Linear algebra

MAC 2103
Module 1
Systems of Linear Equations and
Matrices I

1

Learning Objectives
Upon completing this module, you should be able to:

1. Represent a system of linear equations as an augmented
matrix.
2. Identify whether the matrix is in row-echelon form, reduced
row-echelon form, both, or neither.
3. Solve systems of linear equations by using the Gaussian
elimination and Gauss-Jordan elimination methods.
4. Perform matrix operations of addition, subtraction,
multiplication, and multiplication by a scalar.
5. Find the transpose and the trace of a matrix.

http://faculty.valenciacc.edu/ashaw/
Rev.F09 Click link to download other modules. 2

1

Systems of Linear Equations
and Matrices I
There are three major topics in this module:

Introduction to Systems of Linear Equations
Gaussian Elimination
Matrices and Matrix Operations

Rev.09 Click link to download other modules. 3

A Quick Review
 A linear equation in two variables can be written in
the form ax + by = k, where a, b, and k are
constants, and a and b are not equal to 0.
Note: The power of the variables is always 1.
 Two or more linear equations is called a system of

linear equations because they involve solving more
than one linear equation at once.
 A system of linear equations can have either exactly

one solution (unique), no solution, or infinitely many
solutions.


2

Let’s Look at a System of Two Linear Equations
in Two Variables


Remember How to Use the Elimination Method
to Solve a System of Linear Equations?

Example: Use elimination to solve each system of
equations, if possible. Identify the system as consistent
or inconsistent. If the system is consistent, state whether
the equations are dependent or independent. Support
your results graphically.

a) 3x − y = 7 b) 5x − y = 8 c) x − y = 5
5x + y = 9 −5x + y = −8 x−y=−2


3

Solving a System of Linear Equations Using the
Elimination Method (Cont.)
Solution Eliminate y by adding
the equations.
a)

Find y by substituting
x = 2 in either equation.

The solution is (2, −1). The system is
consistent and the equations are independent.

b) If we add the equations we obtain the
following result.

The equation 0 = 0 is an
identity that is always true.
true.
The two equations are equivalent.
There are infinitely many solutions.
solutions.
{(x, y)| 5x − y = 8}
{(x 5x


4

c) If we subtract the second equation from
the first, we obtain the following result.

The equation 0 = 7 is a
contradiction that is never true.
true.
Therefore, there is no solution, and
solution,
the system is inconsistent.
inconsistent.


Let’s Look at Solving a System of Linear
Equations with Three Variables
Solve the following system.

Solution
Step 1: Eliminate the variable z from equation one and
two and then from equation two and three.

Equation 1 Equation 2

Equation 2 times 6 Equation 3

Add Add


5

Solving a System of Linear Equations with Three
Variables Using the Elimination Method (Cont.)
Step 2: Take the two new equations and eliminate
either variable.

Find x using y = −2.
Do you
remember using
this method
before?


Solving a System of Linear Equations with Three
Variables Using the Elimination Method (Cont.)
Step 3: Substitute x = 1 and y = −2 in any of the
given equations to find z.

Simple?
The solution is (1, −2, 2). Let’s
move on.


6

One More Example
Solve the system. Step 2 The two
equations are
inconsistent
because the sum
of 10x + 9y cannot
Solution be both −3 and 0.
Step 1 Multiply equation one by 2 and add
to equation two. Step 3 is not
necessary - the
system of
equations has no
Subtract equation three from equation two. solution.


How to Represent a System of Linear
Equations in an Augmented Matrix?
Let’s represent the previous system of
linear equations in an Augmented Matrix. Just keep two
items in mind:

1. The constants
must be on the
How? right most column.

Basically, we just need to write down the
coefficients of the variables and the 2. The coefficients
constants in an rectangular array of of the variables
must be in the
numbers. That’s it. " 3 9 6 !3 % same order for
$ '
$ 2 1 !1 !2 ' each equation (or
$ 1 1 1
# 1 '
& each row).

7

How to Solve a System of Linear
Equations Using an Augmented Matrix?
Let’s start by labeling our augmented matrix What method(s)
with r1 (row 1), r2 (row 2), and r3 (row 3). can we use to
Each row corresponds to an equation. accomplish this?
We can use:
r1 " 3 9 6 !3 % 1. Gauss-Jordan
r2 $ 2 1 !1 !2
'
Elimination
$ '
r3 $ 1 1 1 1 ' method to obtain
# &
a reduced row-
echelon form.
What’s next?
2. Gaussian
We want to simplify the augmented matrix Elimination
into either a row-echelon form or a reduced Method to obtain
row-echelon form. a row-echelon
form.

How to Identify a Matrix that is in a Row-Echelon
Form or a Reduced Row-Echelon Form?
Pictures are worth a thousand words. Here What are
are two pictures. Picture 1 shows a reduced those *s?
row-echelon form matrix, and Picture 2 shows
a row-echelon form matrix. * represents any
* can be any
numbers.
numbers. The
! 1 0 0 * $ ! 1 * * * $ row-echelon
# & # &
# 0 1 0 * & # 0 1 * * & form has a
#
" 0 0 1 * &
% #
" 0 0 1 * % & leading 1 with
Picture 1 Picture 2 zero(s) below
it, but it can
See the basic differences? have any
The reduced row-echelon form shown in numbers
Picture 1 has a leading 1 in each row with above it.
zero(s) above it and below it when possible.

8

Properties for a Matrix in
Reduced Row-Echelon Form

The four basic properties:

1. The first nonzero number in a nonzero row has to be a 1.
2. Any row with all zeros is below all nonzero rows.
3. For nonzero rows, the leading 1 in the next row has to be
farther to the right than the leading 1 in the previous row.
4. Each column that has a leading 1 can only have zeros
everywhere else in that column.

Note: A matrix that meets only the first three properties is a matrix
in row-echelon form.

How to Solve a System of Linear Equations
Using an Augmented Matrix? (Cont.)
We want to
Let’s look at our augmented matrix. reduce our
augmented
r1 " 3 9 6 !3 %
$ ' matrix into
r2 $ 2 1 !1 !2 ' something like
$
r3 # 1 1 1 1 '
& this.
We can simplify our augmented matrix into a
reduced row-echelon form - through a step- ! 1 0 0 * $
by-step elimination process. # &
# 0 1 0 * &
Step 1: We want a leading 1 in row 1. We #
" 0 0 1 * &
%
can scale row 1 to accomplish this.
1
3 r1 ! r1 " 1 3 2 !1 %
$ '
r2 $ 2 1 !1 !2 '
r3 $ 1 1 1
# 1 '
&

9


Step 2: We need zeros below our leading 1 in From Step 1:
row 1. How to make 2 and 1 become zeros? r1
" 1 3 2 !1 %
r1 " 1 3 2 !1 % r2 $ 2 1 !1 !2 '
$ '
!2r1 + r2 " r2 $ 0 !5 !5
$ 0 '
' r3 $ 1 1 1
# '
1 &
!r1 + r3 " r3 # 0 !2 !1
$ 2 &'
We want to
reduce our
augmented matrix
Step 3: We need a leading 1 in row 2. How?
into something
like this.
r1 " 1 3 2 !1 %
$ ' ! 1 0 0 * $
! r2 " r2 $ 0 1 1
1
0 ' # &
# 0 1 0 * &
5

r3 # 0 !2 !1
$ 2 '
& # 0 0 1 * &
" %



Step 4: We need a zero below our leading 1 From Step 3:
in row 2. r1 " 1 3 2 !1 %
$ '
r1 " 1 3 2 !1 % r2 $ 0 1 1 0 '
$ ' r3 $ 0 !2 !1 2 '
r2 $ 0 1 1 0 ' # &
2r2 + r3 ! r3 $
# 0 0 1 2 '
& We want to
reduce our
augmented matrix
into something
like this.
Alright, we have a row-echelon form matrix.
Gaussian elimination stops at this step but then ! 1 0 0 * $
# &
requires back-substitution to find the solution. # 0 1 0 * &
#
" 0 0 1 * &
%


10


Step 5: We need zeros above our leading 1 in From Step 4:
row 3 from step 4. " 1 3 2 !1 %
$ '
!2r3 + r1 " r1 " 1 3 0 !5 % $ 0 1 1 0 '
!r3 + r2 " r2 $ ' $ '
$ 0 1 0 !2 ' # 0 0 1 2 &
r3 $
# 0 0 1 2 '
& We want to
reduce our
augmented matrix
Step 6: We need a zero above our leading 1
into a reduced
at row 2. How? row-echelon form.
!3r2 + r1 " r1 " 1 0 0 1 %
$ ' ! 1 0 0 * $
r2 $ 0 1 0 !2 ' # &
r3 $ ' # 0 1 0 * &
# 0 0 1 2 & # 0 0 1 * &
" %
Now, we have a reduced row-echelon form matrix.


What does our matrix say?
Note:
" 1 0 0 1 % 1.x + 0.y + 0.z = 1 ! x = 1
$ ' If you remember,
$ 0 1 0 !2 ' 0.x + 1.y + 0.z = "2 ! y = "2
we have already
$
# 0 0 1 2 '
& 0.x + 0.y + 1.z = 2 ! z = 2 obtained the row-
echelon form in
Can you identify the solution? step 4.
Can we stop there
We have just obtained the solution of the and find the
system of linear equations by using the solutions for the
Gauss-Jordan Elimination Method. system of Linear
The Gauss-Jordan Elimination method has Equations? We
will look at this
reduced the augmented matrix into its
situation next.
reduced row-echelon form.

11


Let’s say we stop at Step 4. Then, we will
have the following equations to solve: Note:
! 1 1 1 1 $ 1.x + 1.y + 1.z = 1 ! x + y + z = 1 This method is
# & 0.x + 1.y + 3.z = 4 ! y + 3z = 4 the so called
# 0 1 3 4 & Gaussian
#
" 0 0 1 2 &
% 0.x + 0.y + 1.z = 2 ! z = 2
Elimination
In this case, we can solve the system of Method with
equations by using back-substitution. back-
substitution.
Step 1: Substitute z = 2 to the second
equation, we will obtain y = 4 - 3 (2) = -2
Step 2: Substitute z = 2 and y = -2 to the first
equation, we will obtain x = 1.


Matrix Notation and Terminology
• A matrix is a rectangular array of numbers.
Here is an
• The numbers in the array are called the example of size
entries in the matrix. 2 X 3 matrix, a
• The size of the matrix is described in matrix with two
terms of the number of rows and the rows and three
number of columns. columns.
• The entry that occurs in row i and column j ! 10 2 1 $
of a matrix A will be denoted by aij. # &
# 0 5 8 &
" %
• An example of a 3 x 3 matrix will have the
following entries: ! a a a $
# &
11 12 13

A = # a21 a22 a23 & = ! aij $for i, j = 1, 2, 3.
" %
# a a32 a33 &
" 31 %

12

Matrix Notation and Terminology (Cont.)

• Column Matrix: A matrix with only one column.!
#
4 $
&
Example: 2 x 1 matrix " 1 %

• Row Matrix: A matrix with only one row.
Example: 1 x 3 matrix ! 4 3 1 #
" $
• Square Matrix: A matrix with the same number of
rows and columns. Example: 2 x 2 matrix
• Two matrices are defined to be equal if they have the
same size and their corresponding entries are equal.
Example: a = 1, , b = 2, ,c = 3, and ,d = 4 ,

! a b $ ! 1 2 $
# &=# &
" c d % " 3 4 %

Matrix Operations

Let A, B, and C be matrices.
! 3 0 $ " 1 !5 % " 2 3 4 %
A=# & B=$ ' C=$ '
" 5 7 % # 0 !2 & $ 0 1 !1 '
# &

Addition: If A and B are the same size, then A + B is
the matrix obtained by adding the entries of B to the
entries of A.
Example: A + B = ! aij # + !bij # = ! aij + bij #
" $ " $ " $

! 3 0 $ ! 1 '5 $ ! 3 + 1 0 + ('5) $ ! 4 '5 $
# &+# &=# &=# &
" 5 7 % " 0 '2 % # 5 + 0 7 + ('2)
" & " 5 5 %
%


13

Matrix Operations (Cont.)
Let A, B, and C be matrices.

! 3 0 $ " 1 !5 % " 2 3 4 %
A=# & B=$ ' C=$ '
" 5 7 % # 0 !2 & $ 0 1 !1 '
# &
Subtraction: If A and B are the same size, then A - B is
the matrix obtained by subtracting the entries of B
from the entries of A.
Example: A - B = ! aij # % !bij # = ! aij % bij #
" $ " $ " $

! 3 0 $ ! 1 '5 $ ! 3 ' 1 0 ' ('5) $ ! 2 5 $
# &'# &=# &=# &
" 5 7 % " 0 '2 % # 5 ' 0 7 ' ('2)
" & " 5 9 %
%


Multiplication: If B is an m x r and C is an r x n, then
the product BC is the m x n matrix. To find the entry
in row m and column n of BC, we multiply the
corresponding entries from the row and column
together, and then add up the resulting products.
Example: r
! #
BC = !bij # ! c jk # = & % bij c jk ' = [ dik ] = D
" $" $
" j =1 $
" 1 !5 % " 2 3 4 %
BC = $ '$ '
# 0 !2 & $ 0 1 !1 '
# &
" (1)(2) + (!5)(0) (1)(3) + (!5)(1) (1)(4) + (!5)(!1) % " 2 !2 9 %
=$ '=$ '=D
$ (0)(2) + (!2)(0) (0)(3) + (!2)(1) (0)(4) + (!2)(!1) ' $ 0 !2 2 '
# & # &


14

Scalar Multiple: If C is any matrix and s is any scalar,
then the product of sC is the matrix obtained by
multiplying each entry of the matrix by s.

Example: sC = ! sc jk #
" $
" 2 3 4 % " (2)(2) (2)(3) (2)(4) %
2C = 2 $ '=$ '
$ 0 1 !1 ' $ (2)(0) (2)(1) (2)(!1) '
# & # &
" 4 6 8 %
=$ '
$ 0 2 !2
# '
&


What is a Linear Combination?
! 3 0 $ " 1 !5 % ! 2 0 $
A=# & B=$ ' E=# &
" 5 7 % # 0 !2 & " 0 1 %

Linear Combination: If A, B, and E are matrices, then
3A - B + 2E is called a linear combination.
Example:
3A ! B + 2E = " 3aij $ + " !bij $ + " 2eij $ = " 3aij ! bij + 2eij $
# % # % # % # %
" 3 0 $ " 1 !5 $ " 2 0 $
= 3& ' + (!1) & ' + 2& '
# 5 7 % # 0 !2 % # 0 1 %
" 9 0 $ " !1 5 $ " 4 0 $ " 12 5 $
=& '+& '+& '=& '
# 15 21 % # 0 2 % # 0 2 % # 15 25 %

15

What is the Transpose of a Matrix?
Transpose of a matrix: If A is any m x n matrix, then the
transpose, denoted by AT, is defined to be the n x m
matrix that results from interchanging the rows and
columns of A.

Example: ! 2 1 #
% & ! 2 4 6 8 #
4 3 & AT = ! a ji # = % &
A = ! aij # = %
" $ % " $
6 5 & % 1 3 5 9 &
% & " $
" 8 9 $
2x4
4x2


What is the Trace of a Matrix?
Trace of a matrix: If A is any square matrix, then the trace
of A, denoted by tr(A), is defined to be the sum of the
entries on the main diagonal of A. If A is not a square
matrix, then the trace of A is undefined.

Example:
! 2 1 1 2 $
# &
4 3 1 4
A=# & = ! aij $ for i, j = 1, 2, 3, 4.
# 6 5 7 2 & " %
# 8 9 1 0 &
" %

4
tr(A) = ! aii = 2 + 3 + 7 + 0 = 12
i =1


16

What have we learned?
We have learned to:

1. Represent a system of linear equations as an augmented
matrix.
2. Identify whether the matrix is in row-echelon form, reduced
row-echelon form, both, or neither.
3. Solve systems of linear equations by using the Gaussian
elimination and Gauss-Jordan elimination methods.
4. Perform matrix operations of addition, subtraction,
multiplication, and multiplication by a scalar.
5. Find the transpose and the trace of a matrix.


Credit
Some of these slides have been adapted/modified in part/whole from the
text or slides of the following textbooks:
•Anton, Howard: Elementary Linear Algebra with Applications, 9th Edition
•Rockswold, Gary: Precalculus with Modeling and Visualization, 3th Edition


17

MAC 2103
Module 2
Systems of Linear Equations and
Matrices II

1

Learning Objectives

Upon completing this module, you should be able to :

1. Find the inverse of a square matrix.
2. Determine whether a matrix is invertible.
3. Construct and identify elementary matrices;
represent A and A-1 as a product of elementary
matrices.
4. Solve systems of linear equations by using the
inverse matrix.
5. Identify diagonal, triangular and symmetric matrices.

1

Systems of Linear Equations
and Matrices II

There are four major topics in this module:

Inverses
Elementary Matrices
Systems of Equations and Invertibility
Diagonal, Triangular, and Symmetric Matrices


Inverse of a Square Matrix
Let A represent a square matrix as What happens if
follows: ad - bc = 0 ?
! a b $ The matrix is not
A=# & invertible; it has
" c d % no inverse.
The inverse of matrix A can be obtained
If matrix A has no
as follows: inverse, then A
1 " d !b % is said to be
A !1 = $ ' singular.
ad ! bc # !c a &

One important condition: ad ! bc " 0
This will let us know whether the matrix is
invertible or not.

2

Example: Finding an Inverse

Let A be a square matrix as follows:
" 1 !1 % 1 " d !b %
A=$ ' A !1 = $ '
# !3 6 & ad ! bc # !c a &
The inverse of matrix A is:
Note:
!1 1 " 6 1 %
A = $ ' The matrix is
(1)(6) ! (!1)(!3) # 3 1 & invertible
because ad - bc
1 " 6 1 % produces a
= $ '
6! 3# 3 1 & nonzero value.
How do we know
1" 6 1 % " 2 %
1
the resulting
= '=$ '
3
$ matrix is the
3# 3 1 & $ 1 1
'
# 3
& inverse of A?
=

Example: Finding an Inverse (Cont.)
How do we know the resulting matrix is the inverse of A?

" 2 1 % " 1 !1 %
A =$ !1 3
' A=$ '
$ 1 1
' # !3 6 &
# 3
&
Multiply the two matrices:
The product of a matrix and its inverse matrix is the identity
matrix I. Notice that the inverse matrix of an inverse matrix is
the original matrix.
" 2 1 %" %
" 1 !1 % " 2 1 % A A=$
!1 3
' $ 1 !1 '
AA !1 = $ '$ '
3
$ 1 3 ' # !3 6 &
1
# !3 6 & $ 1 3 ' # &
1
# &
" 1 0 % " 1 0 % !1 !1 !1
=$ !1 !1 !1 =$ ' = I = A (A )
' = I = (A ) A # 0 1 &
# 0 1 &

3

Can We Use the Gauss-Jordan Elimination
Method to Find the Inverse?

The answer is YES. How?
1. Reduce A to
This is actually a better method. However, the identity
the previous method is practical for a 2 x 2 matrix I by
elementary row
matrix.
operations.
2. Apply the same
Let’s use the Gauss-Jordan elimination elementary row
method to find the inverse of matrix A; and operations to the
identify how many elementary row operations identity matrix I
that we need to produce the inverse matrix by to produce the
converting A to I. inverse.


Method to Find the Inverse? (Cont.)
Step I: Adjoin the I to the right side of A. Step I: Adjoin
the I to the
" 1 !1 1 0 % right side of A
$ ' = [A | I ] as follows:
# !3 6
$ 0 1 &'
[A|I]
Step II: Apply elementary row operations
until the left side is reduced to the identity Step II: Apply
matrix I and the right side will be the inverse. elementary
row operations
until the left
r1 " 1 !1 1 0 % side is reduced
Label r1 (row 1)
$ '
and r2 (row 2).
r2 # !3 6
$ '
0 1 & to I and the
right side will
be the inverse.
Rev.F09
http://faculty.valenciacc.edu/ashaw/ [ I | A⁻¹ ]
Click link to download other modules. 8

4

Method to Find the Inverse? (Cont.)
We want a zero
below the leading r1 # 1 "1 1 0 &
% (
1 at r1:
3r1 + r2 ! r2 $ 0 3
% (
3 1 '

We want a r1 # 1 "1 1 0 &
leading 1 at r2: % (
3 r2 ! r2 % 0
1
1 1 1 (
$ 3
'
We want a zero
" 2 1 %
above the r2 + r1 ! r1 $ 1 0 3
'
leading 1 at r2:
r2 $ 0 1 1 1
'
We have just used the Gauss-Jordan
# 3
&
elimination method to obtain the " 2 1 %
A =$
!1
'
3
inverse; it takes three elementary row = [I| A-1 ],
operations to convert A to I. $ 1 1
'
# 3
&

Elementary Row Operations and
Elementary Matrices
By definition, an n x n matrix is called an
elementary matrix if it can be obtained from the Note: AI n = A
n x n identity matrix In by performing a single where n is the size
elementary row operation. in a n x n matrix.
A 2 x 2 matrix is called an elementary matrix if it
can be obtained from the 2 x 2 identity matrix I2 If A is an m x n
by performing a single elementary row matrix, then we
operation. will have the
A 3 x 3 matrix is called an elementary matrix if it following:
can be obtained from the 3 x 3 identity matrix I3 AI n = A
by performing a single elementary row
operation. ! 1 0 0 $ Im A = A
# &
I3 = # 0 1 0 &
# 0 0 1 &
" %

5

Elementary Matrices (Cont.)
If you remember from our previous problem, in the first elementary
row operation, we add 3 times the first row to the second row.

r1 # 1 "1 &
% (
3r1 + r2 ! r2 $ 0 3 '
The corresponding first elementary matrix is obtained by adding 3
times the first row of I2 to the second row. This is a row replacement
for the second row.
r1 ! 1 0 $
# & = I2
r2 " 0 1 %

r1 " 1 0 %
$ ' = E1
3r1 + r2 ! r2 # 3 1 &

In the second elementary row operation, we
multiply the second row by 1/3. Theorem 1.5.1
If the elementary
r1 # 1 "1 & matrix E results

r2 ! r2 % 0 1 (
1
from performing a
3 $ ' certain row
The corresponding second elementary matrix is operation on I m
obtained by multiplying the second row of I 2 by and if A is an m x n
1/3. This is a scaling of the second row. matrix, then EA is
the matrix that
r1 ! 1 0 $ results when this
# & = I2 same row
r2 " 0 1 % operation is
performed on A.
r1 " 1 0 %
$ ' = E2
1
r2 ! r2 $ 0 1 '
3
# 3
&

6

In the third elementary row operation, we add 1
times the second row to the first row. So far, we have
constructed three
r2 + r1 ! r1 " 1 0 % elementary matrices E3 ,E2
$ ' and E1, which perform
r2 # 0 1 & elementary row operations
The corresponding third elementary matrix is by multiplication on the left.
obtained by adding 1 times the second row of By multiplication of these
I2 to the first row. This is a row replacement. elementary matrices, we
obtain
r1 ! 1 0 $ [E3E2E1A | E3E2E1I]
# & = I2 =[ I | A-1] from [A | I].
r2 " 0 1 %

r2 + r1 ! r1 " 1 1 %
$ ' = E3
r2 # 0 1 &

Based on the three elementary row operations performed in our
previous problem, we can represent our Gauss-Jordan elimination as
multiplications on the left by elementary matrices:

r1 # 1 "1 1 0 &
% ( = [E1 A | E1 I ]
3r1 + r2 ! r2 $ 0 3
% (
3 1 '

r1 # 1 "1 1 0 &
% ( = [E2 E1 A | E2 E1 I ]
1
r2 ! r2 % 0 1 1 1 (
3
$ 3
'
" 2 1 %
r2 + r1 ! r1 $ 1 0 3
' = [E3 E2 E1 A | E3 E2 E1 I ] = [I | A (1 ]
r2 $ 0 1 1 1
'
# 3
&

7

How to Represent A-1 and A as a
Product of Elementary Matrices?
Now, we can represent our inverse matrix as
follows:
A-1 = E3E2E1I2 = E3E2E1

Let’s check it:

' ! 1 1 $ ! 1 0 $* ! 1 0 $
E3E2E1 = ) # 0 1 & # 0 1 &, # 3 1 &
)" 3 &, "
( %#" %+ %
! 1 1 $! $ ! 2 1 $
=#
3
&# 1 0 & = # 3
&
& =A
-1
# 0 1
&" 3 1 % # 1 1
" 3
% " 3
%


How to Represent A-1 and A as
a Product of Elementary Matrices? (Cont.)
Since E3E2E1A = I2 and
Theorem 1.5.3
A-1 = E3E2E1I2 = E3E2E1 If A is an n x n matrix,
we can obtain then the following
!1 !1 !1
A = (A-1)-1 = (E3E2E1)-1 = E1 E2 E3 statements are
equivalent:
AA-1 = E1!1E2 E3 E3 E2 E1
!1 !1 (a) A is invertible
!
!
!1 !1 !1 !1 (b)Ax = 0 has only
= E E (I )E2 E1 = E E E2 E1
1 2 1 2 the trivial solution
= E1!1 (I )E1 = E1!1E1 = I (c) The reduced row-
Thus, A can be represented as a product of echelon form of A
elementary matrices since the inverse of a is In .
elementary matrix is an elementary matrix of the (d) A is expressible
same type. There are three types, namely, as a product of
scaling, rows interchange, and row replacement. elementary
matrices.

8

How to Represent A-1 and A as a
Product of Elementary Matrices? (Cont.)
We know from our example that
Theorem 1.5.2
Every elementary
E3E2E1 ' ! 1 1 $ ! 1 0 $* ! 1 0 $ = A-1 matrix is invertible,
= )# # &,
) " 0 1 & # 0 1 &, # 3 1 &
%"
and the inverse is
( 3
%+ " % also an elementary
matrix.

( " 1 0 % " 1 0 %+ " 1 !1 %
Now, we see that E1!1E2 E3
!1 !1
= *$ '$ '- $ '
) # !3 1 & # 0 3 &, # 0 1 &
" 1 0 % " 1 !1 % " 1 !1 %
=$ '$ '=$ '=A
# !3 3 & # 0 1 & # !3 6 &

!1 !1 !1
where E1 , E2 and E3 are all elementary matrices. It is an easy
!1
check to see that E j E j = I for j = 1, 2, and 3.

Using the Inverse Matrix?

Let’s look at a system. Theorem 1.6.2
x1 ! x2 = !1 " 1 !1 % If A is an invertible n x
A=$ ' n matrix, then for !
!3x1 + 6x2 = 3 # !3 6 & each n x 1 matrix b ,
the system of !
!
! ! x1 $ ! " !1 % equations Ax = b
x=# & b=$ ' has exactly one
# x2 &
" % # 3 & solution, namely,
!
!
x = A !1b .
How to solve using the inverse matrix?
Step 1: Use the Gauss-Jordan elimination to
solve for the inverse. " 2 1 % Remember: This was
A =$
!1
'
3
solved at the beginning.
$ 1 3 '
1
# &

9

Using the Inverse Matrix? (Cont.)
!
Step 2: Solve for x by using A-1 as in
Theorem 1.6.2 as follows:
! !
x = A !1b
! 2 1 $! $
=#
3
& # '1 &
# 1 1
&" 3 %
" 3
%
! (2)('1) + ( 1 )(3) $
=# &
3

# (1)('1) + ( 3 )(3)
1
&
" %
! '1 $
=# & Thus, x1 = -1 and x2 = 0
" 0 %

How to Identify a Diagonal Matrix?

Diagonal Matrix: A square matrix in which all the
entries off the main diagonal are zero.
Examples:

" 3 0 0 % " 10 0 0 % Note:
$ ' $ ' A diagonal matrix
$ 0 8 0 ' $ 0 !1 0 ' is invertible if and
$ 0 0 !7 '
# & $
# 0 0 0 '
& only if all of its
diagonal entries
! $ are nonzero.
1 0 0 0
# & Can you identify
# 0 1 0 0 & which one of the
# 0 0 1 0 & examples is
noninvertible?
# 0 0 0 1 &
" %

10

How to Identify a Triangular Matrix?
Upper Triangular Matrix: A square matrix in Note:
which all the entries below the main A triangular
diagonal are zero. matrix is
Lower Triangular Matrix: A square matrix in invertible if and
which all the entries above the main only if all of its
diagonal are zero. diagonal entries
are nonzero -
Triangular Matrix: A square matrix that is just like the
either an upper triangular matrix or a lower diagonal matrix.
triangular matrix.
! * * * $ ! * 0 0 $
Examples: # 0 &
* * &
# &
# # * * 0 &
# 0 0 * &
" % # * * * &
" %
Upper Triangular Matrix Lower Triangular Matrix

Recall: Transpose of a Matrix
Transpose of a matrix: If A is any m x n matrix, then the
transpose, denoted by AT, is defined to be the n x m
matrix that results from interchanging the rows and
columns of A.

Example: ! 2 1 #
% & ! 2 4 6 8 #
4 3 & AT = ! a ji # = % &
A = ! aij # = %
" $ % " $
6 5 & % 1 3 5 9 &
% & " $
" 8 9 $
2x4
4x2


11

How to Identify a Symmetric Matrix?
Symmetric Matrix: A square matrix A is Theorem 1.7.2
called symmetric if A = AT. If A and B are
symmetric matrices
with the same size,
Examples: and if s is a scalar,
then:
" 1 !2 5 % 1. AT is symmetric.
$ '
$ !2 1 0 ' 2. A + B and A - B are
symmetric.
$ 5 0 1 '
# & 3. sA is symmetric.
" 6 0 0 0 %
$ ' Theorem 1.7.3
$ 0 1 0
3 0 ' If A is an invertible
$ 0 0 5 0 ' symmetric matrix,
$ ' than A-1 is
# 0 0 0 !2 & symmetric.

What Have We Learned?
We have learned to:

1. Find the inverse of a square matrix.
2. Determine whether a matrix is invertible.
3. Construct and identify elementary matrices;
represent A and A-1 as a product of elementary
matrices.
4. Solve systems of linear equations by using the
inverse matrix.
5. Identify diagonal, triangular and symmetric
matrices.

12

Credit
Some definitions and theorems have been adapted/modified in part/whole
from the following textbook:
•Anton, Howard: Elementary Linear Algebra with Applications, 9th Edition


13

MAC 2103
Module 3
Determinants

1

Learning Objectives


1. Determine the minor, cofactor, and adjoint of a matrix.
2. Evaluate the determinant of a matrix by cofactor
expansion.
3. Determine the inverse of a matrix using the adjoint.
4. Solve a linear system using Cramer’s Rule.
5. Use row reduction to evaluate a determinant.
6. Use determinants to test for invertibility.
7. Find the eigenvalues and eigenvectors of a matrix.


1

Determinants


Determinants by Cofactor Expansion
Evaluating Determinants by Row Reduction
Properties of the Determinant


What is a Determinant?
A determinant is a real number associated with a square matrix.

a b
=
c d

Determinants are commonly used to test if a matrix is
invertible and to find the area of certain geometric figures.

2

How to Determine if a Matrix is Invertible?

The following is often used to determine if a square matrix is invertible.


Example
Determine if A-1 exists by computing the determinant of
the matrix A.
a) b)

Solution
!5 9
a) det(A) = 4 !1 = (!5)(!1) ! (4)(9) = !31
A-1 does exist

9 3
b) det(A) = = (9)(!1) ! (!3)(3) = 0
!3 !1
A-1 does not exist


3

What are Minors and Cofactors?
We know we can find the determinants of 2 x 2 matrices; but can we find the
determinants of 3 x 3 matrices, 4 x 4 matrices, 5 x 5 matrices, ...?

In order to find the determinants of larger square matrices, we need to
understand the concept of minors and cofactors.


Example of Finding Minors and Cofactors
Find the minor M11 and cofactor A11
for matrix A.
Solution
To obtain M11 begin by crossing out the first row and
column of A.
The minor is equal to
det B = −6(5) − (−3)(7)
(−
= −9

(−
Since A11 = (−1)1+1M11, A11 can
be computed as follows:
A11 = (−1)2 (−9) = −9
(−

4

How to Find the Determinant of Any
Square Matrix?
Once we know how to obtain a cofactor, we can find the determinant of any
square matrix. You may pick any row or column, but the calculation is easier if
some elements in the selected row or column equal 0.

n n

!a A ij ij
or !a A ij ij
i =1 j =1
for any column j for any row i

Example of Finding the Determinant
by Cofactor Expansion

Find det A, if

Solution To find the determinant of A, we can select any
row or column. If we begin expanding about the first
column of A, then
det A = a11A11 + a21A21 + a31A31.
Now, try to
A11 = −9 from the previous example
find the
A21 = −12 and A31 = 24 determinant
of A by
det A = a11A11 + a21A21 + a31A31 expanding
the first row
= (−8)(−9) + (4)(−12) + (2)(24) of A.
= 72

5

How to Find the Adjoint of a Matrix?

The adjoint of a matrix can be found by taking the
transpose of the matrix of cofactors from A.
In our previous example, we have found the cofactors
A11, A21, A31. If we continue to solve for the rest of the
cofactors for matrix A, namely A12, A22, A32 , A13, A23, and
A33 , then we will have a 3 x 3 matrix of cofactors from A
as follows:
! A11 A12 A13 $
# &
# A21 A22 A23 &
# A A32 A33 &
" 31 %

How to Find the Adjoint of a Matrix? (Cont.)

The transpose of this 3 x 3 matrix of cofactors from A is
called the adjoint of A, and it is denoted by Adj(A).

! A11 A21 A31 $
# &
Adj(A) = # A12 A22 A32 &
# A A23 A33 &
" 13 %

What are we going to do with this Adj(A)? We can use it
to help us find the A-1 if A is an invertible matrix.


6

How to Find A-1 Using the Adjoint of a Matrix?

Theorem 2.1.2: If A is an invertible matrix, then
1
A !1 = Adj(A)
det(A)

Note:
1. The square matrix A is invertible if and only if det(A) is not
zero.
2. If A is an n x n triangular matrix, then det(A) is the product of
the entries on the main diagonal of the matrix (Theorem
2.1.3.)

What is Cramer’s Rule?

Cramer’s Rule is a method that utilizes determinants to solve systems of linear
equations. This rule can be extended to a system of n linear equations in n
unknowns as long as the determinant of the matrix is non-zero.

7

Example of Using Cramer’s Rule to
Solve the Linear System
Use Cramer’s rule to solve
the linear system.
Solution In this system a1 = 1, b1 = 4, c1 = 3, a2 = 2, b2
= 9 and c2 = 5


Example of Using Cramer’s Rule to
Solve the Linear System (cont.)

E = 7, F = −1 and D = 1

The solution is

Note that Gaussian elimination with backward substitution is usually
more efficient than Cramer’s Rule.


8

What Are the Limitations on the Method of
Cofactors and Cramer’s Rule?

The main limitations are as follow:

1. A substantial number of arithmetic operations are needed to
compute determinants of large matrices.
2. The cofactor method of calculating the determinant of an n x
n matrix, n > 2, generally involves more than n! multiplication
operations.
3. Time and cost required to solve linear systems that involve
thousands of equations in real-life applications.

Next, we are going to look at a more efficient method to find the
determinant of a general square matrix.

Evaluating Determinants by Reducing the
Matrix to Row-Echelon Form
Just keep these in
Let A be a square matrix. (See Theorem 2.2.3) mind when A is a
(a) If B is the matrix that results from scaling square matrix:
by a scalar k, then 1. det(A)=det(AT).
det(B) = k det(A). 2. If A has a row of
zeros or a column
(b) If B is the matrix that results from either
of zeros, then
rows interchange or columns interchange, det(A)=0.
then
3. If A has two
det(B) = - det(A). proportional rows
(c) If B is the matrix that results from row or two proportional
replacement, then columns, then
det(A)=0.
det(B) = det(A).

9

How to Evaluate the Determinant by Row
Reduction?
Let’s look at a square matrix A.
! 0 3 1 $
# &
A=# 1 1 2 &
# 3 2 4 &
" %
We can find the determinant by reducing it
into row-echelon form.
Step 1: We want a leading 1 in row 1. We
can interchange row 1 and row 2 to
accomplish this.
1 1 2 1 1 2
det(A) = 0 3 1 = ! 0 3 1
3 2 4 3 2 4


Reduction? (Cont.)
Step 2: We want a leading 1 in row 2. We From Step 1:
can take a common factor of 3 from row 2 to
accomplish this (scaling). 1 1 2
1 1 2 det(A) = ! 0 3 1
det(A) = !3 0 1 1
3 3 2 4
3 2 4
Step 3: We want a zero at both row 2 and
row 3 below the leading 1 in row 1. We can
add -3 times row 1 to row 3 to accomplish
this (row replacement).
1 1 2
det(A) = !3 0 1 1
3

0 !1 !2

10

Reduction? (Cont.)
Step 4: We want a zero below the leading 1 From Step 3:
in row 2. We can add row 2 to row 3 to
accomplish this (row replacement). 1 1 2
1 1 2 det(A) = !3 0 1 1
3
1
det(A) = !3 0 1 3
0 !1 !2
!5
0 0 3
Remember: If A is an n x
n triangular matrix, then
Step 5: We want a leading 1 in row 3. We det(A) is the product of
take a common factor of -5/3 from row 3 to the entries on the main
accomplish this (scaling). diagonal of the matrix.
1 1 2
" !5 % " !5 %
det(A) = (!3) $ ' 0 1 1 = (!3) $ ' (1) = 5
# 3& 3
# 3&
0 0 1

Let’s Look at Some Useful
Basic Properties of Determinants
• Let A and B be n x n matrices and k is any
Question:
scalar. Then,
Is det(A+B) =
det(kA) = k n det(A) det(A) + det(B) ?
det(AB) = det(A)det(B)
Remember: If A is
an n x n triangular
• If A is invertible, then matrix, then
1 det(A) is the
det(A !1 ) = product of the
det(A) entries on the
This is because A-1A=I, det(A-1A) =det(I) =1; main diagonal of
det(A-1) det(A) = 1, and so the matrix.
1
det(A !1 ) = , det(A) " 0.
det(A)http://faculty.valenciacc.edu/ashaw/

11

What are Eigenvalues and EigenVectors?
An eigenvector of an n x n matrix A!is a nontrivial
! !
(nonzero) vector x such that Ax = ! x , where ! is
a scalar called an eigenvalue.

Linear systems of this form can be rewritten as follows:
!
! !
! x " Ax = 0
! ! !
(! I " A) x = Bx = 0 !
The system has a nontrivial solution x if and only if
det(! I " A) = det(B) = 0.
This is the so called characteristic equation of A and ! !
therefore B has no inverse, and the linear system Bx =0
has infinitely many solutions.

Example
! !
Express the following linear system in the form (! I " A) x = 0.
x1 + 2x2 = ! x1 Find the characteristic equation, eigenvalues
2x1 + x2 = ! x2 and eigenvectors corresponding to each of the
eigenvalues.
The linear system can be written in matrix form as
! 1 2 $ ! x1 $ ! x1 $
! 1 2 $ ! ! x1 $
# & # & = '# &
with A = # &, x = # &
" 2 1 % # x2 &
" % # x2 &
" % " 2 1 % # x2 &
" %
" x1 % " 1 2 % " x1 % " 0 %
!$ '($ '$ '=$ '
$ x2 ' # 2 1 & $ x2 ' # 0 &
# & # &
" 1 0 % " x1 % " 1 2 % " x1 % " 0 %
!$ '$ '($ '$ '=$ '
# 0 1 & $ x2 ' # 2 1 & $ x2 ' # 0 &
# & # &

12

Example (Cont.)
) " ! 0 % " 1 2 %, " x1 % " 0 %
+ $ 0 ! ' ( $ 2 1 '. $ x ' = $ 0 '
*# & # &- $ 2 ' #
# & &
" ! ( 1 (2 % " x1 % " 0 %
$ '$ '=$ '
# (2 ! ( 1 & $ x2 ' # 0 &
# &
! !
which is of the form (! I " A) x = 0.

# ! " 1 "2 &
Thus, !I " A = % (.
$ "2 ! " 1 '

Can you tell what is the characteristic equation for A?


Example (Cont.)
The characteristic equation for A is

det(! I " A) = 0
! " 1 "2
=0
"2 ! " 1

or (! " 1)(! " 1) " ("2)("2) = 0
(! " 1)2 " 4 = 0
! 2 " 2! + 1 " 4 = 0
! 2 " 2! " 3 = 0
(! " 3)(! + 1) = 0

13

Example (Cont.)
Thus, the eigenvalues of A are: !1 = 3, !2 = "1
! !
By definition, x is an eigenvector of A! if and only if x
!
is a nontrivial solution of (! I " A) x = 0.
that is # ! " 1 "2 & # x1 & # 0 &
% (% (=% (
$ "2 ! " 1 ' % x2 ( $ 0 '
$ '
If ! = 3 , then we have
" 2 !2 % " x1 % " 0 %
$ '$ '=$ '
# !2 2 & $ x2 ' # 0 &
# &
Thus, we can form the augmented matrix and solve
by Gauss Jordan Elimination.


Example (Cont.)
Let’s form the augmented matrix and solve by
Gauss Jordan Elimination.
r1 " 2 !2 0 %
$ '
r2 # !2 2 0 &
1
2 r1 ( r1 " 1 !1 0 %
$ '
r2 # !2 2 0 &

r1 " 1 !1 0 %
$ '
2r1 + r2 ( r2 # 0 0 0 &

Thus, x1 ! x2 = 0
x1 = x2 = t a free variable, t !("#, #)

14

Example (Cont.)
Solving this system yields: x1 = t
x2 = t

So the eigenvectors corresponding to !1 = 3
are the nontrivial solutions of the form !
! x1 $ ! t $ ! 1 $
x1 = # &= # &=t# &
# x2 & " t %
" % " 1 %
Similarly, if ! = "1 , then we have
" !2 !2 % " x1 % " 0 %
$ '$ '=$ '
# !2 !2 & $ x2 ' # 0 &
# &
" !2x1 ! 2x2 % " 0 %
$ '=$ '
$ !2x1 ! 2x2
# ' # 0 &
&

Example (Cont.)
Let’s form the augmented matrix and solve by
Gauss Jordan Elimination.
r1 " !2 !2 0 %
$ '
r2 # !2 !2 0 &

! 1 r1 ( r1 " 1 1
2 0 %
$ '
r2 # !2 !2 0 &

r1 " 1 1 0 %
$ '
2r1 + r2 ( r2 # 0 0 0 &
x + x2 = 0
Thus, 1
x1 = !x2 = t
x1 = t, x2 = !t,t "(!#, #)

15

Example (Cont.)
Solving this system yields: x1 = t
x2 = !t

So the eigenvectors corresponding to !2 = "1
are the nontrivial solutions of the form

! ! x1 $ ! t $ ! 1 $
x2 = # &=# &=t# &
# x2 & " 't %
" % " '1 %


What have we learned?
We have learned to:

1. Determine the minor, cofactor, and adjoint of a matrix.
2. Evaluate the determinant of a matrix by cofactor expansion.
3. Determine the inverse of a matrix using the adjoint.
4. Solve a linear system using Cramer’s Rule.
5. Use row reduction to evaluate a determinant.
6. Use determinants to test for invertibility.
7. Find the eigenvalues and eigenvectors of a matrix.


16

Credit
Some of these slides have been adapted/modified in part/whole from the
text or slides of the following textbooks:
• Anton, Howard: Elementary Linear Algebra with Applications, 9th Edition
• Rockswold, Gary: Precalculus with Modeling and Visualization, 3th Edition


17

MAC 2103
Module 4
Vectors in 2-Space and 3-Space I

1

Learning Objectives

In this module, we apply our earlier ideas specifically to vectors
in 2-space, ℜ2, (in the xy-plane) in two dimensions and to
vectors in 3-space, ℜ3,(in the xyz-space) in three
dimensions.


1

Learning Objectives (Cont.)
1. Determine the components of a vector in ℜ2 and ℜ3.
2. Perform vector addition, subtraction, and scalar multiplication in
ℜ2 and ℜ3.
3. Find the norm of a vector and the distance between points in ℜ2
and ℜ3.
4. Find the dot product of two vectors in ℜ2 and ℜ3.
5. Use the dot product to find the angle between two vectors in ℜ2
and ℜ3.
6. Find the projection of a vector onto another vector in ℜ2 and ℜ3,
and express the original vector as a sum of two orthogonal
vectors.
7. Find the distance between a point and a line in ℜ2 and ℜ3.


Vectors in ℜ2 and ℜ3


Introduction to Vectors (Geometric)
Norm of a Vector; Vector Operations
Dot Product; Projections


2

What are Vectors in ℜ2 and ℜ3?
• Vectors can be represented as directed line segments
or arrows in ℜ2 and ℜ3.
• The direction of the arrow specifies the direction of the
vector.
• A vector that starts from an initial point A and !!
!"
terminates at a point B can be represented as AB .
• A vector is usually denoted in lowercase boldface type
(like v) in the textbook or with an arrow above it when
! """
we write it by hand. For example: v = v = AB.
!

!v "
!
A B
!
!
#v
$! B
A !

What are Vectors in ℜ2 and ℜ3? (Cont.)
• The magnitude of the vector is
the length of the vector.
• The vector of length zero is
called the zero vector.
• Vectors with the same
magnitude and same direction
are equal to each other.
• A vector v in standard position
has its starting point at the
origin. The coordinates (v1, v2)
of the terminal point of v are Note: The negative of vector v
called the components of v. is defined to be the vector that
! has the same magnitude as v
v = v = (v1 , v2 ) but is oppositely directed.

3

If s is any scalar, then a vector of the form sv
is called a scalar multiple of v.
!
sv = sv = s(v1 , v2 ) = (sv1 , sv2 )
For example, if v = (2,-7) and s =- 5, then

!
!5 v = !5(v1 , v2 ) = (!5v1 , !5v2 ) = (!10, 35)



• If v and u are any two vectors in standard
position, then the sum and difference of the
two vectors is also a vector. It’s also a vector
in standard position.
! !
v + u = v + u = (v1 , v2 ) + (u1 ,u2 ) = (v1 + u1 , v2 + u2 )
! !
v ! u = v ! u = (v1 , v2 ) ! (u1 ,u2 ) = (v1 ! u1 , v2 ! u2 )


4

What are the Components of a Vector in ℜ3?

A !v " B
!
!
!!!
"
If the initial point of AB is A(x1,y1,z1) and the terminal !!!
!!!
" "
point of AB is B(x2,y2,z2) in ℜ3, then the components of AB
can be obtained by subtracting the coordinates of the
initial point from the coordinates of the terminal point.
! """ !
v = v = AB = (x2 ! x1 , y2 ! y1 , z2 ! z1 )
!!!
"
Example: Suppose the initial point of AB is A(1,-2,5)
and terminal point"is! B(-1,4,9), then the components of
! ""
the vector v = v = AB = (!2, 6, 4) . We see that the vector
!!!
"
AB is equal to the vector v in standard position.

Example
Suppose !
u = (!5,1, 6)
!
v = (1, 0, !8)
! !
Find the components of 7u ! 2 v.
! ! ! !
7u ! 2 v = 7u + (!2)v = 7(!5,1, 6) + (!2)(1, 0, !8)
= ((7)(!5) + (!2)(1),(7)(1) + (!2)(0),(7)(6) + (!2)(!8))
= (!37, 7, 58)
Note: In chapter 1, we would represent these vectors as column matrices:

" !5 % " 1 %
!
v = $ 0 ' = " 1 0 !8 %
T
! $ ' T
u = $ 1 ' = " !5 1 6 %
# & $ ' # &
$
# 6 '& $ !8 '
# &

5

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