needs work need to add –hw quizzes chapter 13 matrices and determinants
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Needs Work
• Need to add – HW Quizzes
Chapter 13
Matrices and Determinants
13.1 Matrices and Systems of Equations
row
nmrows
mnmmm
n
n
n
aaaa
a
a
a
aaa
aaa
aaa
A
321
3
2
1
333231
232221
131211
A matrix is a rectangular array of numbers. We subscript entries to tell
their location in the array
Matrices are
identified by their
size.
14
51 20513
3
1
6
2
0974
9852
7531
4212
44
A matrix that has the same number of rows as columns is called a square matrix.
main diagonal
44434241
34333231
24232221
14131211
aaaa
aaaa
aaaa
aaaa
A
174
242
3523
zyx
zyx
zyx
741
412
523
ACoefficient matrix
If you have a system of equations and just pick off the coefficients and put them in a matrix it is called a coefficient matrix.
174
242
3523
zyx
zyx
zyx If you take the coefficient matrix and then add a last column with the constants, it is called the augmented matrix. Often the constants are separated with a line.
1741
2412
3523#AAugmented matrix
We are going to work with our augmented matrix to get it in a form that will tell us the solutions to the system of equations. The three things above are the only things we can do to the matrix but we can do them together (i.e. we can multiply a row by something and add it to another row).
Operations that can be performed without altering the solution set of a linear system
1. Interchange any two rows
2. Multiply every element in a row by a nonzero constant
3. Add elements of one row to corresponding elements of another row
#100
##10
###1
After we get the matrix to look like our goal, we put the variables back in and use back substitution to get the solutions.
We use elementary row operations to make the matrix look like the one below. The # signs just mean there can be any number here---we don’t care what.
#100
##10
###1
Suse row operations to obtain row echelon form:
1762
353
12
zyx
zyx
zyx
1762
3153
1121
The augmented matrix
Work on this column first. Get the 1 and then use it as a “tool” to get zeros below it with row operations.
We already have the 1 where we need it.
We’ll take row 1 and multiply it by -3 and add to row 2 to get a 0. The notation for this step is -3r1 + r2 we write it by the row we replace in the matrix (see next screen).
1762
0210
1121
1762
3153
1121
1762
0210
1121
-3r1 + r2
-3r1 -3 -6 -3 -3 + r2 3 5 1 3
0 -1 -2 0
Now we’ll use -2 times row 1 added to row 3 to get a 0 there.
-2r1 + r3
1520
0210
1121
-2r1 -2 -4 -2 -2 + r3 2 6 7 1
0 2 5 -1
Now our first column is like our goal.
1520
0210
1121
- 1r2
-2r2 0 -2 -4 0 + r3 0 2 5 -1
0 0 1 -1
-2r2 + r3
1520
0210
1121
We’ll use row 2 with the 1 as a tool to get a 0 below it by multiplying it by -2 and adding to row 3
#100
##10
###1Now we’ll move to the second column and do row operations to get it to look like our goal.
We need a 1 in the second row second column so we’ll multiply row 2 by a -1
1100
0210
1121
the second column is like we need it now
#100
##10
###1
Now we’ll move to the third column and we see for our goal we just need a 1 in the third row third column and we have it so we’ve achieved the goal and it’s time for back substitution. We put the variables and = signs back in.
1100
0210
1121Substitute -1 in for z in second equation to find y
1
02
12
z
zy
zyxx
colu
mn
y co
lum
n
z co
lum
n
equal signs 012 y
2y
Substitute -1 in for z and 2 for y in first equation to find x.
1122 x
2x
Solution is: (-2, 2, -1)
Solution is: (-2, 2, -1)
1762
353
12
zyx
zyx
zyx
This is the only (x, y, z) that make ALL THREE equations true. Let’s check it.
1172622
312523
11222
These are all true.
Geometrically this means we have three planes that intersect at a point, a unique solution.
#100
#010
#001
This method requires no back substitution. When you put the variables back in, you have the solutions.
To obtain reduced row echelon form, you continue to do more row operations to obtain the goal below.
Let’s try this method on the problem we just did. We take the matrix we ended up with when doing row echelon form:
1762
353
12
zyx
zyx
zyx
Let’s get the 0 we need in the second column by using the second row as a tool.
-2r2+r1
Now we’ll use row 3 as a tool to work on the third column to get zeros above the 1.
#100
#010
#001
1100
0210
1121
1100
0210
1301
-2r3+r2
3r3+r1
1100
0210
2001
1100
2010
2001
Notice when we put the variables and = signs back in we have the solution
1,2,2 zyx
Let’s try another one:
1237
0432
6223
zyx
zyx
zyxThe augmented matrix:
1237
0432
6223
If we subtract the second row from the first we’ll get the 1 we need for the first column.
r1-r2
We’ll now use row 1 as our tool to get 0’s below it.
-2r1+r2
1237
0432
6211
1237
12850
6211
-7r1+r3
4316100
12850
6211We have the first column like our goal. On the next screen we’ll work on the next column.
#100
##10
###1
1237
0432
6223
zyx
zyx
zyx
If we multiply the second row by a -1/5 we’ll get the one we need in the second column.
We’ll now use row 2 as our tool to get 0’s below it.
-1/5r2
10r2+r3
4316100
12850
6211
Wait! If you put variables and = signs back in the bottom equation is 0 = -19 a false statement!
43161005
12
5
810
6211
190005
12
5
810
6211
#100
##10
###1INCONSISTENT - NO SOLUTION
534
132
465
zyx
zyx
zyx
5134
1132
4165 One more:
#100
##10
###1
r1-r3
-2r1+r2
-4r1+r3
5134
1132
1231
9990
3330
1231
1/3r2
9990
1110
1231
-9r2+r3
0000
1110
1231
Oops---last row ended up all zeros. Put variables and = signs back in and get 0 = 0 which is true. This is the dependent case. We’ll figure out solutions on next slide.
5134
1132
1231
0000
1110
1231
x y z
zz
zy
zx
1
2
zz
zy
zx
1
2
0000
1110
2101
Let’s go one step further and get a 0 above the 1 in the second column
3r2+r1
Infinitely many solutions where z is any real number
No restriction on z
put variables back in
solve for x & y
zz
zy
zx
1
2
Infinitely many solutions where z is any real number
534
132
465
zyx
zyx
zyx
What this means is that you can choose any real number for z and put it in to get the x and y that go with it and these will solve the equation. You will get as many solutions as there are values of z to put in (infinitely many).
Let’s try z = 1. Then y = 2 and x = 3
512334
112332
412635
works in all 3
Let’s try z = 0. Then y = 1 and x = 2
501324
101322
401625
The solution can be written: (z + 2, z + 1, z)
HW #13.1Pg 572 1-4, 6-7, 9-11, 15
HW Quiz 13.1Wednesday, April 19, 2023
Row 1, 3, 5
1. 4
2. 6
3. 10
4. 15
Row 2, 4
1. 2
2. 4
3. 6
4. 10
Chapter 13Matrices and Determinants
Section 13.2
Addition and Subtraction of Matrices
To Add and Subtract matrices
To find the additive inverse of a matrix
To compare matrices they must have the same dimensions and have the same entries in the same
positions
You can only add or subtract matrices when they have exactly the same dimensions
4. The operation is not possible
The additive inverse of a matrix can be obtained by replacing each element by its additive inverse.
Finding the Additive Inverse of a Matrix
Find the additive inverse of the matrix
12 2 15
16 0 9
9 13 3
13 3 2 10
Find the Additive Inverse of each Matrix
2 5
1 3
5 0
2 1
4 3
4 3 2
1 5 4
2 7 6
3 0 10 8
Subtracting by finding the Additive Inverse
Subtract by finding the Additive Inverse
2 1
4 2
2 3
1 3
Exercises for Example 4
Subtract by finding the Additive Inverse
HW # 13.2
Pg 575 1-32
Chapter 13Matrices and Determinants
Section 13.3
Cramer’s Rule
Objective: Evaluate a 2 x 2 Determinant
A B
Your Turn Hidden
Objective: Solve a system of 2 equations and 2 variables using Cramer’s Rule
Objective: Evaluate a 3 x 3 Determinant
Your Turn Hidden
C
Objective: Solve a system of 3 equations and 3 variables using Cramer’s Rule
ED
Your Turn Hidden
HW #13.3Pg 580 1-33 odd, 34-42
Chapter 13Matrices and Determinants
Section 13.4
Multiplying Matrices
4 X 3 3 X 5 4 X 5
A B AB
MULTIPLYING TWO MATRICES
4 rows4 rows
3 columns3 columns
3 rows3 rows
5 columns5 columns
4 X 3 3 X 5 4 X 5
MULTIPLYING TWO MATRICES
4 rows4 rows
5 columns5 columns
4 rows4 rows
5 columns5 columns
A B AB
If A is a 4 X 3 matrix and B is a 3 X 5 matrix, then the product AB is a 4 X 5 matrix.
MULTIPLYING TWO MATRICES
m X n n X p m X p
A B AB
MULTIPLYING TWO MATRICES
m rowsm rows
n columnsn columns
n rowsn rows
p columnsp columns
m X n n X p m X p
MULTIPLYING TWO MATRICES
m rowsm rows
p columnsp columns
m rowsm rows
p columnsp columns
A B AB
If A is an m X n matrix and B is an n X p matrix, then the product AB is an m X p matrix.
MULTIPLYING TWO MATRICES
Finding the Product of Two Matrices
– 2 3 1 – 4 6 0
– 1 3– 2 4
Find AB if A = and B =
Use a similar procedure to write the other entries of the product.
Because A is a 3 X 2 matrix and B is a 2 X 2 matrix, the product AB is defined and is a 3 X 2 matrix.
To write the entry in the first row and first column of AB, multiply corresponding entries in the first row of A and the first column of B. Then add.
SOLUTION
(– 2)(– 1) + (3)(– 2) (– 2)(3) + (3)(4)
(1)(– 1) + (– 4)(– 2) (1)(3) + (– 4)(4)
(6)(– 1) + (0)(– 2) (6)(3) + (0)(4)
3 X 2 2 X 2 3 X 2
A B AB
– 2 3
1 – 4
6 0
– 1 3
– 2 4
Finding the Product of Two Matrices
3 X 2 2 X 2 3 X 2
A B AB
Finding the Product of Two Matrices
(– 2)(– 1) + (3)(– 2) (– 2)(3) + (3)(4)
(1)(– 1) + (– 4)(– 2) (1)(3) + (– 4)(4)
(6)(– 1) + (0)(– 2) (6)(3) + (0)(4)
– 2 3
1 – 4
6 0
– 1 3
– 2 4
3 X 2 2 X 2 3 X 2
A B AB
Finding the Product of Two Matrices
(– 2)(– 1) + (3)(– 2) (– 2)(3) + (3)(4)
(1)(– 1) + (– 4)(– 2) (1)(3) + (– 4)(4)
(6)(– 1) + (0)(– 2) (6)(3) + (0)(4)
– 2 3
1 – 4
6 0
– 1 3
– 2 4
3 X 2 2 X 2 3 X 2
A B AB
Finding the Product of Two Matrices
– 4 6
7 – 13
– 6 18
(– 2)(– 1) + (3)(– 2) (– 2)(3) + (3)(4)
(1)(– 1) + (– 4)(– 2) (1)(3) + (– 4)(4)
(6)(– 1) + (0)(– 2) (6)(3) + (0)(4)
3 x 2 2 x 2
AB will be 3 x 2
2 x 2 2 x 2
AB will be 2 x 2
2 x 2 2 x 2
BA will be 2 x 2
Properties of Matrix Arithmetic
• For any matrices A, B, C of dimensions appropriate for them to be added or multiplied.– Commutative Property of Addition
• A + B = B + A
– Associative Property• A + (B + C) = (A + B) + C
• A(BC) = (AB)C
Properties of Matrix Arithmetic
• For any matrices A, B, C of dimensions appropriate for them to be added or multiplied.– Additive Identity
• There exists a unique matrix O such that A + 0 = 0 + A = A
– Additive Inverse• There Exists a unique matrix –A such that
A + (-A) = -A + A = 0
Properties of Matrix Arithmetic
• For any matrices A, B, C of dimensions appropriate for them to be added or multiplied.– Distributive Property
• A(B + C) = AB + AC
Properties of Matrix Arithmetic
• For any real numbers k and mk(A + B) = kA + kB(k + m)A = kA + mA(km)A = k(mA)
HW #13.4
Pg 587-588 1-39 Odd, 40-46
HW Quiz 13.4Wednesday, April 19, 2023
Row 1, 3, 5
Write the answer to
1. 9
2. 15
3. 17
4. 39
Row 2, 4, 6
Write the answer to
1. 9
2. 15
3. 17
4. 39
1 34 6 14 71.3 1 3 2
14 7
4 6 1 02.
3 1 0 1
3 1 28 8 83 1 01 3 6
3. 1 1 28 8 8
1 1 1 2 2 48 8 8
1 0
0 1
4 6
3 1
1 0 0
0 1 0
0 0 1
Chapter 13Matrices and Determinants
Section 13.5
Inverses of Matrices
To write a matrix equation equivalent to a system of matrices
To determine when two matrices are multiplicative inverses and find the multiplicative inverse of a 2 x 2 matrix
Two n x n matrices are inverses of each other if their product (in both orders) is the n x n identity matrix.
AA-1 = A-1A = 1
To determine if two matrices A and B are inverses of each other you need to make sure AB = BA = I
Determine if A and B are multiplicative inverses of each other
3 4
2 6A
3 25 51 35 10
B
No
2 3
3 6C
Determine if C and D are multiplicative inverses of each other
2 1
21
3
D
Yes
Tell whether the matrices are multiplicative inverses of each other.
Computing an Inverse Matrix 2 x 2
Let’s Prove it!
Use the shortcut
Find the inverse of the given matrix
2 1 1
1 2 3
4 1 2
7 1 1
15 5 152 1
03 33 2 1
5 5 5
HW #13.5
Pg 591-592 1-19, 21-25 Odd
13.6
Inverses and Systems
Writing Linear Systems as Matrix Equations
Consider the system
Let A = Let X = Let B =
Write the equation AX = B using the above matrices
Coefficient Matrix
Matrix of Variables
Matrix of Constants
For a linear system of equations written as a matrix equation the matrix A is the coefficient matrix of the system, X is the matrix of variables, and B is the matrix of constants.
Write the system of linear equations as a matrix equation
4. 2 3 4
2 1
4 1
x y z
x y z
x y z
2 1 1
3 2 0
x
y
3 4 4
4 5 7
x
y
6 5 3
3 2 3
x
y
1 2 3 4
2 1 1 1
4 1 1 1
x
y
z
SOLUTION OF A LINEAR SYSTEM Let AX = B represent a system of linear equations. If the determinant of A is nonzero, then the linear system has exactly one solution, which is X = A-1B.
SOLUTION OF A LINEAR SYSTEM Let AX = B represent a system of linear equations. If the determinant of A is nonzero, then the linear system has exactly one solution, which is X = A-1B.
SOLUTION OF A LINEAR SYSTEM Let AX = B represent a system of linear equations. If the determinant of A is nonzero, then the linear system has exactly one solution, which is X = A-1B.
Solve system of linear equations.
4. 2 3 4
2 1
4 1
x y z
x y z
x y z
2 1 1
3 2 0
x
y
3 4 4
4 5 7
x
y
6 5 3
3 2 3
x
y
1 2 3 4
2 1 1 1
4 1 1 1
x
y
z
HW 13.6Pg 596-597 1-19 Odd
13-7 Using Matrices
Application
Two stores sell the exact same brand and style of a dresser, a night stand, and a bookcase. Matrix A gives the retail prices (in dollars) for the items. Matrix B gives the number of each item sold at each store in one month.
Calculate AB and interpret the entries of AB
Your Turn
HW #13.7Pg 600-601 1-7
Chapter 13 Test Review
Part 1
Helpful Hints
• What does it mean when the determinant of a matrix is 0– In terms of a system of Linear Equations– In Terms of a Matrix
• Rules of Multiplication/Addition/Equality• Scalar Multiplication• Shortcut for finding the inverse of a 2 x 2
matrix
Solve
Evaluate the determinant of the matrix.
Use Cramer’s rule to solve the system of equations.
Use Augmented Matrices to solve the system of equations.
Use the matrix equation AX = B to solve the system.
When does a matrix fail to have an inverse?
Study all the challenge problems in the book
Know how to multiply and add matrices
Chapter 13 Test Review
Part 2
This test will have only 1 part
HW #R-13Pg 609 1-14
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