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Organizing Data Into Matrices
Lesson 4-1
Algebra 2
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?
U.S. Passenger Vehicles and
Light Trucks Imports
And Exports (millions)
4.678 million units
How many were imported in 2000?
6.964 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.
Check Skills You’ll Need
Lesson
Main
Lesson
4-1
Feature
Organizing Data Into Matrices
Lesson 4-1
Algebra 2
Additional Examples
Write the dimensions of each matrix.
a.
7 –4
12 9
The matrix has 2 rows and 2 columns and is
b.
0 6 15
The matrix has 1 row and 3 columns and is
therefore a 2  2 matrix.
therefore a 1  3 matrix.
Quick Check
Lesson
Main
Lesson
4-1
Feature
Organizing Data Into Matrices
Lesson 4-1
Algebra 2
Additional Examples
Identify each matrix element.
K =
3 –1 –8 5
1 8 4 9
8 –4 7 –5
a. k12
b. k32
a. K =
3 –1 –8 5
1 8 4 9
8 –4 7 –5
c. k23
b. K =
d. k34
3 –1 –8 5
1 8 4 9
8 –4 7 –5
k12 is the element in the first row
and second column.
k32 is the element in the third row
and second column.
Element k12 is –1.
Element k32 is –4.
Lesson
Main
Lesson
4-1
Feature
Organizing Data Into Matrices
Lesson 4-1
Algebra 2
Additional Examples
(continued)
K =
3 –1 –8 5
1 8 4 9
8 –4 7 –5
a. k12
b. k32
c. K =
3 –1 –8 5
1 8 4 9
8 –4 7 –5
c. k23
d. K =
d. k34
3 –1 –8 5
1 8 4 9
8 –4 7 –5
k23 is the element in the second
row and third column.
k34 is the element in the third row
and the fourth column.
Element k23 is 4.
Element k34 is –5.
Lesson
Main
Lesson
4-1
Quick Check
Feature
Organizing Data Into Matrices
Lesson 4-1
Algebra 2
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
Ed
X = Loss
X
X
Jo
Lew
X
X
X X
X
Let each row represent the number of wins and losses and each
column represent a student.
Ed Jo Lew
Wins
Losses
Lesson
Main
5
2
6
1
3
4
Quick Check
Lesson
4-1
Feature
Organizing Data Into Matrices
Lesson 4-1
Algebra 2
Additional Examples
Refer to the table.
U.S. Passenger Car Imports
And Exports (millions)
a. Write a matrix N to represent
the information.
Use 2  3 matrix.
1996
1998
2000
Imports
4.678
5.185
6.964
Exports
1.295
1.331
1.402
Source: U.S. Department of Commerce.
Each column represents a
different year.
N = Import
Exports
Lesson
Main
1996
1998
2000
4.678
1.295
5.185
1.331
6.964
1.402
Lesson
4-1
Each row represents
imports and exports.
Feature
Organizing Data Into Matrices
Lesson 4-1
Algebra 2
Additional Examples
U.S. Passenger Car Imports
And Exports (millions)
(continued)
b. Which element represents
exports for 2000?
1996
1998
2000
Imports
4.678
5.185
6.964
Exports
1.295
1.331
1.402
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
Lesson
Main
Lesson
4-1
Feature
Organizing Data Into Matrices
Lesson 4-1
Algebra 2
Lesson Quiz
1. Write the dimensions of the matrix. M =
8
9
–1
4 0
3 –5
2 6
1
0
1
2. Identify the elements m24, m32, and m13 of the matrix M in
question 1.
34
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.
Deposits Withdrawals
A
$450
$370
B
C
D
$475
$364
$420
Lesson
Main
$289
$118
$400
Deposits
Withdrawals
Lesson
4-1
A
450
370
B
475
289
C
364
118
D
420
400
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 2
Check Skills You’ll Need
(For help, go to Skills Handbook page 845.)
Simplify the elements of each matrix.
1.
10 + 4
–2 + 4
0 + 4
–5 + 4
2.
5 – 2
–1 – 2
3 – 2
0 – 2
3.
–2 + 3
1 – 3
0 – 3
–5 + 3
4.
3 + 1
–2 + 0
4 + 9
5 + 7
5.
8 – 4
9 – 1
–5 – 1
6 – 9
6.
2 + 4
4 – 3
6 – 8
5 + 2
Check Skills You’ll Need
Lesson
Main
Lesson
4-2
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 2
Check Skills You’ll Need
Solutions
1.
10 + 4
–2 + 4
0 + 4
–5 + 4
=
14
2
2.
5 – 2
–1 – 2
3 – 2
0 – 2
=
3
–3
1
–2
3.
–2 + 3
1 – 3
0 – 3
–5 + 3
=
1
–2
–3
–2
4.
3 + 1
–2 + 0
4 + 9
5 + 7
=
4
–2
13
12
5.
8 – 4
9 – 1
–5 – 1
6 – 9
=
4
8
–6
–3
6.
2 + 4
4 – 3
6 – 8
5 + 2
=
6
1
–2
7
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Main
4
–1
Lesson
4-2
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 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).
Matinee
Theater C
1
198
2
201
A
350
375
Evening
C
54
58
A
439
386
a. Write two 2  2 matrices to represent matinee and evening sales.
Theater 1
Theater 2
Lesson
Main
Matinee
C
A
198 350
201 375
Theater 1
Theater 2
Lesson
4-2
Evening
C
A
54 439
58 386
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 2
Additional Examples
(continued)
b. Find the combined sales for the two showings.
198
201
=
350
375
Theater 1
Theater 2
+
54
58
439
386
C
252
259
A
789
761
=
198 + 54
201 + 58
350 + 439
375 + 386
Quick Check
Lesson
Main
Lesson
4-2
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 2
Additional Examples
Find each sum.
a.
9
–4
0
6
+
=
9+0
–4 + 0
=
9
–4
0
0
0
0
b.
0+0
6+0
0
6
–8
1
3
–5
+
–3 8
5 –1
=
3 + (–3) –8 + 8
–5 + 5 1 + (–1)
=
0
0
0
0
Quick Check
Lesson
Main
Lesson
4-2
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 2
Additional Examples
A =
4 8
–2 0
and B =
7 –9
. Find A – B.
4 5
Method 1: Use additive inverses.
+
–7 9
–4 –5
=
4 + (–7)
–2 + (–4)
8+9
0 + (–5)
=
–3 17
–6 –5
A – B = A + (–B) =
Lesson
Main
4 8
–2 0
Write the additive inverses
of the elements of the
second matrix.
Add corresponding
elements
Simplify.
Lesson
4-2
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 2
Additional Examples
(continued)
Method 2: Use subtraction.
A–B =
4 8
–2 0
=
4–7
–2 – 4
=
–3 17
–6 –5
–
7 –9
4 5
8 – (–9)
0–5
Subtract corresponding
elements
Simplify.
Quick Check
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Main
Lesson
4-2
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 2
Additional Examples
Solve X –
X –
X –
2 5
3 –1
8 0
+
2 5
3 –1
8 0
2 5
3 –1
8 0
2 5
3 –1
8 0
=
=
10 –3
–4 9
6 –9
=
10 –3
–4 9
6 –9
10 –3
–4 9
6 –9
+
Quick Check
for the matrix X.
2 5
3 –1
8 0
Add
2 5
3 –1
8 0
to each side of
the equation.
X =
Lesson
Main
12 2
–1 8
14 –9
Lesson
4-2
Simplify.
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 2
Additional Examples
Determine whether the matrices in each pair are equal.
a. M =
M =
8 + 9 5
–6
–1 ;
0
0.7
8 + 9 5
–6
–1
0
0.7
;
N =
N =
17
4 – 10
5
–2 + 1
0
– 7
9
17
4 – 10
5
–2 + 1
0
–7
9
7
Both M and N have three rows and two columns, but –
=/ 0.7.
9
M and N are not equal matrices.
Lesson
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Lesson
4-2
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 2
Additional Examples
(continued)
b. P =
P =
3
40
3
40
–4
;
–3
–4
;
–3
27
9
– 16
4
8
0.2
– 12
4
27
9
– 16
4
8
0.2
– 12
4
Q =
Q =
Both P and Q have two rows and two columns, and their corresponding
Quick Check
elements are equal. P and Q are equal matrices.
Lesson
Main
Lesson
4-2
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 2
Additional Examples
Solve the equation
2m – n
–3
8
–4m + 2n
=
15
8
m+ n
–30
2m – n
–3
8
–4m + 2n
=
15
8
m+ n
–30
2m – n = 15
–3 = m + n
for m and n.
–4m + 2n = –30
Since the two matrices are equal, their corresponding elements are equal.
Lesson
Main
Lesson
4-2
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 2
Additional Examples
(continued)
Solve for m and n.
2m – n = 15
m + n = –3
3m = 12
m = 4
4 + n = –3
n = –7
Add the equations.
Solve for m.
Substitute 4 for m.
Solve for n.
The solutions are m = 4 and n = –7.
Quick Check
Lesson
Main
Lesson
4-2
Feature
Adding and Subtracting Matrices
Lesson 4-2
Algebra 2
Lesson Quiz
Find each sum or difference.
2
8
–9
6
–7 14
1. 0 –12 +
6 –5
6 –17
2.
–3
4
–5
3. What is the additive identity for 2  4 matrices?
4. Solve the equation for x and y.
–2x
–1
18
–3x + 4y
=
5 x + y
x – 2y
–16
1
8
4
–3 –2
9 5
4 –6
–
0 3
–5 3
–9 10
0 0 0 0
0 0 0 0
x = –9, y = –7
5. Are the following matrices equal?
3
0.5
6 0.50
2
2
2
no, –
=/ –0.6
–2 ;
3
0.4
–0.6
5
3
6. Solve X –
4
1
Lesson
Main
3
5
=
2
0
7
6
for the matrix X.
Lesson
4-2
6
1
10
11
Feature
Matrix Multiplication
Lesson 4-3
Algebra 2
Check Skills You’ll Need
(For help, go to Lesson 4-2.)
Find each sum.
1.
3
2
5
8
2.
–4
7
3.
–1 3 4
0 –2 –5
+
+
–4
7
3
2
5
8
+
+
–4
7
3
2
+
+
5
8
–4
7
+
–4
7
–1 3 4
–1 3 4
+
+
0 –2 –5
0 –2 –5
–1 3 4
0 –2 –5
Check Skills You’ll Need
Lesson
Main
Lesson
4-3
Feature
Matrix Multiplication
Lesson 4-3
Algebra 2
Check Skills You’ll Need
Solutions
1.
3 5
2 8
2.
–4
7
3.
+
3 5
2 8
3 5
2 8
–4
7
+
3+3+3
2+2+2
=
–4
7
+
–4
7
–4 + (–4) + (–4) + (–4) + (–4)
7+7+7+7+7
=
–20
35
+
–4
7
+
–1 3 4
0 –2 –5
=
4(–1)
4(0)
Lesson
Main
+
+
4(3)
4(–2)
5+5+5
8+8+8
=
–4
0
Lesson
4-3
9
6
15
24
=
–1 3 4
–1 3 4
+
+
0 –2 –5
0 –2 –5
4(4)
4(–5)
=
–1 3 4
0 –2 –5
12 16
–8 –20
Feature
Matrix Multiplication
Lesson 4-3
Algebra 2
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.
Store
M1
M2
M3
A $38,500 $40,000 $44,600
B $39,000 $37,800 $43,700
1.08
=
38500
39000
40000
37800
1.08(38500)
1.08(39000)
Lesson
Main
44600
43700
1.08(40000)
1.08(37800)
1.08(44600)
1.08(43700)
Lesson
4-3
Multiply each element
by 1.08.
Feature
Matrix Multiplication
Lesson 4-3
Algebra 2
Additional Examples
(continued)
= A
B
M1
41580
42120
M2
43200
40824
M3
48168
47196
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
Lesson
Main
Lesson
4-3
Feature
Matrix Multiplication
Lesson 4-3
Algebra 2
Additional Examples
Find the sum of –3M + 7N for
M=
–3
6
2
0
–3M + 7N = –3
Lesson
Main
–5
2
and N =
2
0
–3
6
=
–6
9
0 –18
=
–41
14
+ 7
+
–1 .
9
–5
2
–35
14
2
45
–1
9
–7
63
Quick Check
Lesson
4-3
Feature
Matrix Multiplication
Lesson 4-3
Algebra 2
Additional Examples
Solve the equation –3Y + 2
–3Y + 2
–3Y +
6
–12
12
–24
=
27 –18
30
6
=
27 –18
30
6
–3Y =
27 –18
30
6
–3Y =
15 –36
54 –24
9
15
18
30
Y = –1
3
Lesson
Main
6
–12
9
15
=
27 –18
.
30
6
Scalar multiplication.
–
12
–24
18
30
12 18
–24 30
from each side.
Subtract
Simplify.
15 –36
54 –24
Lesson
4-3
=
–5 12
–18 8
Multiply each side
by – 1 and simplify.
3
Feature
Matrix Multiplication
Lesson 4-3
Algebra 2
Additional Examples
(continued)
Check:
–3Y + 2
–3
6
–12
9
15
=
27 –18
30
6
–5
–18
12
8
+2
6
–12
9
15
27 –18
30
6
Substitute.
15
54
–36
–24
+
12
–24
18
30
27 –18
30
6
Multiply.
27
30
–18
6
27 –18
30
6
Simplify.
Lesson
Main
=
Lesson
4-3
Quick Check
Feature
Matrix Multiplication
Lesson 4-3
Algebra 2
Additional Examples
Find the product of
–2 5
3 –1
and
4 –4
.
2 6
Multiply a11 and b11. Then multiply a12 and b21. Add the products.
–2
3
5
–1
4
2
–4
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
3
5
–1
4
2
–4
6
–2
3
5
–1
4
2
–4
6
Lesson
Main
=
=
2
2 38
(–2)(4) + (5)(6) = 38
(3)(4) + (–1)(2) = 10
Lesson
4-3
Feature
Matrix Multiplication
Lesson 4-3
Algebra 2
Additional Examples
(continued)
–2
3
5
–1
4
2
The product of
–4
6
=
–2 5
3 –1
2 38
(3)(–4) + (–1)(6) = –18
10
and
4 –4
2 6
is
2 38
.
10 –18
Quick Check
Lesson
Main
Lesson
4-3
Feature
Matrix Multiplication
Lesson 4-3
Algebra 2
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
Shirts Jeans
A =
$18
$22
Number of Items Sold
Shirts 109
B =
Jeans 76
Multiply each price by the number of items sold and add the products.
18
22
109
76
= (18)(109) + (22)(76)
=
3634
The store’s income for the day from the sales of shirts and jeans was $3634.
Lesson
Main
Lesson
4-3
Feature
Matrix Multiplication
Lesson 4-3
Algebra 2
Additional Examples
Use matrices P =
3 –1 2
5 9 0
0 1 8
and Q =
6 5 7 0
2 0 3 1 .
1 –1 5 2
Determine whether products PQ and QP are defined or undefined.
Find the dimensions of each product matrix.
PQ
(3

3) (3
QP

4)
equal
3  4
product
matrix
(3
4) (3

3)
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
Lesson
Main

Lesson
4-3
Feature
Matrix Multiplication
Lesson 4-3
Algebra 2
Lesson Quiz
Use matrices A, B, C, and D.
A =
2 3 –1
0 –5 4
B =
–7
2
1 0
6 –6
C =
2
9
4
D =
–3
2
1. Find 8A.
16 24 –8
0 –40 32
4. Is BD defined or undefined?
3. Find CD.
–6 4 –2
–27 18 –9
–12 8 –4
undefined
5. What are the dimensions of (BC)D?
Lesson
Main
27
–29
2. Find AC.
–1
2
Lesson
4-3

3
Feature
Geometric Transformations with Matrices
Lesson 4-4
Algebra 2
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. y = x + 2; left 4 units
2. ƒ(x) = 1 x + 2; up 5 units
3. g(x) = |x|; right 3 units
4. y = x; down 2 units
5. y = 1 |x – 3|; down 2 units
6. ƒ(x) = –2|x|; right 2 units
3
2
Check Skills You’ll Need
Lesson
Main
Lesson
4-4
Feature
Geometric Transformations with Matrices
Lesson 4-4
Algebra 2
Check Skills You’ll Need
Solutions
2. ƒ(x) = 1 x + 2;
1. y = x + 2;
left 4 units:
y=x+6
2
up 5 units:
ƒ(x) = 1 x + 7;
2
3. g(x) = |x|
right 3 units:
g(x) = |x – 3|
Lesson
Main
4. y = x;
down 2 units:
y=x–2
Lesson
4-4
Feature
Geometric Transformations with Matrices
Lesson 4-4
Algebra 2
Check Skills You’ll Need
Solutions (continued)
5. y = 1 |x – 3|;
6. ƒ(x) = –2|x|
right 2 units:
ƒ(x) = –2|x + 4|
3
down 2 units:
y = 1 |x – 3| – 2
3
Lesson
Main
Lesson
4-4
Feature
Geometric Transformations with Matrices
Lesson 4-4
Algebra 2
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.
Vertices of
Translation
Vertices of
the Triangle
Matrix
the image
A
1
–2
B C
3 2
1 3
+
–3
1
–3
1
A B C
–3
–2
=
1
–1
0
2
–1
4
Subtract 3 from each
x-coordinate.
Add 1 to each
y-coordinate.
The coordinates of the vertices of the image are
A (–2, –1), B (0, 2), C (–1, 4).
Quick Check
Lesson
Main
Lesson
4-4
Feature
Geometric Transformations with Matrices
Lesson 4-4
Algebra 2
Additional Examples
The figure in the diagram is to be reduced by a factor of 2 .
3
Find the coordinates of the vertices of the reduced figure.
Write a matrix to represent the coordinates of the
vertices.
2
3
A
B
C
D
E
0
3
2 3 –1 –2
2 –2 –3 0
=
A B
0 4
3
2 4
3
C
D E
2 –2 –4
3 3
– 4 –2 0
3
Multiply.
4
The new coordinates are A (0, 2), B ( 4 , 4 ), C (2, – ),
3
3 3
2
D (– , –2), and E (– 4 , 0).
3
3
Lesson
Main
Lesson
4-4
Quick Check
Feature
Geometric Transformations with Matrices
Lesson 4-4
Algebra 2
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
1 0
2 3 4
2 3 4
=
0 –1
–1 0 –2
1 0 2
b. y-axis
–1 0
2 3 4
–2 –3 –4
=
0 1
–1 0 –2
–1 0 –2
c. y = x
0 1
2 3 4
–1 0 –2
=
1 0
–1 0 –2
2 3 4
d. y = –x
0 –1
2 3 4
1
0
2
=
–1 0
–1 0 –2
–2 –3 –4
Lesson
Main
Lesson
4-4
Feature
Geometric Transformations with Matrices
Lesson 4-4
Algebra 2
Additional Examples
(continued)
a. x-axis
1 0
0 –1
2
–1
3 4
0 –2
=
2
1
3
0
4
2
2
–1
3 4
0 –2
=
–2 –3
–1 0
b. y-axis
–1 0
0 1
–4
–2
c. y = x
1 0
0 1
2
–1
3 4
0 –2
=
–1
2
0 –2
3 4
d. y = –x
0 –1
–1 0
2
–1
Lesson
Main
3 4
3 –2
=
1
–2
0
–3
Lesson
4-4
2
–4
Quick Check
Feature
Geometric Transformations with Matrices
Lesson 4-4
Algebra 2
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
0 –1
2 3 4
1 0 2
=
1 0
–1 0 –2
2 3 4
b. 180
–1 0
2 3 4
–2 –3 –4
=
0 –1
–1 0 –2
1 0 2
c. 270
0 1
–1 0
d. 360
1 0
0 1
2
–1
2
–1
Lesson
Main
3 4
0 –2
3 4
0 –2
=
–1 0 –2
–2 –3 –4
=
2
–1
3 4
0 –2
Lesson
4-4
Quick Check
Feature
Geometric Transformations with Matrices
Lesson 4-4
Algebra 2
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. –1 2 1
–7 –7 –7
–8 –5 –6
+
=
1 2 –2
3 3 3
4 5 1
2. Write a matrix equation that represents a dilation of triangle ABC with a
scale factor of 5.
–1 2 1
–5 10
5
5
=
1 2 –2
5 10 –10
3. Use matrix multiplication to reflect triangle ABC in the
line y = –x. Then draw the preimage and image on
the same coordinate plane.
0 –1
+
–1 0
Lesson
Main
–1
1
2 1
2 –2
=
Lesson
4-4
–1 –2 2
1 –2 –1
Feature
2
X
2 Matrices, Determinants, and Inverses
Lesson 4-5
Algebra 2
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
Lesson
Main
Lesson
4-5
Feature
2
X
2 Matrices, Determinants, and Inverses
Lesson 4-5
Algebra 2
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
Lesson
Main
Lesson
4-5
Feature
2
X
2 Matrices, Determinants, and Inverses
Lesson 4-5
Algebra 2
Additional Examples
Show that matrices A and B are multiplicative inverses.
–1
1
3
7
A =
AB =
3
7
B =
–1
1
0.1
–0.7
0.1
–0.7
0.1
0.3
0.1
0.3
=
(3)(0.1) + (–1)(–0.7)
(7)(0.1) + (1)(–0.7)
=
1
0
(3)(0.1) + (–1)(0.3)
(7)(0.1) + (1)(0.3)
0
1
AB = I, so B is the multiplicative inverse of A.
Lesson
Main
Lesson
4-5
Quick Check
Feature
2
X
2 Matrices, Determinants, and Inverses
Lesson 4-5
Algebra 2
Additional Examples
Evaluate each determinant.
a. det
7
–5
8
–9
7
–5
b. det
4
5
–3
6
=
4
5
–3
6
= (4)(6) – (–3)(5) = 39
c. det
a
b
–b
a
=
a
b
–b
a
= (a)(a) – (–b)(b) = a2 + b2
=
8
–9
= (7)(–9) – (8)(–5) = –23
Quick Check
Lesson
Main
Lesson
4-5
Feature
2
X
2 Matrices, Determinants, and Inverses
Lesson 4-5
Algebra 2
Additional Examples
Determine whether each matrix has an inverse. If it does,
find it.
a.
X =
12
9
4
3
Find det X.
ad – bc = (12)(3) – (4)(9)
Simplify.
= 0
Since det X = 0, the inverse of X does not exist.
b.
Y =
6
25
5
20
Find det Y.
ad – bc = (6)(20) – (5)(25)
Simplify.
= –5
Since the determinant =/ 0, the inverse of Y exists.
Lesson
Main
Lesson
4-5
Feature
2
X
2 Matrices, Determinants, and Inverses
Lesson 4-5
Algebra 2
Additional Examples
(continued)
Y–1
1
=
det Y
20
–25
–5
6
1
det Y
20
–25
–5
6
=
= – 1
5
=
–4
5
Lesson
Main
20
–25
1
–1.2
–5
6
Change signs.
Switch positions.
Use the determinant to
write the inverse.
Substitute –5 for the
determinant.
Multiply.
Quick Check
Lesson
4-5
Feature
2
X
2 Matrices, Determinants, and Inverses
Lesson 4-5
Algebra 2
Additional Examples
Solve
9
4
25
11
X =
3
–7
for the matrix X.
The matrix equation has the form AX = B. First find A–1.
A–1
1
=
ad – bc
d
–c
–b
a
1
=
(9)(11) – (25)(4)
=
–11
4
Use the definition
of inverse.
11
–4
–25
9
25
–9
Substitute.
Simplify.
Use the equation X = A–1B.
X =
–11
4
Lesson
Main
25
–9
3
–7
Substitute.
Lesson
4-5
Feature
2
X
2 Matrices, Determinants, and Inverses
Lesson 4-5
Algebra 2
Additional Examples
(continued)
(–11)(3) + (25)(–7)
(4)(3) + (–9)(–7)
=
Check:
=
Multiply and
simplify.
3
–7
Use the original
equation.
–208
75
3
–7
Substitute.
9(–208) + 25(75)
4(–208) + 11(75)
3
–7
Multiply and simplify.
3
–7
3
–7
9
4
9
4
Lesson
Main
–208
75
25
11
25
11
X =
=
Lesson
4-5
Quick Check
Feature
2
X
2 Matrices, Determinants, and Inverses
Lesson 4-5
Algebra 2
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.
From
To A
To B
A
0.84
0.16
Lesson
Main
B
0.07
0.93
Write the percents as decimals.
Lesson
4-5
Feature
2
X
2 Matrices, Determinants, and Inverses
Lesson 4-5
Algebra 2
Additional Examples
(continued)
b. Predict the number of households that will be using distance carrier B
next year.
Use A
Use B
0.84
0.16
25000
20000
0.07
0.93
Write the information in a matrix.
25000
20000
=
22,400
22,600
22,600 households will use carrier B.
Lesson
Main
Lesson
4-5
Feature
2
X
2 Matrices, Determinants, and Inverses
Lesson 4-5
Algebra 2
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.
First find the determinant of
0.84
0.16
0.07
0.93
0.84
0.16
0.07
.
0.93
= 0.77
Multiply the inverse matrix by the information matrix in part (b).
Use a calculator and the exact inverse.
1
0.77
0.93
–.016
–0.07
0.84
25,000
20,000
28,377
16,623
About 28,400 households used carrier A.
Lesson
Main
Lesson
4-5
Quick Check
Feature
2
X
2 Matrices, Determinants, and Inverses
Lesson 4-5
Algebra 2
Lesson Quiz
1. Is
5
–2
2
1
1
2
the inverse of
no; Answers may vary. Sample
is not the 2  2 identity matrix.
2. Find the determinant of
–12
–16
3. Find the inverse of
–2 4
3 –7 .
4. Solve the equation
20
1
Lesson
Main
35
2
2
? How do you know?
5
1
2
5
–2
2
5
5
.
4
2
–1
is
1 0
, which
0 –1
32
–3.5 –2
–1.5 –1
X =
Lesson
4-5
–3
7
for X.
–50.2
28.6
Feature
3
X
3 Matrices, Determinants, and Inverses
Lesson 4-6
Algebra 2
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
–1
4
3
3
–3
0
–2
–4
4.
2
–1
4
3
3
–3
0
–2
–4
2.
2
–1
4
3
3
–3
0
–2
–4
5.
2
–1
4
3
3
–3
0
–2
–4
3.
2
–1
4
3
3
–3
0
–2
–4
6.
2
–1
4
3
3
–3
0
–2
–4
Check Skills You’ll Need
Lesson
Main
Lesson
4-6
Feature
3
X
3 Matrices, Determinants, and Inverses
Lesson 4-6
Algebra 2
Check Skills You’ll Need
Solutions
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
Lesson
Main
Lesson
4-6
Feature
3
X
3 Matrices, Determinants, and Inverses
Lesson 4-6
Algebra 2
Additional Examples
Evaluate the determinant of X =
8 –4
–2 9
1 6
3
5
0
8 –4
–2 9
1 6
3
5 .
0
= [(8)(9)(0) + (–2)(6)(3) + (1)(–4)(5)]
– [(8)(6)(5) + (–2)(–4)(0) + (1)(9)(3)]
Use the
definition.
= [0 + (–36) + (–20)] – [240 + 0 + 27]
Multiply.
= –56 – 267 = –323.
Simplify.
The determinant of X is –323.
Quick Check
Lesson
Main
Lesson
4-6
Feature
3
X
3 Matrices, Determinants, and Inverses
Lesson 4-6
Algebra 2
Additional Examples
Enter matrix T into your graphing calculator. Use the matrix
submenus to evaluate the determinant of the matrix.
T =
4 2
–2 –1
1 3
3
5
6
The determinant of the matrix is –65.
Quick Check
Lesson
Main
Lesson
4-6
Feature
3
X
3 Matrices, Determinants, and Inverses
Lesson 4-6
Algebra 2
Additional Examples
Determine whether the matrices are multiplicative inverses.
a. C =
0.5 0
0 0
0 1
0.5 0
0 0
0 1
0
0.5
1
0
0.5 , D =
1
2 0 0
0 2 1
0 2 0
2 0 0
0 2 1
0 2 0
=
1 0 0
0 1 0
0 4 1
Since CD =/ I, C and D are not multiplicative inverses.
Lesson
Main
Lesson
4-6
Feature
3
X
3 Matrices, Determinants, and Inverses
Lesson 4-6
Algebra 2
Additional Examples
(continued)
b. A =
0
0
1
0
0
1
0 1
1 0 , B =
0 –1
0 1
1 0
0 –1
1 0 1
0 1 0
1 0 0
1 0 1
0 1 0
1 0 0
=
1 0 0
0 1 0
0 0 1
Since AB = I, A and B are multiplicative inverses.
Quick Check
Lesson
Main
Lesson
4-6
Feature
3
X
3 Matrices, Determinants, and Inverses
Lesson 4-6
Algebra 2
Additional Examples
2
0
1
Solve the equation.
Let A =
2
0
1
Find A–1.
0
X = –4
1
X =
0
1
0
0 1
1 8
0 –2
1
4
0
X =
–1
8
–2
1
4 .
0
–1
8
–2
Use the equation X = A–1C.
Multiply.
–2
–4
3
Lesson
Main
0
1
0
Quick Check
Lesson
4-6
Feature
3
X
3 Matrices, Determinants, and Inverses
Lesson 4-6
Algebra 2
Additional Examples
Use the alphabet table and
the encoding matrix.
matrix K =
0.5
0.25 0.25
0.25 –0.5
0.5 .
0.5
1
–1
a. Find the decoding matrix K–1.
K–1 =
0
2
2
Lesson
Main
2
–2.5
–1.5
1
–0.75
–1.25
Use a graphing calculator.
Lesson
4-6
Feature
3
X
3 Matrices, Determinants, and Inverses
Lesson 4-6
Algebra 2
Additional Examples
(continued)
b. Decode
0
2
2
2
–2.5
–1.5
11.25
5.75
1.5
1
–0.75
–1.25
16.75
17
–12
24.5
5.5 .
15
11.25 16.75
5.75
17
1.5
–12
Zero indicates a space holder.
24.5
5.5
15
=
13
7
12
22
0
23
26
24
22
Use the
decoding
matrix
from
part (a).
Multiply.
The numbers 13 22 26 7 0 24 12 23 22 correspond to the letters
NEAT CODE.
Quick Check
Lesson
Main
Lesson
4-6
Feature
3
X
3 Matrices, Determinants, and Inverses
Lesson 4-6
Algebra 2
Lesson Quiz
1. Use pencil and paper to evaluate the determinant of
–2 –4 2
3 1 0
5 –6 –2
–66
.
2. Determine whether the matrices are multiplicative inverses.
1 1 –1
–1 0 1
0 –1 1
;
1
1
1
0
1
1
1
0
1
yes
3. Solve the equation for M.
–1 –1 1
1 2 –1
0 –1 1
Lesson
Main
M =
–1
–4
3
4
–5
–2
Lesson
4-6
Feature
Inverse Matrices and Systems
Lesson 4-7
Algebra 2
Check Skills You’ll Need
(For help, go to Lesson 3-6.)
Solve each system.
1.
5x + y = 14
4x + 3y = 20
2.
x – y – z = –9
3x + y + 2z = 12
x = y – 2z
3.
–x + 2y + z = 0
y = –2x + 3
z = 3x
Check Skills You’ll Need
Lesson
Main
Lesson
4-7
Feature
Inverse Matrices and Systems
Lesson 4-7
Algebra 2
Check Skills You’ll Need
Solutions
1.
5x + y = 14
4x + 3y = 20
2.
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).
Lesson
Main
Lesson
4-7
x – y – z = –9
3x + y + 2z = 12
x = y – 2z
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).
Feature
Inverse Matrices and Systems
Lesson 4-7
Algebra 2
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
Lesson
Main
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).
Lesson
4-7
Feature
Inverse Matrices and Systems
Lesson 4-7
Algebra 2
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.
Matrix equation:
–3 –4 5
–2 7 0
–5 1 –1
x
y
z
=
11
–6
20
Coefficient matrix
Variable matrix
Constant matrix
–3 –4 5
–2 7 0
–5 1 –1
x
y
z
11
–6
20
Lesson
Main
Lesson
4-7
Quick Check
Feature
Inverse Matrices and Systems
Lesson 4-7
Algebra 2
Additional Examples
Solve the system.
2x + 3y = –1
x – y = 12
2 3
1 –1
A–1 =
x
y
=
x
y
1
5
1
5
A–1B
Lesson
Main
=
–1
12
Write the system as a matrix equation.
3
5
–2
5
=
Find A–1.
1
5
1
5
3
5
–2
5
–1
12
=
Lesson
4-7
7
–5
Solve for the
variable matrix.
Feature
Inverse Matrices and Systems
Lesson 4-7
Algebra 2
Additional Examples
(continued)
The solution of the system is (7, –5).
2x + 3y = –1
Check:
2(7) + 3(–5)
–1
x – y = 12
(7) – (–5)
14 – 15 = –1
12
7 + 5 = 12
Use the
original
equations.
Substitute.
Simplify.
Quick Check
Lesson
Main
Lesson
4-7
Feature
Inverse Matrices and Systems
Lesson 4-7
Algebra 2
Additional Examples
Solve the system
Step 1:
7 3
–2 1
1 –4
7x + 3y + 2z = 13
–2x + y – 8z = 26 .
x – 4y +10z = –13
Write the system as
a matrix equation.
2
–8
10
x
y
z
=
Step 2:
13
26
–13
Store the coefficient
matrix as matrix A
and the constant
matrix as matrix B.
The solution is (9, –12, –7).
Quick Check
Lesson
Main
Lesson
4-7
Feature
Inverse Matrices and Systems
Lesson 4-7
Algebra 2
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.
Relate: 3 sheets and 5 towels cost $137.50.
4 sheets and 2 towels cost $118.00.
Define: Let x = the price of one sheet.
Let y = the price of one towel.
Write:
3
4
5
2
x
y
=
137.50
118.00
Use a graphing calculator. Store the coefficient matrix as matrix A and
the constant matrix as matrix B.
The price of a sheet is $22.50. The price of a towel is $14.00.
Lesson
Main
Lesson
4-7
Quick Check
Feature
Inverse Matrices and Systems
Lesson 4-7
Algebra 2
Additional Examples
Write the coefficient matrix for each system. Use it to
determine whether the system has exactly a unique solution.
a.
4x – 2y = 7
–6x + 3y = 5
A =
4 –2
; det A =
–6 3
4 –2
= 4(3) – (–2)(–6) = 0
–6 3
Since det A = 0, the matrix does not have an inverse and the system
does not have a unique solution.
Lesson
Main
Lesson
4-7
Feature
Inverse Matrices and Systems
Lesson 4-7
Algebra 2
Additional Examples
(continued)
b.
12x + 8y = –3
3x – 7y = 50
A =
12 8
; det A =
3 –7
12 8
= 12(–7) – 8(–3) = –60
3 –7
Since det A =/ 0, the matrix has an inverse and the system has a unique
solution.
Quick Check
Lesson
Main
Lesson
4-7
Feature
Inverse Matrices and Systems
Lesson 4-7
Algebra 2
Lesson Quiz
Write each system as a matrix equation. Then solve the system.
1.
2.
3x + 2y = –6
2x – 3y = 61
3 2
2 –3
2x + 4y + 5z = –3
7x + 9y + 4z = 19
–3x + 2y + 8z = 0
2
7
–3
x
y
=
4
9
2
5
4
8
–6
; (8, –15)
61
x
y
z
=
–3
19
0
; (–10, 13, –7)
Determine whether each system has a unique solution.
7x – 2y = 15
no
3. –28x + 8y = 7
4.
20x + 5y = 33
–32x + 8y = 47
Lesson
Main
yes
Lesson
4-7
Feature
Augmented Matrices and Systems
Lesson 4-8
Algebra 2
Check Skills You’ll Need
(For help, go to Lessons 4-5 and 4-6.)
Evaluate the determinant of each matrix.
1.
–1
0
2
3
4.
0
4
–1
1 –3
5 –1
0 1
2.
0
–1
1
3
5.
3 4
–1 2
0 –1
5
0
1
3.
2
–1
1
5
6.
0 2 –1
3 4 0
–2 –1 5
Check Skills You’ll Need
Lesson
Main
Lesson
4-8
Feature
Augmented Matrices and Systems
Lesson 4-8
Algebra 2
Check Skills You’ll Need
Solutions
1. det
–1
0
2
= (–1)(3) – (2)(0) = –3 – 0 = –3
3
2. det
0
–1
1
= (0)(3) – (1)(–1) = 0 – (–1) = 1
3
3. det
2
–1
1
= (2)(5) – (1)(–1) = 10 – (–1) = 11
5
4. det
0
4
–1
1 –3
5 –1 = [(0)(5)(–1) + (4)(0)(–3) + (–1)(1)(–1)] – [(0)(0)(–1) +
0 1
(4)(1)(1) + (–1)(5)(–3)] = [0 + 0 + 1] – [0 + 4 + 15] = 1 – 19 = –18
Lesson
Main
Lesson
4-8
Feature
Augmented Matrices and Systems
Lesson 4-8
Algebra 2
Check Skills You’ll Need
Solutions (continued)
5. det
3 4
–1 2
0 –1
5
0 = [(3)(2)(1) + (–1)(–1)(5) + (0)(4)(0)] – [(3)(–1)(0) +
1
(–1)(4)(1) + (0)(2)(5)] = [6 + 5 + 0] – [0 + (–4) + 0] = 11 – (–4) = 15
6. det
0 2 –1
3 4 0 =[(0)(4)(5) + (3)(–1)(–1) + (–2)(2)(0)] – [(0)(–1)(0) +
–2 –1 5
(3)(2)(5) + (–2)(4)(–1)] = [0 + 3 + 0] – [0 + 30 + 8] = 3 – 38 = –35
Lesson
Main
Lesson
4-8
Feature
Augmented Matrices and Systems
Lesson 4-8
Algebra 2
Additional Examples
Use Cramer’s rule to solve the system
7x – 4y = 15
.
3x + 6y = 8
Evaluate three determinants. Then find x and y.
D =
7 –4
3 6
= 54
Dx =
15 –4
= 122
8 6
Dx
61
x =
=
27
D
The solution of the system is
Dy =
y =
7
3
15
= 11
8
Dy
11
=
54
D
61 11
,
.
27 54
Quick Check
Lesson
Main
Lesson
4-8
Feature
Augmented Matrices and Systems
Lesson 4-8
Algebra 2
Additional Examples
Find the y-coordinate of the solution of the
–2x + 8y + 2z = –3
–6x +
2z = 1 .
–7x – 5y + z = 2
system
D =
Dy =
–2 8
–6 0
–7 –5
–2 –3
–6 1
–7 2
2
2
1
= –24
Evaluate the determinant.
2
2 = 20
1
Replace the y-coefficients with the
constants and evaluate again.
Dy
20
5
y =
=–
=–
D
24
6
Find y.
The y-coordinate of the solution is –
5
.
6
Lesson
Main
Lesson
4-8
Quick Check
Feature
Augmented Matrices and Systems
Lesson 4-8
Algebra 2
Additional Examples
Write an augmented matrix to represent the
system
System of equations
–7x + 4y = –3
x + 8y = 9
–7x + 4y
x + 8y
x-coefficients
Augmented matrix
= –3
= 9
y-coefficients
–7
1
constants
–3
9
4
8
Draw a vertical bar to separate the
Quick Check
coefficients from constants.
Lesson
Main
Lesson
4-8
Feature
Augmented Matrices and Systems
Lesson 4-8
Algebra 2
Additional Examples
Write a system of equations for the augmented
matrix
9 –7
2 5
Augmented matrix
9
2
–1 .
–6
–7
5
x-coefficients
System of equations
–1
–6
y-coefficients
9x – 7y
2x + 5y
constants
= –1
= –6
Quick Check
Lesson
Main
Lesson
4-8
Feature
Augmented Matrices and Systems
Lesson 4-8
Algebra 2
Additional Examples
Use an augmented matrix to solve the system
x – 3y = –17
4x + 2y = 2
1 –3
4 2
–17
2
1 –3
–17
0 14
70
1 –3
0 1
–17
5
Lesson
Main
Write an augmented matrix.
–4(1 –3 –17)
4
2
2
0
14 70
1
14 (0
0
14
1
Multiply Row 1 by –4 and add it to Row 2.
Write the new augmented matrix.
70) Multiply Row 2 by 1 .
14
5
Write the new augmented matrix.
Lesson
4-8
Feature
Augmented Matrices and Systems
Lesson 4-8
Algebra 2
Additional Examples
(continued)
1 –3
0 1
1
0
0
1
–17
5
–2
5
1
14 (0
0
14
1
70)
5
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.
The solution to the system is (–2, 5).
Check:
x – 3y = –17
4x + 2y = 2
(–2) – 3(5) –17 4(–2) + 2(5) 2
–2 – 15 –17
–8 + 10 2
–17 = –17
2=2
Lesson
Main
Lesson
4-8
Use the original equations.
Substitute.
Multiply.
Quick Check
Feature
Augmented Matrices and Systems
Lesson 4-8
Algebra 2
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
Step 1: Enter the
augmented matrix
as matrix A.
Step 2: Use the rref feature
of your graphing
calculator.
The solution is (7, –9, –2).
Lesson
Main
Lesson
4-8
Feature
Augmented Matrices and Systems
Lesson 4-8
Algebra 2
Additional Examples
(continued)
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
Lesson
Main
Lesson
4-8
Feature
Augmented Matrices and Systems
Lesson 4-8
Algebra 2
Lesson Quiz
1. Use Cramer’s Rule to solve the system.
3x + 2y = –2
(–12, 17)
5x + 4y = 8
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.
–7x + 3y + 9z = 12
–7 3 9
–7 3 12
5x
+ 3z = 8
5 0 3 , Dz =
5 0 8
D =
4x – 6y + z = –2
4 –6 1
4 –6 –2
3. Solve the system by using an augmented matrix.
5x + y = 1
(2, –9)
3x – 2y = 24
4. Solve the system by using an augmented matrix.
4x + y – z = 7
–2x + 2y + 5z = 3
(–1, 8, –3)
7x – 3y – 9z = –4
Lesson
Main
Lesson
4-8
Feature