–2 –3 –4

d. y = –x

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-4


a. x-axis

1 00 –1

2 3 4–1 0 –2

=2 3 41 0 2

b. y-axis

–1 0 0 1

2 3 4–1 0 –2

=–2 –3 –4–1 0 –2

c. y = x

1 00 1

2 3 4–1 0 –2

=–1 0 –2 2 3 4

0 –1–1 0

2 3 4–1 3 –2

= 1 0 2–2 –3 –4

d. y = –x

Quick Check

Additional Examples

4-4

FeatureLesson

Algebra 2Algebra 2

LessonMain

Lesson 4-4


Additional Examples

4-4

Rotate the triangle from Additional Example 3 as indicated.

Graph the triangle ABC and each image on the same coordinate plane.a. 90

c. 270

b. 180

d. 360

0 –11 0

2 3 4–1 0 –2

=1 0 22 3 4

–1 0 0 –1

2 3 4–1 0 –2

=–2 –3 –4 1 0 2

0 1–1 0

2 3 4–1 0 –2

=–1 0 –2–2 –3 –4

1 00 1

2 3 4–1 0 –2

= 2 3 4–1 0 –2

Quick Check

FeatureLesson

Algebra 2Algebra 2

LessonMain

1. Write a matrix equation that represents a translation of triangle ABC 7 units left and 3 units up.

2. Write a matrix equation that represents a dilation of triangle ABC with a

scale factor of 5.

3. Use matrix multiplication to reflect triangle ABC in the

line y = –x. Then draw the preimage and image on

the same coordinate plane.

For these questions, use triangle ABC with vertices A(–1, 1), B(2, 2), and C(1, –2).

Lesson 4-4


–1 2 1 1 2 –2

+ =–7 –7 –7 3 3 3

–8 –5 –6 4 5 1

–1 2 1 1 2 –2

5 =–5 10 5 5 10 –10

0 –1–1 0

+ =–1 2 1 1 2 –2

–1 –2 2 1 –2 –1

Lesson Quiz

4-4

FeatureLesson

Algebra 2Algebra 2

LessonMain

1a. 3(4) b. 2(6) c. 3(4) – 2(6)

2a. 3(–4) b. 2(–6) c. 3(–4) – 2(–6)

3a. –3(–4) b. 2(–6) c. –3(–4) – 2(–6)

4a. –3(4) b. –2(–6) c. –3(4) – (–2)(–6)

Lesson 4-5


2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses

Simplify each group of expressions.



4-5

FeatureLesson

Algebra 2Algebra 2

LessonMain

1a. 3(4) = 121b. 2(6) = 121c. 3(4) – 2(6) = 12 – 12 = 0

2a. 3(–4) = –122b. 2(–6) = –122c. 3(–4) – 2(–6) = –12 – (–12) = –12 + 12 = 0

3a. –3(–4) = 123b. 2(–6) = –123c. –3(–4) – 2(–6) = 12 – (–12) = 12 + 12 = 24

4a. –3(4) = –124b. –2(–6) = 124c. –3(4) – (–2)(–6) = –12 – 12 = –24

Solutions

Lesson 4-5



4-5

FeatureLesson

Algebra 2Algebra 2

LessonMain

Show that matrices A and B are multiplicative inverses.

A = B =

Lesson 4-5


3 –17 1

0.1 0.1–0.7 0.3

AB = 3 –17 1

0.1 0.1–0.7 0.3

= (3)(0.1) + (–1)(–0.7) (3)(0.1) + (–1)(0.3) (7)(0.1) + (1)(–0.7) (7)(0.1) + (1)(0.3)

= 1 00 1

AB = I, so B is the multiplicative inverse of A. Quick Check

Additional Examples

4-5

FeatureLesson

Algebra 2Algebra 2

LessonMain

Evaluate each determinant.

a. det

b. det

c. det

Lesson 4-5


7 8–5 –9

4 –35 6

a –bb a

= = (7)(–9) – (8)(–5) = –23 7 8–5 –9

= = (4)(6) – (–3)(5) = 394 –35 6

= = (a)(a) – (–b)(b) = a2 + b2a –bb a

Quick Check

Additional Examples

4-5

FeatureLesson

Algebra 2Algebra 2

LessonMain

Determine whether each matrix has an inverse. If it does, find it.

Lesson 4-5


Find det X.

ad – bc = (12)(3) – (4)(9) Simplify.

= 0

12 4 9 3

Since det X = 0, the inverse of X does not exist.

Find det Y.

ad – bc = (6)(20) – (5)(25) Simplify.

= –5

6 525 20

Since the determinant 0, the inverse of Y exists.=/

a. X =

b. Y =

Additional Examples

4-5

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-5


= – 20 –5 Substitute –5 for the–25 6 determinant.

15

= Multiply.–4 1 5 –1.2

Y–1 = 20 –5 Change signs.–25 6 Switch positions.

1 det Y

Quick Check

20 –5 Use the determinant to–25 6 write the inverse.

= 1 det Y

Additional Examples

4-5

FeatureLesson

Algebra 2Algebra 2

LessonMain

Solve X = for the matrix X.

The matrix equation has the form AX = B. First find A–1.

Lesson 4-5


9 254 11

3–7

A–1 = 1ad – bc

d –b–c a

Use the definition of inverse.

= 1(9)(11) – (25)(4)

11 –25–4 9

Substitute.

=–11 25 4 –9

Simplify.

Use the equation X = A–1B.

X =–11 25 4 –9

Substitute. 3–7

Additional Examples

4-5

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-5


= =(–11)(3) + (25)(–7) (4)(3) + (–9)(–7)

Multiply andsimplify.

–208 75

Check:X =

9 254 11

Use the originalequation.

3–7

9 254 11

Substitute. 3–7

–208 75

Multiply and simplify. 3–7

9(–208) + 25(75)4(–208) + 11(75)

= 3–7

3–7 Quick Check

Additional Examples

4-5

FeatureLesson

Algebra 2Algebra 2

LessonMain

In a city with a stable group of 45,000 households, 25,000

households use long distance carrier A, and 20,000 use long distance

carrier B. Records show that over a 1-year period, 84% of the

households remain with carrier A, while 16% switch to B. 93% of the

households using B stay with B, while 7% switch to A.

a. Write a matrix to represent the changes in long distance carriers.

Lesson 4-5


0.84 0.070.16 0.93

To ATo B

FromA B

Write the percents as decimals.

Additional Examples

4-5

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

b. Predict the number of households that will be using distance carrier B next year.

Lesson 4-5


2500020000

Use AUse B

Write the information in a matrix.

2500020000

0.84 0.070.16 0.93

22,40022,600

=

22,600 households will use carrier B.

Additional Examples

4-5

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-5


First find the determinant of . 0.84 0.070.16 0.93

0.84 0.070.16 0.93

= 0.77

About 28,400 households used carrier A.

Multiply the inverse matrix by the information matrix in part (b).Use a calculator and the exact inverse.

28,37716,623

25,00020,000

0.93 –0.07–.016 0.84

10.77

c. Use the inverse of the matrix from part (a) to find, to the nearest hundred households, the number of households that used carrier A last year.

Quick Check

Additional Examples

4-5

FeatureLesson

Algebra 2Algebra 2

LessonMain

1. Is the inverse of ? How do you know?

2. Find the determinant of .

3. Find the inverse of .

4. Solve the equation X = for X.

Lesson 4-5


5 2–2 1

1 22 5

–12 5–16 4

–2 4 3 –7

20 35 1 2

–3 7

no; Answers may vary. Sample is , which

is not the 2 2 identity matrix.

1 22 5

5 2–2 –1

1 00 –1

32

–3.5 –2–1.5 –1

–50.2 28.6

Lesson Quiz

4-5

FeatureLesson

Algebra 2Algebra 2

LessonMain

Find the product of the circled elements in each matrix.

1. 2. 3.

4. 5. 6.

Lesson 4-6



2 3 0–1 3 –2 4 –3 –4

2 3 0–1 3 –2 4 –3 –4

2 3 0–1 3 –2 4 –3 –4

2 3 0–1 3 –2 4 –3 –4

2 3 0–1 3 –2 4 –3 –4

2 3 0–1 3 –2 4 –3 –4



4-6

FeatureLesson

Algebra 2Algebra 2

LessonMain

1. (2)(3)(–4) = 6(–4) = –24

2. (0)(–1)(–3) = 0(–3) = 0

3. (3)(–2)(4) = (–6)(4) = –24

4. (2)(–2)(–3) = (–4)(–3) = 12

5. (3)(–1)(–4) = (–3)(–4) = 12

6. (0)(3)(4) = (0)(4) = 0

Solutions

Lesson 4-6



4-6

FeatureLesson

Algebra 2Algebra 2

LessonMain

Evaluate the determinant of X = .

Lesson 4-6


8 –4 3–2 9 5 1 6 0

= [(8)(9)(0) + (–2)(6)(3) + (1)(–4)(5)] Use the – [(8)(6)(5) + (–2)(–4)(0) + (1)(9)(3)] definition.

8 –4 3–2 9 5 1 6 0

= [0 + (–36) + (–20)] – [240 + 0 + 27] Multiply.

= –56 – 267 = –323. Simplify.

The determinant of X is –323.Quick Check

Additional Examples

4-6

FeatureLesson

Algebra 2Algebra 2

LessonMain

Enter matrix T into your graphing calculator. Use the matrix

submenus to evaluate the determinant of the matrix.

The determinant of the matrix is –65.

Lesson 4-6


T = 4 2 3–2 –1 5 1 3 6

Quick Check

Additional Examples

4-6

FeatureLesson

Algebra 2Algebra 2

LessonMain

Determine whether the matrices are multiplicative inverses.

a. C = , D =

Lesson 4-6


0.5 0 00 0 0.50 1 1

2 0 00 2 10 2 0

0.5 0 00 0 0.50 1 1

2 0 00 2 10 2 0

1 0 00 1 00 4 1

=

Since CD I, C and D are not multiplicative inverses. =/

Additional Examples

4-6

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-6


b. A = , B = 0 0 10 1 01 0 –1

1 0 10 1 01 0 0

1 0 00 1 00 0 1

= 0 0 10 1 01 0 –1

1 0 10 1 01 0 0

Since AB = I, A and B are multiplicative inverses.

Quick Check

Additional Examples

4-6

FeatureLesson

Algebra 2Algebra 2

LessonMain

Solve the equation.

Lesson 4-6


2 0 10 1 41 0 0

–1 8–2

X =

Let A = .2 0 10 1 41 0 0

Find A–1.

X = 0 0 1–4 1 8 1 0 –2

–1 8–2

Use the equation X = A–1C.Multiply.

X =–2–4 3 Quick Check

Additional Examples

4-6

FeatureLesson

Algebra 2Algebra 2

LessonMain

Use the alphabet table and

the encoding matrix.

matrix K = .

Lesson 4-6


0.5 0.25 0.250.25 –0.5 0.50.5 1 –1

a. Find the decoding matrix K–1.

K–1 = Use a graphing calculator.0 2 12 –2.5 –0.752 –1.5 –1.25

Additional Examples

4-6

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-6


b. Decode . Zero indicates a space holder.11.25 16.75 24.5 5.75 17 5.5 1.5 –12 15

=0 2 12 –2.5 –0.752 –1.5 –1.25

11.25 16.75 24.5 5.75 17 5.5 1.5 –12 15

13 22 26 7 0 24 12 23 22

Use thedecodingmatrixfrompart (a).Multiply.

The numbers 13 22 26 7 0 24 12 23 22 correspond to the letters NEAT CODE.

Quick Check

Additional Examples

4-6

FeatureLesson

Algebra 2Algebra 2

LessonMain

1. Use pencil and paper to evaluate the determinant of

2. Determine whether the matrices are multiplicative inverses.

3. Solve the equation for M.

Lesson 4-6


–2 –4 2 3 1 0 5 –6 –2

.

1 1 –1–1 0 1 0 –1 1

;1 0 11 1 01 1 1

–1 –1 1 1 2 –1 0 –1 1

M =–1–4 3

–66

yes

4–5–2

Lesson Quiz

4-6

FeatureLesson

Algebra 2Algebra 2

LessonMain

Lesson 4-7

Inverse Matrices and SystemsInverse Matrices and Systems


1.

2.

3.

5x + y = 144x + 3y = 20

x – y – z = –93x + y + 2z = 12x = y – 2z

–x + 2y + z = 0y = –2x + 3z = 3x

Solve each system.



4-7

FeatureLesson

Algebra 2Algebra 2

LessonMain

1. 2.

Solutions

Lesson 4-7


5x + y = 144x + 3y = 20

Solve the first equation for y:y = –5x + 14Substitute this into the secondequation:4x + 3(–5x + 14) = 20

4x – 15x + 42 = 20 –11x + 42 = 20 –11x = –22 x = 2Use the first equation with x = 2: 5(2) + y = 14 10 + y = 14 y = 4The solution is (2, 4).

x – y – z = –93x + y + 2z = 12x = y – 2z

Use the first equation withx = y – 2z (third equation):(y – 2z) – y – z = –9 –3z = –9 z = 3Use the second equation with x =y – 2z (third equation) and z = 3: 3(y – 2z) + y + 2z = 12 3(y – 2(3) + y + 2(3) = 12 3y – 18 + y + 6 = 12 4y – 12 = 12 4y = 24 y = 6Use the third equation with y = 6And z = 3:x = 6 – 2(3) = 6 – 6 = 0The solution is (0, 6, 3).


4-7

FeatureLesson

Algebra 2Algebra 2

LessonMain


Lesson 4-7


Use the first equation withy = –2x + 3 (second equation)and z = 3x (third equation):–x + 2(–2x + 3) + 3x = 0 –x – 4x + 6 + 3x = 0 2x + 6 = 0 –2x = –6 x = 3

–x + 2y + z = 0y = –2x + 3z = 3x

Use the second equation with x = 3:y = –2(3) + 3 = –6 + 3 = –3Use the third equation with x = 3:z = 3(3) = 9The solution is (3, –3, 9).

3.


4-7

FeatureLesson

Algebra 2Algebra 2

LessonMain

Write the system

as a matrix equation.

Lesson 4-7


–3x – 4y + 5z = 11–2x + 7y = –6–5x + y – z = 20

Then identify the coefficient matrix, the variable matrix, and the constant matrix.

Matrix equation: =–3 –4 5–2 7 0–5 1 –1

xyz

11–6 20

Coefficient matrix

–3 –4 5–2 7 0–5 1 –1

xyz

Variable matrix

11–6 20

Constant matrix

Quick Check

Additional Examples

4-7

FeatureLesson

Algebra 2Algebra 2

LessonMain

Solve the system.

Lesson 4-7


2x + 3y = –1 x – y = 12

2 31 –1

xy

–112

= Write the system as a matrix equation.

A–1 = Find A–1.

15

35

15

25

–

= A–1B = = Solve for the variable matrix.

xy

15

35

15

25

–

–112

7–5

Additional Examples

4-7

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-7


The solution of the system is (7, –5).

Check: 2x + 3y = –1 x – y = 12 Use theoriginalequations.

2(7) + 3(–5) –1 (7) – (–5) 12 Substitute.

14 – 15 = –1 7 + 5 = 12 Simplify.

Quick Check

Additional Examples

4-7

FeatureLesson

Algebra 2Algebra 2

LessonMain

Solve the system .

Step 1: Write the system asa matrix equation.

Lesson 4-7


7x + 3y + 2z = 13–2x + y – 8z = 26 x – 4y +10z = –13

Step 2: Store the coefficientmatrix as matrix Aand the constantmatrix as matrix B. 7 3 2

–2 1 –8 1 –4 10

13 26–13

xyz

=

The solution is (9, –12, –7).

Quick Check

Additional Examples

4-7

FeatureLesson

Algebra 2Algebra 2

LessonMain

A linen shop has several tables of sheets and towels on special sale. The sheets are all priced the same, and so are the towels. Mario bought 3 sheets and 5 towels at a cost of $137.50. Marco bought 4 sheets and 2 towels at a cost of $118.00. Find the price of each item.

Lesson 4-7


Define: Let x = the price of one sheet.

Let y = the price of one towel.

Write: =3 54 2

xy

137.50118.00

Use a graphing calculator. Store the coefficient matrix as matrix A and the constant matrix as matrix B.

Relate: 3 sheets and 5 towels cost $137.50.

4 sheets and 2 towels cost $118.00.

The price of a sheet is $22.50. The price of a towel is $14.00.Quick Check

Additional Examples

4-7

FeatureLesson

Algebra 2Algebra 2

LessonMain

Write the coefficient matrix for each system. Use it to

determine whether the system has exactly a unique solution.

Lesson 4-7


a. 4x – 2y = 7–6x + 3y = 5

A = ; det A = = 4(3) – (–2)(–6) = 0 4 –2–6 3

4 –2–6 3

Since det A = 0, the matrix does not have an inverse and the system does not have a unique solution.

Additional Examples

4-7

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-7


b. 12x + 8y = –3 3x – 7y = 50

A = ; det A = = 12(–7) – 8(–3) = –60 12 8 3 –7

12 8 3 –7

Since det A 0, the matrix has an inverse and the system has a unique solution.

=/

Quick Check

Additional Examples

4-7

FeatureLesson

Algebra 2Algebra 2

LessonMain

1.

2.

Determine whether each system has a unique solution.

3.

4.

Lesson 4-7


yes

3x + 2y = –62x – 3y = 61

2x + 4y + 5z = –3 7x + 9y + 4z = 19–3x + 2y + 8z = 0

7x – 2y = 15–28x + 8y = 7

20x + 5y = 33–32x + 8y = 47

= ; (8, –15)3 22 –3

–661

xy

2 4 5 7 9 4–3 2 8

–319 0

xyz

= ; (–10, 13, –7)

no

Write each system as a matrix equation. Then solve the system.

Lesson Quiz

4-7

FeatureLesson

Algebra 2Algebra 2

LessonMain

1. 2. 3.

Lesson 4-8

Augmented Matrices and SystemsAugmented Matrices and Systems

(For help, go to Lessons 4-5 and 4-6.)

4. 5. 6.

–1 2 0 3

0 1–1 3

2 1–1 5

0 1 –3 4 5 –1–1 0 1

3 4 5–1 2 0 0 –1 1

0 2 –1 3 4 0–2 –1 5

Evaluate the determinant of each matrix.



4-8

FeatureLesson

Algebra 2Algebra 2

LessonMain

1. det = (–1)(3) – (2)(0) = –3 – 0 = –3

2. det = (0)(3) – (1)(–1) = 0 – (–1) = 1

3. det = (2)(5) – (1)(–1) = 10 – (–1) = 11

4. det = [(0)(5)(–1) + (4)(0)(–3) + (–1)(1)(–1)] – [(0)(0)(–1) +

(4)(1)(1) + (–1)(5)(–3)] = [0 + 0 + 1] – [0 + 4 + 15] = 1 – 19 = –18

Solutions

Lesson 4-8


–1 2 0 3

0 1–1 3

2 1–1 5

0 1 –3 4 5 –1–1 0 1


4-8

FeatureLesson

Algebra 2Algebra 2

LessonMain

5. det = [(3)(2)(1) + (–1)(–1)(5) + (0)(4)(0)] – [(3)(–1)(0) +

(–1)(4)(1) + (0)(2)(5)] = [6 + 5 + 0] – [0 + (–4) + 0] = 11 – (–4) = 15

6. det =[(0)(4)(5) + (3)(–1)(–1) + (–2)(2)(0)] – [(0)(–1)(0) +

(3)(2)(5) + (–2)(4)(–1)] = [0 + 3 + 0] – [0 + 30 + 8] = 3 – 38 = –35


Lesson 4-8


3 4 5–1 2 0 0 –1 1

0 2 –1 3 4 0–2 –1 5


4-8

FeatureLesson

Algebra 2Algebra 2

LessonMain

Use Cramer’s rule to solve the system .

Lesson 4-8


Evaluate three determinants. Then find x and y.

7x – 4y = 153x + 6y = 8

D = = 547 –43 6

Dx = = 12215 –4 8 6

Dy = = 117 153 8

x = = Dx

D6127

1154

Dy

Dy = =

The solution of the system is , .6127

1154

Quick Check

Additional Examples

4-8

FeatureLesson

Algebra 2Algebra 2

LessonMain

Find the y-coordinate of the solution of the

system .

Lesson 4-8


–2x + 8y + 2z = –3–6x + 2z = 1–7x – 5y + z = 2

D = = –24 Evaluate the determinant.–2 8 2–6 0 2–7 –5 1

Dy = = 20 Replace the y-coefficients with theconstants and evaluate again.

–2 –3 2–6 1 2–7 2 1

y = = – = – Find y.2024

Dy

D56

The y-coordinate of the solution is – .56

Quick Check

Additional Examples

4-8

FeatureLesson

Algebra 2Algebra 2

LessonMain

Write an augmented matrix to represent the

system

Lesson 4-8


–7x + 4y = –3 x + 8y = 9

System of equations –7x + 4y = –3 x + 8y = 9

x-coefficients y-coefficients constants

Augmented matrix –7 4 –3 1 8 9

Draw a vertical bar to separate the coefficients from constants. Quick Check

Additional Examples

4-8

FeatureLesson

Algebra 2Algebra 2

LessonMain

Write a system of equations for the augmented

matrix .

Lesson 4-8


9 –7 –12 5 –6

Augmented matrix 9 –7 –1 2 5 –6

x-coefficients y-coefficients constants

System of equations 9x – 7y = –12x + 5y = –6

Quick Check

Additional Examples

4-8

FeatureLesson

Algebra 2Algebra 2

LessonMain

Use an augmented matrix to solve the system

Lesson 4-8


x – 3y = –174x + 2y = 2

1 –3 –174 2 2

Write an augmented matrix.

Multiply Row 1 by –4 and add it to Row 2.Write the new augmented matrix.

1 –3 –17

0 14 70

–4(1 –3 –17) 4 2 2 0 14 70

1141 –3 –17

0 1 5

Multiply Row 2 by .

Write the new augmented matrix.

(0 14 70) 0 1 5

114

Additional Examples

4-8

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-8


1141 –3 –17

0 1 5 (0 14 70) 0 1 5

1 0 –20 1 5

1 –3 –173(0 1 5) 1 0 –2

Multiply Row 2 by 3 and add it to Row 1.Write the final augmented matrix.

The solution to the system is (–2, 5).

Check: x – 3y = –17 4x + 2y = 2 Use the original equations. (–2) – 3(5) –17 4(–2) + 2(5) 2 Substitute. –2 – 15 –17 –8 + 10 2 Multiply. –17 = –17 2 = 2 Quick Check

Additional Examples

4-8

FeatureLesson

Algebra 2Algebra 2

LessonMain

Use the rref feature on a graphing calculator to solve the

system

Lesson 4-8


4x + 3y + z = –1–2x – 2y + 7z = –10. 3x + y + 5z = 2

Step 1: Enter theaugmented matrixas matrix A.

Step 2: Use the rref featureof your graphingcalculator.

The solution is (7, –9, –2).

Additional Examples

4-8

FeatureLesson

Algebra 2Algebra 2

LessonMain

(continued)

Lesson 4-8


Partial Check: 4x + 3y + z = –1 Use the original equation.

4(7) + 3(–9) + (–2) –1 Substitute.

28 – 27 – 2 –1 Multiply.

–1 = –1 Simplify.

Quick Check

Additional Examples

4-8

FeatureLesson

Algebra 2Algebra 2

LessonMain

1. Use Cramer’s Rule to solve the system.

2. Suppose you want to use Cramer’s Rule to find the value of z in the following system. Write the determinants you would need to evaluate.

3. Solve the system by using an augmented matrix.

4. Solve the system by using an augmented matrix.

Lesson 4-8


(–1, 8, –3)

3x + 2y = –25x + 4y = 8

–7x + 3y + 9z = 12 5x + 3z = 8 4x – 6y + z = –2

4x + y – z = 7–2x + 2y + 5z = 3 7x – 3y – 9z = –4

5x + y = 13x – 2y = 24

(–12, 17)

D = , Dz =–7 3 9 5 0 3 4 –6 1

–7 3 12 5 0 8 4 –6 –2

(2, –9)

Lesson Quiz

4-8

Documents

Organizing Data Into Matrices