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– HW Quizzes
Chapter 13
Matrices and Determinants
13.1 Matrices and Systems of
Equations
A matrix is a rectangular array of
numbers. We subscript entries to tell
their location in the array
rows
a
a
a

11
12
13
row
a
a
a
21
22
23

A  a31 a32 a33


 
am1 am 2 am3
m n
 a1n 

 a2 n 
 a3n 

 
 amn 
Matrices
are
identified
by their
size.
1 5 3 1 5 0 2
4 1
 2
6 
 
1 
 
  3
4 4
 2  1  2 4
 1 3

5
7


 2 5  8 9 


7
9 0
4
A matrix that has the same number of rows as
columns is called a square matrix.
 a11 a12
a
a
21
22

A
a31 a32

a41 a42
a13
a23
a33
a43
a14 

a24 
a34 

a44 
3x  2 y  5 z  3
 2 x  y  4 z  2
x  4 y  7z  1
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.
 3 2 5 


1
4
Coefficient matrix A   2


 1
4  7
3x  2 y  5 z  3
 2 x  y  4 z  2
x  4 y  7z  1
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.
3
 3 2 5


#
1
4

2
Augmented matrix A   2


 1
4  7 1 
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
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).
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.
1
0

0
#
#
1
0
#
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.
Suse row operations to obtain row
echelon form:
We already
have the 1
where we
need it.
1 2 1 1
3 5 1 3


2 6 7 1
The augmented matrix
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).
x  2y  z 1
3x  5 y  z  3
2x  6 y  7z  1
Work on this column first.
Get the 1 and then use it
as a “tool” to get zeros
below it with row
operations.
1
0

0
#
#
1
0
#
1
#

#
#
11
-3r1 + r2 03

22
2 11 11

51 1 2 30
6 77 11
-2r1 + r3
11
00

02
22
11
26
11
22
57
1
0

11
Now our first column is
like our goal.
-3r1
+ r2
-3 -6 -3 -3
3
5
1
3
0
-1 -2
0
-2r1
-2 -4 -2 -2
+ r3
2
6
7
0
2
5 -1
1
Now we’ll use -2 times row 1 added to row 3 to get a 0 there.
- 1r2
1
1
1 2
0  1  2 0 


0 2
5  1
We need a 1 in the second
row second column so we’ll
multiply row 2 by a -1
-2r2
+ r3
0 -2 -4
0
0
2
5 -1
0
0
1 -1
-2r2 + r3
11 22 11 11 
00 11 22 00 


00 20 51 1
1
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
the second column is
like we need it now
Now we’ll move to the second
column and do row operations to
get it to look like our goal.
1
0

0
#
#
1
0
#
1
#

#
#
z column
y column
x column
xx222y z11  1
y y 
22z100
y2
z  1
equal signs
1 2 1 1  x  2
0 1 2 0 


0 0 1 1
Substitute -1 in for z in
second equation to find y
Substitute -1 in for z and 2 for
y in first equation to find x.
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.
Solution is: (-2, 2, -1)
1
0

0
#
#
1
0
#
1
#

#
#
x  2y  z 1
3x  5 y  z  3
2x  6 y  7z  1
Solution is: (-2, 2, -1)
This is the only (x, y, z) that
make ALL THREE equations
true. Let’s check it.
 2  22   1  1
3 2  52   1  3
2 2  62  7 1  1
These are all true.
Geometrically this means
we have three planes that
intersect at a point, a
unique solution.
To obtain reduced row echelon form, you continue to do
more row operations to obtain the goal below.
1
0

0
0
1
0
0
0
1
#

#
#
This method requires no back substitution.
When you put the variables back in, you have
the solutions.
Let’s try this method on the
problem we just did. We take the
matrix we ended up with when
doing row echelon form:
-2r
3r32+r11
-2r3+r2
11
00

000

0
13
022
x  2y  z 1
3x  5 y  z  3
2x  6 y  7z  1
112



02
00 x  2, y  2, z  1
000 1
11 1
11 Let’s get the 0 we need in
002
11
Notice when we put the variables the second column by
and = signs back in we have the using the second row as
a tool.
solution
Now we’ll use row 3 as a tool to
work on the third column to get
zeros above the 1.
1
0

0
0
1
0
0
0
1
#

#
#
Let’s try another one:
The augmented matrix:
3  2 2 6 
2  3 4 0 


7  3 2  1
11
0
-2r1+r2 2

-7r1+r3 7
07
r1-r2
11
553
10
33
222 66 


848 
12
12
0  
216
 
2 143
If we subtract the second
row from the first we’ll get
the 1 we need for the first
column.
3x  2 y  2 z  6
2x  3 y  4z  0
7 x  3 y  2 z  1
We’ll now use row 1 as our
tool to get 0’s below it.
We have the first column
like our goal. On the
next screen we’ll work
on the next column.
1
0

0
#
#
1
0
#
1
#
#
#
3x  2 y  2 z  6
6 
1 1  2
0  5 8  12 


0  10 16  43
-1/5r2
10r2+r3
11 11 22 66 

8 12
12 
8
00 11  5

5 
0  10 16
5
5

43

0 0
0
2x  3 y  4z  0
7 x  3 y  2 z  1
We’ll now use row 2 as our
tool to get 0’s below it.
 19
INCONSISTENT - NO SOLUTION
If we multiply the second
row by a -1/5 we’ll get the
one we need in the second
column.
Wait! If you put
variables and = signs
back in the bottom
equation is 0 = -19 a
false statement!
1
0

0
#
#
1
0
#
1
#

#
#
5  6 1
2  3 1

4  3  1
r1-r3 1  3
2  3

4  3
4
1
5
2
One more:
 1

1 1
 1 5 
5x  6 y  z  4
2x  3y  z  1
4x  3y  z  5
1  3 2 11


1/3r2 0 1

1
1
1


-9r2+r3 0 9
0 09 09 

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.
-2r1+r2
-4r1+r3
1  3 2 11
02 33 13 13 


04 93  19 59 
1
0

0
#
#
1
0
#
1
#

#
#
1  3 2  1
0 1  1 1 


0 0
0 0 
put variables
back in
solve for x & y
Let’s go one step further and get a 0
above the 1 in the second column
x
3r2+r1
y
z
1 0  1 2
0 1  1 1 


0 0 0 0
x z2
y  z 1
zz
No restriction on z
x z2
y  z 1
zz
Infinitely many solutions where z is any real number
5x  6 y  z  4
2x  3y  z  1
4x  3y  z  5
532  612  01  4
232  312  01  1
432  312 01  5
works in all 3
What this means is that you can
The solution can be
choose any real number for z and written: (z + 2, z + 1, z)
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
Let’s try z = 0. Then y = 1 and x = 2
x z2
y  z 1
zz
Infinitely many solutions where z is any real number
HW #13.1
Pg 572 1-4, 6-7, 9-11, 15
HW Quiz 13.1
Wednesday, April 6, 2016
Row 1, 3, 5
1. 4
2. 6
3. 10
4. 15
Row 2, 4
1. 2
2. 4
3. 6
4. 10
Chapter 13
Matrices 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 


3 
 9 13
 13
3 2 10
Find the Additive Inverse of each Matrix
 5 0 
 2 1


 4 3 
 2 5
1 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 3
 2 1

 


 1 3
4 2
Exercises for Example 4
Subtract by finding the Additive Inverse
HW # 13.2
Pg 575 1-32
Chapter 13
Matrices and Determinants
Section 13.3
Cramer’s Rule
Objective: Evaluate a 2 x 2 Determinant
A
B
Objective: Solve a system of 2 equations and 2 variables using
Cramer’s Rule
Objective: Evaluate a 3 x 3 Determinant
C
Objective: Solve a system of 3 equations and 3 variables using
Cramer’s Rule
D
E
HW #13.3
Pg 580 1-33 odd, 34-42
Chapter 13
Matrices and Determinants
Section 13.4
Multiplying Matrices
MULTIPLYING TWO MATRICES
A

B

AB
4 X 3

3 X 5

4X5
4 rows
3 rows
3 columns
5 columns
MULTIPLYING TWO MATRICES
A

B

AB
4 X 3

3 X 5

4X5
4 rows
4 rows
5 columns
5 columns
MULTIPLYING TWO MATRICES
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
A

B

m X n

n X p

m rows
n rows
n columns
p columns
AB
mXp
MULTIPLYING TWO MATRICES
A

B

AB
m X n

n X p

mXp
m rows
m rows
p columns
p columns
MULTIPLYING TWO MATRICES
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.
Finding the Product of Two Matrices
Find AB if
–2 3
A = 1 –4
6
0
and
B=
–1
–2
SOLUTION
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.
Use a similar procedure to write the other entries of the
product.
3
4
Finding the Product of Two Matrices
A

B

AB
3X2

2X2

3X2
–2
3
1
–4
6
0
–1
3
–2
4
(– 2)(– 1) + (3)(– 2)
(– 2)(3) + (3)(4)
(1)(–1) + (– 4)(–2)
(1)(3) + (– 4)(4)
(6)(– 1) + (0)(– 2)
(6)(3) + (0)(4)
Finding the Product of Two Matrices
A

B

AB
3X2

2X2

3X2
–2
3
1
–4
6
0
–1
3
–2
4
(– 2)(– 1) + (3)(– 2)
(– 2)(3) + (3)(4)
(1)(–1) + (– 4)(–2)
(1)(3) + (– 4)(4)
(6)(– 1) + (0)(– 2)
(6)(3) + (0)(4)
Finding the Product of Two Matrices
A

B

AB
3X2

2X2

3X2
–2
3
1
–4
6
0
–1
3
–2
4
(– 2)(– 1) + (3)(– 2)
(– 2)(3) + (3)(4)
(1)(–1) + (– 4)(–2)
(1)(3) + (– 4)(4)
(6)(– 1) + (0)(– 2)
(6)(3) + (0)(4)
Finding the Product of Two Matrices
A

B

AB
3X2

2X2

3X2
(– 2)(– 1) + (3)(– 2)
(– 2)(3) + (3)(4)
(1)(– 1) + (– 4)(– 2)
(1)(3) + (– 4)(4)
(6)(– 1) + (0)(– 2)
(6)(3) + (0)(4)

–4
6
7
– 13
–6
18
3x2
2x2
AB will be 3 x 2
2x2
2x2
AB will be 2 x 2
2x2
2x2
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 m
k(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 13.4
Wednesday, April 6, 2016
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 3 
1 0 
 4 6  14 7 
1. 

 


0
1
3
1
3

2





14 7 
1 2 
 3
 8

8
8
3 1 0  

1
3 6
3. 1 1 2   


 8
8 8
1 1 1   2 2
4


8
 8 8
 4 6  1 0   4 6 

2. 




3
1
3
1
0
1





1 0 0 
 0 1 0 


0 0 1 
Chapter 13
Matrices 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 2 
 5
 3 4
5
B
A


2 6
  1  3 
 5
10 
No
Determine if C and D are multiplicative inverses of each other
 2 3
C 

3
6


 2 1
D
2
 1


3
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
 15

 2
 3
 3

 5
1 1
5 15 

1
0
3
2
1

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
Constants
Matrix of
Variables
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
 2 1 x  1 
 3 2  y    0 

   
 6 5  x   3 
 3 2  y    3 

   
 3 4  x   4 
 4 5  y    7 

   
4. x  2 y  3 z  4
2 x  y  z  1
4x  y  z  1
 1 2 3  x   4 
 2 1 1  y    1

   
 4 1 1  z  1 

   
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.
 2 1 x  1 
 3 2  y    0 

   
 6 5  x   3 
 3 2  y    3 

   
 3 4  x   4 
 4 5  y    7 

   
4. x  2 y  3 z  4
2 x  y  z  1
4x  y  z  1
 1 2 3  x   4 
 2 1 1  y    1

   
 4 1 1  z  1 

   
HW 13.6
Pg 596-597 1-19 Odd
13-7 Using Matrices
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 #13.7
Pg 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-13
Pg 609 1-14