organizing data into matrices
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
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
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
Organizing Data Into MatricesOrganizing Data Into Matrices
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
Organizing Data Into MatricesOrganizing Data Into Matrices
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
Organizing Data Into MatricesOrganizing Data Into Matrices
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
Organizing Data Into MatricesOrganizing Data Into Matrices
= 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
Organizing Data Into MatricesOrganizing Data Into Matrices
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
Organizing Data Into MatricesOrganizing Data Into Matrices
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
Organizing Data Into MatricesOrganizing Data Into Matrices
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
(For help, go to Skills Handbook page 845.)
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
Check Skills You’ll Need
4-2
FeatureLesson
Algebra 2Algebra 2
LessonMain
1. =
2. =
3. =
4. =
5. =
6. =
Solutions
Lesson 4-2
Adding and Subtracting MatricesAdding and Subtracting Matrices
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
Check Skills You’ll Need
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
Adding and Subtracting MatricesAdding and Subtracting Matrices
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
Adding and Subtracting MatricesAdding and Subtracting Matrices
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
Adding and Subtracting MatricesAdding and Subtracting Matrices
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
Adding and Subtracting MatricesAdding and Subtracting Matrices
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
Adding and Subtracting MatricesAdding and Subtracting Matrices
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
Adding and Subtracting MatricesAdding and Subtracting Matrices
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
Adding and Subtracting MatricesAdding and Subtracting Matrices
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
Adding and Subtracting MatricesAdding and Subtracting Matrices
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
Adding and Subtracting MatricesAdding and Subtracting Matrices
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
Adding and Subtracting MatricesAdding and Subtracting Matrices
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
Adding and Subtracting MatricesAdding and Subtracting Matrices
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
Check Skills You’ll Need
Check Skills You’ll Need
4-3
FeatureLesson
Algebra 2Algebra 2
LessonMain
Solutions
Lesson 4-3
Matrix MultiplicationMatrix Multiplication
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
Check Skills You’ll Need
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
Matrix MultiplicationMatrix Multiplication
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
Matrix MultiplicationMatrix Multiplication
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
Matrix MultiplicationMatrix Multiplication
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
Matrix MultiplicationMatrix Multiplication
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
Matrix MultiplicationMatrix Multiplication
–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
Matrix MultiplicationMatrix Multiplication
–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
Matrix MultiplicationMatrix Multiplication
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
Matrix MultiplicationMatrix Multiplication
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
Matrix MultiplicationMatrix Multiplication
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
Matrix MultiplicationMatrix Multiplication
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
(For help, go to Lesson 2-6.)
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
Check Skills You’ll Need
Check Skills You’ll Need
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
Geometric Transformations with MatricesGeometric Transformations with Matrices
3. g(x) = |x|
right 3 units:
g(x) = |x – 3|
4. y = x;down 2 units:y = x – 2
12
12
Check Skills You’ll Need
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
Geometric Transformations with MatricesGeometric Transformations with Matrices
13
13
Check Skills You’ll Need
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
Geometric Transformations with MatricesGeometric Transformations with Matrices
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
Geometric Transformations with MatricesGeometric Transformations with Matrices
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
Geometric Transformations with MatricesGeometric Transformations with Matrices
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
Geometric Transformations with MatricesGeometric Transformations with Matrices
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
Geometric Transformations with MatricesGeometric Transformations with Matrices
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
Geometric Transformations with MatricesGeometric Transformations with Matrices
–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
(For help, go to Skills Handbook page 845.)
2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses
Simplify each group of expressions.
Check Skills You’ll Need
Check Skills You’ll Need
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
2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses
Check Skills You’ll Need
4-5
FeatureLesson
Algebra 2Algebra 2
LessonMain
Show that matrices A and B are multiplicative inverses.
A = B =
Lesson 4-5
2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses
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
2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses
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
2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses
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
2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses
= – 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
2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses
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
2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses
= =(–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
2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses
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
2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses
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
2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses
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
2 X 2 Matrices, Determinants, and Inverses2 X 2 Matrices, Determinants, and Inverses
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
3 X 3 Matrices, Determinants, and Inverses3 X 3 Matrices, Determinants, and Inverses
(For help, go to Skills Handbook page 845.)
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
Check Skills You’ll Need
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
3 X 3 Matrices, Determinants, and Inverses3 X 3 Matrices, Determinants, and Inverses
Check Skills You’ll Need
4-6
FeatureLesson
Algebra 2Algebra 2
LessonMain
Evaluate the determinant of X = .
Lesson 4-6
3 X 3 Matrices, Determinants, and Inverses3 X 3 Matrices, Determinants, and Inverses
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
3 X 3 Matrices, Determinants, and Inverses3 X 3 Matrices, Determinants, and Inverses
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
3 X 3 Matrices, Determinants, and Inverses3 X 3 Matrices, Determinants, and Inverses
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
3 X 3 Matrices, Determinants, and Inverses3 X 3 Matrices, Determinants, and Inverses
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
3 X 3 Matrices, Determinants, and Inverses3 X 3 Matrices, Determinants, and Inverses
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
3 X 3 Matrices, Determinants, and Inverses3 X 3 Matrices, Determinants, and Inverses
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
3 X 3 Matrices, Determinants, and Inverses3 X 3 Matrices, Determinants, and Inverses
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
3 X 3 Matrices, Determinants, and Inverses3 X 3 Matrices, Determinants, and Inverses
–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
(For help, go to Lesson 3-6.)
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.
Check Skills You’ll Need
Check Skills You’ll Need
4-7
FeatureLesson
Algebra 2Algebra 2
LessonMain
1. 2.
Solutions
Lesson 4-7
Inverse Matrices and SystemsInverse Matrices and Systems
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).
Check Skills You’ll Need
4-7
FeatureLesson
Algebra 2Algebra 2
LessonMain
Solutions (continued)
Lesson 4-7
Inverse Matrices and SystemsInverse Matrices and Systems
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.
Check Skills You’ll Need
4-7
FeatureLesson
Algebra 2Algebra 2
LessonMain
Write the system
as a matrix equation.
Lesson 4-7
Inverse Matrices and SystemsInverse Matrices and Systems
–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
Inverse Matrices and SystemsInverse Matrices and Systems
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
Inverse Matrices and SystemsInverse Matrices and Systems
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
Inverse Matrices and SystemsInverse Matrices and Systems
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
Inverse Matrices and SystemsInverse Matrices and Systems
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
Inverse Matrices and SystemsInverse Matrices and Systems
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
Inverse Matrices and SystemsInverse Matrices and Systems
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
Inverse Matrices and SystemsInverse Matrices and Systems
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.
Check Skills You’ll Need
Check Skills You’ll Need
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
Augmented Matrices and SystemsAugmented Matrices and Systems
–1 2 0 3
0 1–1 3
2 1–1 5
0 1 –3 4 5 –1–1 0 1
Check Skills You’ll Need
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
Solutions (continued)
Lesson 4-8
Augmented Matrices and SystemsAugmented Matrices and Systems
3 4 5–1 2 0 0 –1 1
0 2 –1 3 4 0–2 –1 5
Check Skills You’ll Need
4-8
FeatureLesson
Algebra 2Algebra 2
LessonMain
Use Cramer’s rule to solve the system .
Lesson 4-8
Augmented Matrices and SystemsAugmented Matrices and Systems
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
Augmented Matrices and SystemsAugmented Matrices and Systems
–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
Augmented Matrices and SystemsAugmented Matrices and Systems
–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
Augmented Matrices and SystemsAugmented Matrices and Systems
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
Augmented Matrices and SystemsAugmented Matrices and Systems
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
Augmented Matrices and SystemsAugmented Matrices and Systems
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
Augmented Matrices and SystemsAugmented Matrices and Systems
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
Augmented Matrices and SystemsAugmented Matrices and Systems
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
Augmented Matrices and SystemsAugmented Matrices and Systems
(–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