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# Organizing Data Into Matrices - PowerPoint PPT Presentation

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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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

Imports 4.678 5.185 6.964

Exports 1.295 1.331 1.402

Source: U.S. Department of Commerce.

6.964 million units

Check Skills You’ll Need

4-1

Lesson 4-1

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

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

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

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

(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

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

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

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

Refer to the table.

U.S. Passenger Car Imports

And Exports (millions)

199619982000

Imports 4.678 5.185 6.964

Exports 1.295 1.331 1.402

a. Write a matrix N to represent the information.

Use 2  3 matrix.

Source: U.S. Department of Commerce.

4-1

Lesson 4-1

(continued)

U.S. Passenger Car Imports

And Exports (millions)

199619982000

Imports 4.678 5.185 6.964 Exports 1.295 1.331 1.402

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

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

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

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

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

Lesson 4-2

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

201 375

54 439

58 386

198 + 54350 + 439

201 + 58375 + 386

+

=

C A

Theater 1 252 789

Theater 2 259 761

=

Lesson 4-2

(continued)

b. Find the combined sales for the two showings.

Quick Check

4-2

–4 + 0 6 + 0

3 + (–3) –8 + 8

–5 + 5 1 + (–1)

=

=

9 0

–4 6

0 0

0 0

=

=

Lesson 4-2

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

4 8

–2 0

–7 9

–4 –5

A– B = A + (–B) = +

4 + (–7) 8 + 9

–2 + (–4) 0 + (–5)

=

–3 17

–6 –5

Simplify.

=

Lesson 4-2

4 8

–2 0

7 –9

4 5

A = and B = . Find A – B.

4-2

4 8 matrix.

–2 0

7 –9

4 5

A– B = –

4 – 7 8 – (–9)

–2 – 4 0 – 5

Subtract corresponding elements

=

–3 17

–6 –5

Simplify.

=

Lesson 4-2

(continued)

Method 2: Use subtraction.

Quick Check

4-2

2 5 matrix.

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 – + = +

to each side of the equation.

12 2

–1 8

14 –9

Simplify.

X =

Lesson 4-2

Quick Check

2 5

3 –1

8 0

10 –3

–4 9

6 –9

Solve X – = for the matrix X.

4-2

17 5 matrix.

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.

=

/

Lesson 4-2

Determine whether the matrices in each pair are equal.

4-2

8 matrix.

0.2

27

9

16

4

3–4

40–3

P = ; Q =

8

0.2

12

4

Lesson 4-2

(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

2 matrix.m – 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

Lesson 4-2

Solve the equation

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

4-2

3 matrix.m = 12 Add the equations.

m = 4 Solve for m.

Lesson 4-2

(continued)

Solve for m and n.

2m – n = 15

m + n = –3

4 + n = –3 Substitute 4 for m.

n = –7 Solve for n.

The solutions are m = 4 and n = –7.

Quick Check

4-2

0 3 matrix.

–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

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.

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

–4 matrix.

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

Store matrix.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

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

M matrix.1 M2 M3

41580 43200 48168

42120 40824 47196

= A

B

Matrix Multiplication

Lesson 4-3

(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

2 –3 matrix.

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

Find the sum of –3M + 7N for

M = and N = .

2 –3

0 6

–5 –1

2 9

Quick Check

4-3

6 9 matrix.

–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

6 9

–12 15

27 –18

30 6

Solve the equation –3Y + 2 = .

4-3

Check: matrix.

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

(continued)

Quick Check

4-3

–2 5 matrix.

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

–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

–2 5 matrix.

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

(continued)

Quick Check

4-3

109 matrix.

76

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

Matrix Multiplication

Lesson 4-3

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

QP matrix.

(3 4) (3 3)

not equal

Matrix Multiplication

Lesson 4-3

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

–6 4 –2 matrix.

–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

1 matrix.

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 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

Check Skills You’ll Need

4-4

1 matrix.

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

1 matrix.

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

the Triangle Matrix the image

Subtract 3 from each

x-coordinate.

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

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

A B C D E matrix.A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

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

1 0 matrix.

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

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

a. matrix.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

(continued)

d.y = –x

Quick Check

4-4

0 –1 matrix.

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

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

–1 2 1 matrix.

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 matrix.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 matrix.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

3 –1 matrix.

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

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

7 matrix.8

–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

Evaluate each determinant.

7 8

–5 –9

a. det

b. det

c. det

4 –3

5 6

a –b

ba

Quick Check

4-5

= matrix.

/

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

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

20 matrix.–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

(continued)

Y–1 =

Quick Check

4-5

d matrix.–b

–ca

Use the definition of inverse.

1

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

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

Multiply and matrix.

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

(continued)

Quick Check

4-5

From matrix.

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

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

Use matrix.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

(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

0.84 0.07 matrix.

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

(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

5 2 matrix.

–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 matrix.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 matrix.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

8 –4 3 matrix.

–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

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 matrix.X 3 Matrices, Determinants, and Inverses

Lesson 4-6

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

= matrix.

/

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

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

0 0 1 matrix.

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

(continued)

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

Quick Check

4-6

2 0 1 matrix.

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

2 0 1

0 1 4

1 0 0

–1

8

–2

Solve the equation.

X =

Quick Check

4-6

0 2 1 matrix.

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

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

Use the matrix.

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 0 24

12 23 22

=

3 X 3 Matrices, Determinants, and Inverses

Lesson 4-6

(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

4 matrix.

–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

5 matrix.x + 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 Systems matrix.

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 Systems matrix.

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

–3 –4 5 matrix.

–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

–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

2 3 matrix.

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

Solve the system.

2x + 3y = –1

x – y = 12

4-7

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

original

equations.

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

14 – 15 = –1 7 + 5 = 12 Simplify.

Inverse Matrices and Systems

Lesson 4-7

(continued)

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

Quick Check

4-7

7 3 2 matrix.

–2 1 –8

1 –4 10

x

y

z

13

26

–13

=

Inverse Matrices and Systems

Lesson 4-7

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

Relate: matrix. 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

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

4 –2 matrix.

–6 3

4 –2

–6 3

A = ; det A = = 4(3) – (–2)(–6) = 0

Inverse Matrices and Systems

Lesson 4-7

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

= matrix.

/

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

(continued)

12x + 8y = –3

3x – 7y = 50

b.

Quick Check

4-7

3 2 matrix.

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

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

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

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

7 –4 matrix.

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

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

–2 8 2 matrix.

–6 0 2

–7 –5 1

D = = –24 Evaluate the determinant.

–2 –3 2

–6 1 2

–7 2 1

Dy = = 20 Replace 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

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

–7 matrix.x + 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

Write an augmented matrix to represent the

system

–7x + 4y = –3

x + 8y = 9

Quick Check

4-8

9 matrix.x – 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

Write a system of equations for the augmented

matrix .

9 –7 –1

2 5 –6

Quick Check

4-8

1 –3 –17 matrix.

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

Use an augmented matrix to solve the system

x – 3y = –17

4x + 2y = 2

4-8

1 –3 –17 matrix.

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 = 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

Augmented Matrices and Systems

Lesson 4-8

(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

Step 1: matrix. Enter the

augmented matrix

as matrix A.

Step 2: Use the rref feature

calculator.

Augmented Matrices and Systems

Lesson 4-8

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

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

4(7) + 3(–9) + (–2) –1 Substitute.

28 – 27 – 2 –1 Multiply.

–1 = –1 Simplify.

Augmented Matrices and Systems

Lesson 4-8

(continued)

Quick Check

4-8

–7 3 9 matrix.

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