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Organizing Data Into Matrices. Lesson 4-1. Check Skills You’ll Need. (For help, go to Skills Handbook page 842.). Use the table at the right. How many units were imported to the United States in 1996? How many were imported in 2000?. U.S. Passenger Vehicles and Light Trucks Imports

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Organizing data into matrices

Organizing Data Into Matrices

Lesson 4-1

Check Skills You’ll Need

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

Use the table at the right.

  • How many units were imported to the

  • United States in 1996?

  • How many were imported in 2000?

U.S. Passenger Vehicles and Light Trucks Imports

And Exports (millions)

4.678 million units

1996 1998 2000

Imports4.6785.1856.964

Exports1.2951.3311.402

Source: U.S. Department of Commerce.

6.964 million units

Check Skills You’ll Need

4-1


Organizing data into matrices1

Organizing Data Into Matrices

Lesson 4-1

Additional Examples

Write the dimensions of each matrix.

7 –4

12 9

The matrix has 2 rows and 2 columns and is therefore a 2  2 matrix.

a.

The matrix has 1 row and 3 columns and is therefore a 1  3 matrix.

b.

0 6 15

Quick Check

4-1


Organizing data into matrices2

3 –1 –8 5

1 8 4 9

8 –4 7 –5

3 –1 –8 5

1 8 4 9

8 –4 7 –5

a. K =

b. K =

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

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

Organizing Data Into Matrices

Lesson 4-1

Additional Examples

Identify each matrix element.

3 –1 –8 5

1 8 4 9

8 –4 7 –5

K =

a.k12

b.k32

c.k23

d.k34

Element k12 is –1.

Element k32 is –4.

4-1


Organizing data into matrices3

3 –1 –8 5

1 8 4 9

8 –4 7 –5

3 –1 –8 5

1 8 4 9

8 –4 7 –5

c. K =

d. K =

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

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

Organizing Data Into Matrices

Lesson 4-1

Additional Examples

(continued)

3 –1 –8 5

1 8 4 9

8 –4 7 –5

K =

a.k12

b.k32

c.k23

d.k34

Quick Check

Element k23 is 4.

Element k34 is –5.

4-1


Organizing data into matrices4

Ed X X

Jo X

Lew X X X X

Ed Jo Lew

Wins

Losses

5 6 3

2 1 4

Organizing Data Into Matrices

Lesson 4-1

Additional Examples

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.

= Win X = Loss

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

Quick Check

4-1


Organizing data into matrices5

Each column represents a different year.

1996 1998 2000

N = Import

Exports

4.678 5.185 6.964

1.295 1.331 1.402

Each row represents imports and exports.

Organizing Data Into Matrices

Lesson 4-1

Additional Examples

Refer to the table.

U.S. Passenger Car Imports

And Exports (millions)

199619982000

Imports4.6785.1856.964

Exports1.2951.3311.402

a.Write a matrix N to represent the information.

Use 2  3 matrix.

Source: U.S. Department of Commerce.

4-1


Organizing data into matrices6

Organizing Data Into Matrices

Lesson 4-1

Additional Examples

(continued)

U.S. Passenger Car Imports

And Exports (millions)

199619982000

Imports4.6785.1856.964 Exports1.2951.3311.402

b.Which element represents exports for 2000?

Source: 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

4-1


Organizing data into matrices7

Deposits Withdrawals

A$450$370

B$475$289

C$364$118

D$420$400

A B C D

Deposits

Withdrawals

450 475 364 420

370 289 118 400

Organizing Data Into Matrices

Lesson 4-1

Lesson Quiz

8 4 0 1

9 3 –5 0

–1 2 6 1

3  4

1.Write the dimensions of the matrix. M =

2.Identify the elements m24, m32, and m13 of the matrix M in question 1.

0, 2, 0

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.

4-1


Adding and subtracting matrices

Adding and Subtracting Matrices

Lesson 4-2

Check Skills You’ll Need

(For help, go to Skills Handbook page 845.)

Simplify the elements of each matrix.

1.2.

3.4.

5.6.

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

Check Skills You’ll Need

4-2


Adding and subtracting matrices1

Adding and Subtracting Matrices

Lesson 4-2

Check Skills You’ll Need

Solutions

1. =

2. =

3. =

4. =

5. =

6. =

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


Adding and subtracting matrices2

Evening

C A

Theater 1 54 439

Theater 2 58 386

Matinee Evening

Theater C A C A

1 198 350 54 439

2 201 375 58 386

Matinee

C A

Theater 1 198 350

Theater 2 201 375

Adding and Subtracting Matrices

Lesson 4-2

Additional Examples

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).

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

4-2


Adding and subtracting matrices3

198 350

201 375

54 439

58 386

198 + 54350 + 439

201 + 58375 + 386

+

=

C A

Theater 1 252 789

Theater 2 259 761

=

Adding and Subtracting Matrices

Lesson 4-2

Additional Examples

(continued)

b. Find the combined sales for the two showings.

Quick Check

4-2


Adding and subtracting matrices4

9 + 0 0 + 0

–4 + 0 6 + 0

3 + (–3) –8 + 8

–5 + 5 1 + (–1)

=

=

9 0

–4 6

0 0

0 0

=

=

Adding and Subtracting Matrices

Lesson 4-2

Additional Examples

Find each sum.

9 0

–4 6

0 0

0 0

3 –8

–5 1

–3 8

5 –1

a.

+

b.

+

Quick Check

4-2


Adding and subtracting matrices5

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

4 8

–2 0

–7 9

–4 –5

A– B = A + (–B) = +

4 + (–7) 8 + 9

–2 + (–4) 0 + (–5)

Add corresponding elements

=

–3 17

–6 –5

Simplify.

=

Adding and Subtracting Matrices

Lesson 4-2

Additional Examples

4 8

–2 0

7 –9

4 5

A = and B = . Find A – B.

Method 1: Use additive inverses.

4-2


Adding and subtracting matrices6

4 8

–2 0

7 –9

4 5

A– B = –

4 – 7 8 – (–9)

–2 – 4 0 – 5

Subtract corresponding elements

=

–3 17

–6 –5

Simplify.

=

Adding and Subtracting Matrices

Lesson 4-2

Additional Examples

(continued)

Method 2: Use subtraction.

Quick Check

4-2


Adding and subtracting matrices7

2 5

3 –1

8 0

10 –3

–4 9

6 –9

X – =

2 5

3 –1

8 0

2 5

3 –1

8 0

10 –3

–4 9

6 –9

2 5

3 –1

8 0

2 5

3 –1

8 0

X – + = +

Add

to each side of the equation.

12 2

–1 8

14 –9

Simplify.

X =

Adding and Subtracting Matrices

Lesson 4-2

Additional Examples

Quick Check

2 5

3 –1

8 0

10 –3

–4 9

6 –9

Solve X – = for the matrix X.

4-2


Adding and subtracting matrices8

17 5

4 – 10 –2 + 1

0 –

8 + 9 5

–6 –1

0 0.7

a.M = ; N =

79

17 5

4 – 10 –2 + 1

0 –

8 + 9 5

–6 –1

0 0.7

M = ; N =

7

9

7

9

Both M and N have three rows and two columns, but – 0.7.

M and N are not equal matrices.

=

/

Adding and Subtracting Matrices

Lesson 4-2

Additional Examples

Determine whether the matrices in each pair are equal.

4-2


Adding and subtracting matrices9

8

0.2

27

9

16

4

3–4

40–3

P = ; Q =

8

0.2

12

4

Adding and Subtracting Matrices

Lesson 4-2

Additional Examples

(continued)

27

9

16

4

3 –4

40 –3

b.P = ; Q =

12

4

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

Quick Check

4-2


Adding and subtracting matrices10

2m – n –3

8 –4m + 2n

15 m + n

8 –30

= for m and n.

2m – n–3

8 –4m + 2n

15m + n

8 –30

=

2m – n = 15–3 = m + n–4m + 2n = –30

Adding and Subtracting Matrices

Lesson 4-2

Additional Examples

Solve the equation

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

4-2


Adding and subtracting matrices11

3m = 12Add the equations.

m = 4Solve for m.

Adding and Subtracting Matrices

Lesson 4-2

Additional Examples

(continued)

Solve for m and n.

2m – n = 15

m + n = –3

4 + n = –3Substitute 4 for m.

n = –7Solve for n.

The solutions are m = 4 and n = –7.

Quick Check

4-2


Adding and subtracting matrices12

0 3

–5 3

–9 10

=

/

–7 14

6 –17

0 0 0 0

0 0 0 0

2

3

no, – –0.6

6 10

1 11

Adding and Subtracting Matrices

Lesson 4-2

Lesson Quiz

Find each sum or difference.

1. +2. –

–3 1

4 8

–5 4

–3 –2

9 5

4 –6

2 8

0 –12

–9 6

6 –5

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.

–2x –1

5 x + y

18 –3x + 4y

x – 2y –16

x = –9, y = –7

=

3 0.5

6

2

0.50

0.4 –0.6

;

2

5

2

3

4 3

1 5

2 7

0 6

4-2


Matrix multiplication

Matrix Multiplication

Lesson 4-3

Check Skills You’ll Need

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

Find each sum.

1. + +

3 5

2 8

3 5

2 8

3 5

2 8

–4

7

–4

7

–4

7

–4

7

–4

7

2. + + + +

–1 3 4

0 –2 –5

–1 3 4

0 –2 –5

–1 3 4

0 –2 –5

–1 3 4

0 –2 –5

3. + + +

Check Skills You’ll Need

4-3


Matrix multiplication1

–4

7

–4

7

–4

7

–4

7

–4

7

2. + + + + =

–4 + (–4) + (–4) + (–4) + (–4)

7 + 7 + 7 + 7 + 7

–20

35

=

–1 3 4

0 –2 –5

–1 3 4

0 –2 –5

–1 3 4

0 –2 –5

–1 3 4

0 –2 –5

3. + + +

4(–1) 4(3) 4(4)

4(0) 4(–2) 4(–5)

–4 12 16

0 –8 –20

= =

Matrix Multiplication

Lesson 4-3

Check Skills You’ll Need

Solutions

3 5

2 8

3 5

2 8

3 5

2 8

3 + 3 + 3 5 + 5 + 5

2 + 2 + 2 8 + 8 + 8

9 15

6 24

1. + + = =

4-3


Matrix multiplication2

Store M1 M2 M3

A $38,500 $40,000 $44,600

B $39,000 $37,800 $43,700

38500 40000 44600

39000 37800 43700

1.08

1.08(38500) 1.08(40000) 1.08(44600)

1.08(39000) 1.08(37800) 1.08(43700)

= Multiply each element

by 1.08.

Matrix Multiplication

Lesson 4-3

Additional Examples

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.

4-3


Matrix multiplication3

M1 M2 M3

41580 43200 48168

42120 40824 47196

= A

B

Matrix Multiplication

Lesson 4-3

Additional Examples

(continued)

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

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

Quick Check

4-3


Matrix multiplication4

2 –3

0 6

–5 –1

2 9

–3M + 7N = –3 + 7

–6 9

0 –18

–35 –7

14 63

= +

–41 2

14 45

=

Matrix Multiplication

Lesson 4-3

Additional Examples

Find the sum of –3M + 7N for

M = and N = .

2 –3

0 6

–5 –1

2 9

Quick Check

4-3


Matrix multiplication5

6 9

–12 15

27 –18

30 6

–3Y + 2 =

12 18

–24 30

27 –18

30 6

–3Y + = Scalar multiplication.

27 –18

30 6

12 18

–24 30

12 18

–24 30

–3Y = – Subtract

from each side.

15 –36

54 –24

–3Y = Simplify.

Multiply each side

by – and simplify.

15 –36

54 –24

–5 12

–18 8

1

3

Y = – =

1

3

Matrix Multiplication

Lesson 4-3

Additional Examples

6 9

–12 15

27 –18

30 6

Solve the equation –3Y + 2 = .

4-3


Matrix multiplication6

Check:

6 9

–12 15

27 –18

30 6

–3Y + 2 =

–5 12

–18 8

6 9

–12 15

27 –18

30 6

–3+ 2 Substitute.

15 –36

54 –24

12 18

–24 30

27 –18

30 6

+ Multiply.

27 –18

30 6

27 –18

30 6

= Simplify.

Matrix Multiplication

Lesson 4-3

Additional Examples

(continued)

Quick Check

4-3


Matrix multiplication7

–2 5

3 –1

4 –4

2 6

= (–2)(4) + (5)(2) = 2

–2 5

3 –1

4 –4

2 6

2

= (–2)(4) + (5)(6) = 38

–2 5

3 –1

4 –4

2 6

2 38

= (3)(4) + (–1)(2) = 10

Matrix Multiplication

Lesson 4-3

Additional Examples

–2 5

3 –1

4 –4

2 6

Find the product of and .

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

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

4-3


Matrix multiplication8

–2 5

3 –1

4 –4

2 6

2 38

10

= (3)(–4) + (–1)(6) = –18

–2 5

3 –1

4 –4

2 6

2 38

10 –18

The product of and is .

Matrix Multiplication

Lesson 4-3

Additional Examples

(continued)

Quick Check

4-3


Matrix multiplication9

109

76

18 22 = (18)(109) + (22)(76) = 3634

Matrix Multiplication

Lesson 4-3

Additional Examples

Quick Check

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.

Prices Number of Items Sold

Shirts Jeans

Shirts 109

Jeans 76

A = $18 $22B =

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

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

4-3


Matrix multiplication10

QP

(3 4) (3 3)

not equal

Matrix Multiplication

Lesson 4-3

Additional Examples

3 –1 2

5 9 0

0 1 8

6 5 7 0

2 0 3 1

1 –1 5 2

Use matrices P = and Q = .

Determine whether products PQ and QP are defined or undefined.

Find the dimensions of each product matrix.

PQ

(33) (34) 34

product

equal matrix

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

4-3


Matrix multiplication11

–6 4 –2

–27 18 –9

–12 8 –4

16 24 –8

0 –40 32

27

–29

Matrix Multiplication

Lesson 4-3

Lesson Quiz

Use matrices A, B, C, and D.

2 3 –1

0 –5 4

–7 1 0

2 6 –6

A = B =

C = D =

2

9

4

–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?

undefined

2  3

4-3


Geometric transformations with matrices

1

3

Geometric Transformations with Matrices

Lesson 4-4

Check Skills You’ll Need

(For help, go to Lesson 2-6.)

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

1

2

1.y = x + 2; left 4 units2.ƒ(x) = x + 2; up 5 units

3.g(x) = |x|; right 3 units4.y = x; down 2 units

5.y = |x – 3|; down 2 units6.ƒ(x) = –2|x|; right 2 units

Check Skills You’ll Need

4-4


Geometric transformations with matrices1

1

2

1

2

Geometric Transformations with Matrices

Lesson 4-4

Check Skills You’ll Need

1.y = x + 2;

left 4 units:

y = x + 6 

Solutions

2.ƒ(x) = x + 2;

up 5 units:

ƒ(x) = x + 7;

3.g(x) = |x|

right 3 units:

g(x) = |x – 3| 

4.y = x;

down 2 units:

y = x – 2

4-4


Geometric transformations with matrices2

1

3

1

3

Geometric Transformations with Matrices

Lesson 4-4

Check Skills You’ll Need

5.y = |x – 3|;

down 2 units:

y = |x – 3| – 2

Solutions (continued)

6.ƒ(x) = –2|x|

right 2 units:

ƒ(x) = –2|x + 4|

4-4


Geometric transformations with matrices3

Vertices ofTranslationVertices of

the Triangle Matrixthe image

Subtract 3 from each

x-coordinate.

Add 1 to each

y-coordinate.

A B CA B C

1 3 2

–2 1 3

–3 –3 –3

1 1 1

–2 0 –1

–1 2 4

+ =

Geometric Transformations with Matrices

Lesson 4-4

Additional Examples

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 ABC.

The coordinates of the vertices of the image are

A (–2, –1), B (0, 2), C (–1, 4).

Quick Check

4-4


Geometric transformations with matrices4

A B C D EABCDE

4

3

2

3

4

3

0 2 3 –1 –2

3 2 –2 –3 0

0 2 – –

2 – –2 0

2

3

=

Multiply.

4

3

4

3

4

3

4

3

4

3

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

D (– , –2), and E (– , 0).

2

3

4

3

Geometric Transformations with Matrices

Lesson 4-4

Additional Examples

2

3

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

Write a matrix to represent the coordinates of the vertices.

Quick Check

4-4


Geometric transformations with matrices5

1 0

0 –1

2 3 4

–1 0 –2

2 3 4

1 0 2

=

–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

=

0 –1

–1 0

2 3 4

–1 0 –2

1 0 2

–2 –3 –4

=

Geometric Transformations with Matrices

Lesson 4-4

Additional Examples

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

d.y = –x

4-4


Geometric transformations with matrices6

a.x-axis

1 0

0 –1

2 3 4

–1 0 –2

2 3 4

1 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 0

0 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

=

Geometric Transformations with Matrices

Lesson 4-4

Additional Examples

(continued)

d.y = –x

Quick Check

4-4


Geometric transformations with matrices7

0 –1

1 0

2 3 4

–1 0 –2

1 0 2

2 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 0

0 1

2 3 4

–1 0 –2

2 3 4

–1 0 –2

=

Geometric Transformations with Matrices

Lesson 4-4

Additional Examples

Rotate the triangle from Additional Example 3 as indicated. Graph the triangle ABC and each image on the same coordinate plane.

a. 90

b. 180

c. 270

d. 360

Quick Check

4-4


Geometric transformations with matrices8

–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 10 5

5 10 –10

5 =

0 –1

–1 0

–1 2 1

1 2 –2

–1 –2 2

1 –2 –1

+ =

Geometric Transformations with Matrices

Lesson 4-4

Lesson Quiz

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

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.

4-4


2 x 2 matrices determinants and inverses

2 X 2 Matrices, Determinants, and Inverses

Lesson 4-5

Check Skills You’ll Need

(For help, go to Skills Handbook page 845.)

Simplify each group of expressions.

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)

Check Skills You’ll Need

4-5


2 x 2 matrices determinants and inverses1

2 X 2 Matrices, Determinants, and Inverses

Lesson 4-5

Check Skills You’ll Need

Solutions

1a.3(4) = 12

1b.2(6) = 12

1c.3(4) – 2(6) = 12 – 12 = 0

2a.3(–4) = –12

2b.2(–6) = –12

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

3a.–3(–4) = 12

3b.2(–6) = –12

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

4a.–3(4) = –12

4b.–2(–6) = 12

4c.–3(4) – (–2)(–6) = –12 – 12 = –24

4-5


2 x 2 matrices determinants and inverses2

3 –1

7 1

0.1 0.1

–0.7 0.3

AB =

(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 0

0 1

=

2 X 2 Matrices, Determinants, and Inverses

Lesson 4-5

Additional Examples

Show that matrices A and B are multiplicative inverses.

3 –1

7 1

0.1 0.1

–0.7 0.3

A = B =

AB = I, so B is themultiplicative inverse of A.

Quick Check

4-5


2 x 2 matrices determinants and inverses3

78

–5–9

= = (7)(–9) – (8)(–5) = –23

4–3

56

= = (4)(6) – (–3)(5) = 39

a–b

ba

= = (a)(a) – (–b)(b) = a2 + b2

2 X 2 Matrices, Determinants, and Inverses

Lesson 4-5

Additional Examples

Evaluate each determinant.

7 8

–5 –9

a. det

b. det

c. det

4 –3

5 6

a –b

ba

Quick Check

4-5


2 x 2 matrices determinants and inverses4

=

/

12 4

9 3

6 5

25 20

Since the determinant 0, the inverse of Y exists.

2 X 2 Matrices, Determinants, and Inverses

Lesson 4-5

Additional Examples

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

a.X =

Find det X.

ad – bc = (12)(3) – (4)(9) Simplify.

= 0

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

b.Y =

Find det Y.

ad – bc = (6)(20) – (5)(25) Simplify.

= –5

4-5


2 x 2 matrices determinants and inverses5

20–5 Change signs.

–256 Switch positions.

1

det Y

20 –5 Use the determinant to

–25 6 write the inverse.

1

det Y

=

20 –5 Substitute –5 for the

–25 6 determinant.

1

5

= –

–4 1

5 –1.2

= Multiply.

2 X 2 Matrices, Determinants, and Inverses

Lesson 4-5

Additional Examples

(continued)

Y–1 =

Quick Check

4-5


2 x 2 matrices determinants and inverses6

d–b

–ca

Use the definition of inverse.

1

ad – bc

A–1 =

11–25

–49

1

(9)(11) – (25)(4)

Substitute.

=

–11 25

4 –9

Simplify.

=

–11 25

4 –9

3

–7

Substitute.

X =

2 X 2 Matrices, Determinants, and Inverses

Lesson 4-5

Additional Examples

9 25

4 11

3

–7

Solve X = for the matrix X.

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

Use the equation X = A–1B.

4-5


2 x 2 matrices determinants and inverses7

Multiply and

simplify.

(–11)(3) + (25)(–7)

(4)(3) + (–9)(–7)

–208

75

= =

Check:

Use the original

equation.

9 25

4 11

3

–7

X =

9 25

4 11

–208

75

3

–7

Substitute.

9(–208) + 25(75)

4(–208) + 11(75)

3

–7

Multiply and simplify.

3

–7

3

–7

=

2 X 2 Matrices, Determinants, and Inverses

Lesson 4-5

Additional Examples

(continued)

Quick Check

4-5


2 x 2 matrices determinants and inverses8

From

AB

To A

To B

0.84 0.07

0.16 0.93

Write the percents as decimals.

2 X 2 Matrices, Determinants, and Inverses

Lesson 4-5

Additional Examples

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.

4-5


2 x 2 matrices determinants and inverses9

Use A

Use B

25000

20000

Write the information in a matrix.

0.84 0.07

0.16 0.93

25000

20000

22,400

22,600

=

2 X 2 Matrices, Determinants, and Inverses

Lesson 4-5

Additional Examples

(continued)

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

22,600 households will use carrier B.

4-5


2 x 2 matrices determinants and inverses10

0.84 0.07

0.16 0.93

First find the determinant of .

0.84 0.07

0.16 0.93

= 0.77

1

0.77

25,000

20,000

28,377

16,623

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

0.93–0.07

–.016 0.84

2 X 2 Matrices, Determinants, and Inverses

Lesson 4-5

Additional Examples

(continued)

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

About 28,400 households used carrier A.

4-5


2 x 2 matrices determinants and inverses11

5 2

–2 –1

1 0

0 –1

1 2

2 5

no; Answers may vary. Sample is , which is not the 2  2 identity matrix.

–3.5 –2

–1.5 –1

–50.2

28.6

2 X 2 Matrices, Determinants, and Inverses

Lesson 4-5

Lesson Quiz

5 2

–2 1

1 2

2 5

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.

–12 5

–16 4

32

–2 4

3 –7

20 35

1 2

–3

7

4-5


3 x 3 matrices determinants and inverses

3 X 3 Matrices, Determinants, and Inverses

Lesson 4-6

Check Skills You’ll Need

(For help, go to Skills Handbook page 845.)

Find the product of the circled elements in each matrix.

1.2.3.

4.5.6.

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

Check Skills You’ll Need

4-6


3 x 3 matrices determinants and inverses1

3 X 3 Matrices, Determinants, and Inverses

Lesson 4-6

Check Skills You’ll Need

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

4-6


3 x 3 matrices determinants and inverses2

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.

3 X 3 Matrices, Determinants, and Inverses

Lesson 4-6

Additional Examples

8 –4 3

–2 9 5

1 6 0

Evaluate the determinant of X = .

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

= –56 – 267 = –323.Simplify.

The determinant of X is –323.

Quick Check

4-6


3 x 3 matrices determinants and inverses3

3 X 3 Matrices, Determinants, and Inverses

Lesson 4-6

Additional Examples

Enter matrix T into your graphing calculator. Use the matrix submenus to evaluate the determinant of the matrix.

4 2 3

–2 –1 5

1 3 6

T =

The determinant of the matrix is –65.

Quick Check

4-6


3 x 3 matrices determinants and inverses4

=

/

0.5 0 0

0 0 0.5

0 1 1

2 0 0

0 2 1

0 2 0

1 0 0

0 1 0

0 4 1

=

Since CDI, C and D are not multiplicative inverses.

3 X 3 Matrices, Determinants, and Inverses

Lesson 4-6

Additional Examples

Determine whether the matrices are multiplicative inverses.

0.5 0 0

0 0 0.5

0 1 1

2 0 0

0 2 1

0 2 0

a.C = , D =

4-6


3 x 3 matrices determinants and inverses5

0 0 1

0 1 0

1 0 –1

1 0 1

0 1 0

1 0 0

b.A = , B =

0 0 1

0 1 0

1 0 –1

1 0 1

0 1 0

1 0 0

1 0 0

0 1 0

0 0 1

=

3 X 3 Matrices, Determinants, and Inverses

Lesson 4-6

Additional Examples

(continued)

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

Quick Check

4-6


3 x 3 matrices determinants and inverses6

2 0 1

0 1 4

1 0 0

Let A = .

Find A–1.

0 0 1

–4 1 8

1 0 –2

–1

8

–2

X =

Use the equation X = A–1C.

Multiply.

–2

–4

3

X =

3 X 3 Matrices, Determinants, and Inverses

Lesson 4-6

Additional Examples

2 0 1

0 1 4

1 0 0

–1

8

–2

Solve the equation.

X =

Quick Check

4-6


3 x 3 matrices determinants and inverses7

0 2 1

2 –2.5 –0.75

2 –1.5 –1.25

K–1 =Use a graphing calculator.

3 X 3 Matrices, Determinants, and Inverses

Lesson 4-6

Additional Examples

Use the alphabet table and

the encoding matrix.

0.5 0.25 0.25

0.25 –0.5 0.5

0.5 1 –1

matrix K = .

a. Find the decoding matrix K–1.

4-6


3 x 3 matrices determinants and inverses8

Use the

decoding

matrix

from

part (a).

Multiply.

0 2 1

2 –2.5 –0.75

2 –1.5 –1.25

11.25 16.75 24.5

5.75 17 5.5

1.5 –12 15

13 22 26

7 024

12 23 22

=

3 X 3 Matrices, Determinants, and Inverses

Lesson 4-6

Additional Examples

(continued)

11.25 16.75 24.5

5.75 17 5.5

1.5 –12 15

b. Decode .Zero indicates a space holder.

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

Quick Check

4-6


3 x 3 matrices determinants and inverses9

4

–5

–2

3 X 3 Matrices, Determinants, and Inverses

Lesson 4-6

Lesson Quiz

1.Use pencil and paper to evaluate the determinant of

2.Determine whether the matrices are multiplicative inverses.

3.Solve the equation for M.

–2 –4 2

3 1 0

5 –6 –2

.

–66

1 1 –1

–1 0 1

0 –1 1

1 0 1

1 1 0

1 1 1

yes

;

–1 –1 1

1 2 –1

0 –1 1

–1

–4

3

M =

4-6


Inverse matrices and systems

5x + y = 14

4x + 3y = 20

1.

2.

3.

x – y – z = –9

3x + y + 2z = 12

x = y – 2z

–x + 2y + z = 0

y = –2x + 3

z = 3x

Inverse Matrices and Systems

Lesson 4-7

Check Skills You’ll Need

(For help, go to Lesson 3-6.)

Solve each system.

Check Skills You’ll Need

4-7


Inverse matrices and systems1

Inverse Matrices and Systems

Lesson 4-7

Check Skills You’ll Need

1.2.

Solutions

x – y – z = –9

3x + y + 2z = 12

x = y – 2z

5x + y = 14

4x + 3y = 20

Solve the first equation for y:

y = –5x + 14

Substitute this into the second

equation:

4x + 3(–5x + 14)= 20

4x – 15x + 42= 20

–11x + 42= 20

–11x= –22

x= 2

Use the first equation with x = 2:

5(2) + y= 14

10 + y= 14

y= 4

The solution is (2, 4).

Use the first equation with

x = y – 2z (third equation):

(y – 2z) – y – z = –9

–3z = –9

z = 3

Use 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 = 6

Use the third equation with y = 6

And z = 3:

x = 6 – 2(3) = 6 – 6 = 0

The solution is (0, 6, 3).

4-7


Inverse matrices and systems2

Inverse Matrices and Systems

Lesson 4-7

Check Skills You’ll Need

Solutions (continued)

3.

–x + 2y + z = 0

y = –2x + 3

z = 3x

Use the first equation with

y = –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

Use the second equation with x = 3:

y = –2(3) + 3 = –6 + 3 = –3

Use the third equation with x = 3:

z = 3(3) = 9

The solution is (3, –3, 9).

4-7


Inverse matrices and systems3

–3 –4 5

–2 7 0

–5 1 –1

x

y

z

11

–6

20

Matrix equation: =

Coefficient matrix

Variable matrix

Constant matrix

–3 –4 5

–2 7 0

–5 1 –1

x

y

z

11

–6

20

Inverse Matrices and Systems

Lesson 4-7

Additional Examples

–3x – 4y + 5z = 11

–2x + 7y = –6

–5x + y – z = 20

Write the system

as a matrix equation.

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

Quick Check

4-7


Inverse matrices and systems4

2 3

1 –1

x

y

–1

12

= Write the system as a matrix equation.

1

5

3

5

A–1 = Find A–1.

1

5

2

5

1

5

3

5

x

y

–1

12

7

–5

= A–1B = = Solve for the

variable matrix.

1

5

2

5

Inverse Matrices and Systems

Lesson 4-7

Additional Examples

Solve the system.

2x + 3y = –1

x – y = 12

4-7


Inverse matrices and systems5

Check: 2x + 3y = –1 x – y = 12Use the

original

equations.

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

14 – 15 = –1 7 + 5 = 12Simplify.

Inverse Matrices and Systems

Lesson 4-7

Additional Examples

(continued)

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

Quick Check

4-7


Inverse matrices and systems6

7 3 2

–2 1 –8

1 –4 10

x

y

z

13

26

–13

=

Inverse Matrices and Systems

Lesson 4-7

Additional Examples

7x + 3y + 2z = 13

–2x + y – 8z = 26

x – 4y +10z = –13

Solve the system .

Step 1:Write the system as

a matrix equation.

Step 2:Store the coefficient

matrix as matrix A

and the constant

matrix as matrix B.

The solution is (9, –12, –7).

Quick Check

4-7


Inverse matrices and systems7

Relate: 3 sheets and 5 towels cost $137.50.

4 sheets and 2 towels cost $118.00.

3 5

4 2

x

y

137.50

118.00

Write: =

Inverse Matrices and Systems

Lesson 4-7

Additional Examples

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.

Define:Let x = the price of one sheet.

Let y = the price of one towel.

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

Quick Check

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

4-7


Inverse matrices and systems8

4 –2

–6 3

4 –2

–6 3

A = ; det A = = 4(3) – (–2)(–6) = 0

Inverse Matrices and Systems

Lesson 4-7

Additional Examples

Write the coefficient matrix for each system. Use it to determine whether the system has exactly a unique solution.

4x – 2y = 7

–6x + 3y = 5

a.

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

4-7


Inverse matrices and systems9

=

/

12 8

3 –7

12 8

3 –7

A = ; det A = = 12(–7) – 8(–3) = –60

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

Inverse Matrices and Systems

Lesson 4-7

Additional Examples

(continued)

12x + 8y = –3

3x – 7y = 50

b.

Quick Check

4-7


Inverse matrices and systems10

3 2

2 –3

x

y

–6

61

= ; (8, –15)

2 4 5

7 9 4

–3 2 8

x

y

z

–3

19

0

= ; (–10, 13, –7)

Inverse Matrices and Systems

Lesson 4-7

Lesson Quiz

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

1.

2.

Determine whether each system has a unique solution.

3.

4.

3x + 2y = –6

2x – 3y = 61

2x + 4y + 5z = –3

7x + 9y + 4z = 19

–3x + 2y + 8z = 0

7x – 2y = 15

–28x + 8y = 7

no

20x + 5y = 33

–32x + 8y = 47

yes

4-7


Augmented matrices and systems

Augmented Matrices and Systems

Lesson 4-8

Check Skills You’ll Need

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

Evaluate the determinant of each matrix.

1.2.3.

–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

4.5.6.

Check Skills You’ll Need

4-8


Augmented matrices and systems1

Augmented Matrices and Systems

Lesson 4-8

Check Skills You’ll Need

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

–1 2

0 3

0 1

–1 3

2 1

–1 5

0 1 –3

4 5 –1

–1 0 1

4-8


Augmented matrices and systems2

Augmented Matrices and Systems

Lesson 4-8

Check Skills You’ll Need

Solutions (continued)

3 4 5

–1 2 0

0 –1 1

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

0 2 –1

3 4 0

–2 –1 5

4-8


Augmented matrices and systems3

7 –4

3 6

15 –4

8 6

7 15

3 8

D = = 54

Dx = = 122

Dy = = 11

61

27

11

54

Dx

D

Dy

D

x = =

y = =

61

27

11

54

The solution of the system is , .

Augmented Matrices and Systems

Lesson 4-8

Additional Examples

7x – 4y = 15

3x + 6y = 8

Use Cramer’s rule to solve the system .

Evaluate three determinants. Then find x and y.

Quick Check

4-8


Augmented matrices and systems4

–2 8 2

–6 0 2

–7 –5 1

D = = –24Evaluate the determinant.

–2 –3 2

–6 1 2

–7 2 1

Dy = = 20Replace the y-coefficients with the

constants and evaluate again.

Dy

D

20

24

5

6

y = = – = –Find y.

5

6

The y-coordinate of the solution is – .

Augmented Matrices and Systems

Lesson 4-8

Additional Examples

Find the y-coordinate of the solution of the

system .

–2x + 8y + 2z = –3

–6x + 2z = 1

–7x – 5y + z = 2

Quick Check

4-8


Augmented matrices and systems5

–7x + 4y = –3

x + 8y = 9

System of equations

x-coefficients

y-coefficients

constants

–74 –3

189

Augmented matrix

Draw a vertical bar to separate the coefficients from constants.

Augmented Matrices and Systems

Lesson 4-8

Additional Examples

Write an augmented matrix to represent the

system

–7x + 4y = –3

x + 8y = 9

Quick Check

4-8


Augmented matrices and systems6

9x – 7y = –1

2x + 5y = –6

9–7–1

25–6

System of equations

Augmented matrix

x-coefficients

y-coefficients

constants

Augmented Matrices and Systems

Lesson 4-8

Additional Examples

Write a system of equations for the augmented

matrix .

9 –7 –1

2 5 –6

Quick Check

4-8


Augmented matrices and systems7

1 –3 –17

4 2 2

Write an augmented matrix.

1 –3 –17

0 14 70

–4(1 –3 –17)

4 2 2

0 14 70

Multiply Row 1 by –4 and add it to Row 2.

Write the new augmented matrix.

1

14

1

14

Multiply Row 2 by .

Write the new augmented matrix.

1 –3 –17

0 1 5

(0 14 70)

0 1 5

Augmented Matrices and Systems

Lesson 4-8

Additional Examples

Use an augmented matrix to solve the system

x – 3y = –17

4x + 2y = 2

4-8


Augmented matrices and systems8

1 –3 –17

3(0 1 5)

1 0 –2

Multiply Row 2 by 3 and add it to Row 1.

Write the final augmented matrix.

1 0 –2

0 1 5

Check:x – 3y = –17 4x + 2y = 2Use the original equations.

(–2) – 3(5) –17 4(–2) + 2(5) 2 Substitute.

–2 – 15 –17 –8 + 10 2Multiply.

–17 = –17 2 = 2

Augmented Matrices and Systems

Lesson 4-8

Additional Examples

(continued)

1

14

1 –3 –17

0 1 5

(0 14 70)

0 1 5

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

Quick Check

4-8


Augmented matrices and systems9

Step 1:Enter the

augmented matrix

as matrix A.

Step 2:Use the rref feature

of your graphing

calculator.

Augmented Matrices and Systems

Lesson 4-8

Additional Examples

Use the rref feature on a graphing calculator to solve the system

4x + 3y + z = –1

–2x – 2y + 7z = –10.

3x + y + 5z = 2

The solution is (7, –9, –2).

4-8


Augmented matrices and systems10

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

4(7) + 3(–9) + (–2) –1Substitute.

28 – 27 – 2 –1Multiply.

–1 = –1Simplify.

Augmented Matrices and Systems

Lesson 4-8

Additional Examples

(continued)

Quick Check

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Augmented matrices and systems11

–7 3 9

5 0 3

4 –6 1

–7 3 12

5 0 8

4 –6 –2

D = , Dz =

Augmented Matrices and Systems

Lesson 4-8

Lesson Quiz

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.

3x + 2y = –2

5x + 4y = 8

(–12, 17)

–7x + 3y + 9z = 12

5x + 3z = 8

4x – 6y + z = –2

5x + y = 1

3x – 2y = 24

(2, –9)

4x + y – z = 7

–2x + 2y + 5z = 3

7x – 3y – 9z = –4

(–1, 8, –3)

4-8


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