MAT222 - INTRODUCTION TO LINEAR ALGEBRA WITH APPLICATIONS | ROW-REDUCED ECHELON FORM AND VECTOR SPACES.pdf

‣ A lot of matrix mathematics are handled better when using the row-
reduced echelon form. Some of these, include finding the inverses of
matrices, solving a series of equations, finding column and null spaces
and many more.
‣ The row-reduced echelon form was derived from method of finding the
inverse matrix using the gaussian-elimination method. Hence, the row-
reduced echelon form is also known as the Gauss-Elimination method.
‣ In this session, we will not bother to look at finding inverses or solving
equations using the row-reduced echelon form in detail, we will rather
look at the methodology, and then proceed to column and null spaces.
Tuesday, 07 May 2024
Compiled by Josophat Makawa | Student - B.Sc Mathematics | University of Malawi 2
INTRODUCTION

‣ There are two things occurring here, the echelon form and the reduced
echelon form.
‣ In Row echelon form, the matrix has to satisfy the following properties.
1. All non-zero rows are above any rows of all zeros.
2. Each leading entry (i.e. left most nonzero entry) of a row is in a
column to the right of the leading entry of the row above it.
3. All entries in a column below a leading entry are zero.
‣ In general, we will define some of the general forms of matrices in row
echelon form.
ROW-REDUCED ECHELON FORM

‣ Below are the general forms of the row echelon form.
a)
∎ ∗ ∗ ∗ ∗
0 ∎ ∗ ∗ ∗
0 0 0 0 0
0 0 0 0 0
c.)
0 ∎ ∗ ∗ ∗ ∗ ∗
0 0 0 ∎ ∗ ∗ ∗
0 0 0 0 ∎ ∗ ∗
0 0 0 0 0 0 ∎
0 0 0 0 0 0 0
b)
∎ ∗ ∗
0 ∎ ∗
0 0 ∎
0 0 0
NOTE: Make sure that we are only having zeros below
every leading non-zero entry in a column. And
to the left of a leading non-zero entry in a given
row, we also have zeros.
∎ Non-Zero
entry
∗ Zero or non-
zero entry

‣ In row-reduced echelon form, we make more changes to the leading entries
and their respective columns.
‣ So, to the first 3 properties of row echelon form, add the following properties.
4. The leading entry in each nonzero row is 1.
5. Each leading 1 is the only nonzero entry in its column.
‣ In general,
1 ∗ 0 0 ∗ ∗ 0 0 ∗
0 0 1 0 ∗ ∗ 0 0 ∗
0 0 0 1 ∗ ∗ 0 0 ∗
0 0 0 0 0 0 1 0 ∗
0 0 0 0 0 0 0 1 ∗
is in row-reduced echelon form.

‣ In this session we will deal more with the row-reduced echelon form and
we will often meet a number of terms. For example, pivot position, pivot
and pivot column are the most common terms to be familiar with.
▪ pivot position: a position of a leading entry in an echelon form of the
matrix.
▪ pivot: a nonzero number that either is used in a pivot position to
create 0’s or is changed into a leading 1, which in turn is used to
create 0’s in row-reduced echelon form.
▪ pivot column: a column that contains a pivot position.

‣ Let’s try to identify if matrices are in row-reduced echelon form or not.
1.
0 0 0 0
0 1 0 0
0 0 1 −2
2.
1 0 0
0 1 0
0 0 1
3.
1 2 0 0 4
0 0 1 0 9
0 0 0 1 −1
0 0 0 0 0
(this matrix is left for your practice)
This matrix is not in row-reduced echelon form
because a row with all 0’s is above row(s) with some
non-zero entries.
This matrix in row-reduced echelon form because each
column is a pivot column and every leading pivot
(leading entry) is a 1.

FINDING REDUCED ECHELON FORMS OF MATRICES
‣ In the conversion of a matrix into row-reduced echelon form (RREF), we
create an augmented matrix depending on the nature of a calculation we
need to make.
‣ For example, if we want to find the inverse matrix, we augment the given
square matrix (A𝑛) and its identity matrix (I𝑛) in the form A𝑛 I𝑛 . By
converting A𝑛 into RREF, the matrix I𝑛 is converted into the inverse of
A𝑛. In the same way, a system of equations Aത
x = ത
b, we augment the
coefficient matrix A, and the solution vector ത
b, in the form A ത
b . In
RREF we find the column vector ത
x.

‣ To convert a matrix into RREF, we use the following three rules
simultaneously.
1. Interchanging the positions of any two rows.
2. Multiplying by a non-zero scalar
3. Adding or subtracting other rows or adding or subtracting the scalar
multiples of other rows.
‣ We should always note that we can’t multiply by 0 and we also can’t add
or subtract a scalar. We should also note that any rule we choose is
applied on the entire row and not a specific point or pivot.

Example 1: Find the inverse matrix A given that
A =
3 1
4 2
‣ This a 2 × 2 matrix, thus the augmented matrix will look as follows
3 1
4 2
1 0
0 1
= A I
‣ The idea here is to convert A into an identity matrix I, and I into A−1
3 1
4 2
1 0
0 1
←
1
3
𝑅1
We want to make the first point be 1 (pivot).
Since no other row in this column has 1, we
can’t interchange. Hence, a scalar product to
row 1

‣ After multiplying
1
3
to the entire row, we get
1
1
3
4 2
1
3
0
0 1
1
1
3
0
2
3
1
3
0
−
4
3
1
1
1
3
0 1
1
3
0
−2
3
2
← 𝑅2 − 4𝑅1
We must have all zeros below a pivot,
1. Thus, 4𝑅1 will work since we can’t
subtract by a scalar.
←
3
2
𝑅2
We need to create a pivot for row 2 by
converting
2
3
into 1. Scalar multiplication.
← 𝑅1 −
1
3
𝑅2
A pivot column must only contain a
1. We must convert
1
3
to a 0 but we
shouldn’t disturb the pivot for row 1.

1 0
0 1
−
1
3
−
1
2
−2
3
2
= I A−1
Therefore, 𝐴−1 =
−
1
3
−
1
2
−2
3
2
; your job is to test this using 𝐴 ∙ 𝐴−1 = 𝐼.
‣ As mentioned earlier, more of our attention will be on the process of
finding the row-reduced echelon form of any given matrix.
At this point we have found row-
reduced echelon form of A. For
square matrices, the RREF is their
respective identity matrices.

‣ Some basic points that may help in critical situations when we don’t
know where to start from may be;
▪ Change the first entry in row 1 to 1 by either interchanging the
positions with any other row that has 1 in column 1 or multiplying by
a scalar. Once this is a pivot (1), we must change everything in the
column to zeros.
▪ Change the last row to zeros or the last entry to the pivot if possible. If
not, start with the rows below the first row, defining pivots where
possible. Make sure, you finish with the first row unless otherwise,
being in a rush may affect the pivot point in column 1.

‣ Visually, Given A𝑚×𝑛;
𝑎1,1 ⋯ 𝑎1,𝑛
⋮ ⋱ ⋮
𝑎𝑚,1 ⋯ 𝑎𝑚,𝑛
Start with changing 𝑎1,1
to 1 by any rule. Then
change everything in
column 1 to 0’s.
Change the columns that
have a pivot to zeros one
by one where possible.
1 3
Change the last row into
all zeros and stop if and
only if you have a pivot at
any point in that row.
Change to zeros any
entry in any column
that contains a pivot.
Note that this is simply a proposed
idea, it still needs proper analysis of
any given matrix.
4
2

Example 2: Write the matrix 𝐴 =
0 3 −6 6 4 −5
3 −7 8 −5 8 9
3 −9 12 −9 6 15
in RREF.
0 3 −6 6 4 −5
3 −7 8 −5 8 9
3 −9 12 −9 6 15
3 −9 12 −9 6 15
3 −7 8 −5 8 9
0 3 −6 6 4 −5
← 𝑅1 ↔ 𝑅3 We have 0 on a pivot point.
We will interchange with row
3 (row 2 will give fractions)
←
1
3
𝑅1
Convert the pivot point to 1. If we
had interchanged with row 2, this
operation could have brought
fractions. Always pay attention.

1 −3 4 −3 2 5
3 −7 8 −5 8 9
0 3 −6 6 4 −5
1 −3 4 −3 2 5
0 2 −4 4 2 −6
0 3 −6 6 4 −5
1 −3 4 −3 2 5
0 1 −2 2 1 −3
0 3 −6 6 4 −5
← 𝑅2 − 3𝑅1
Making zeros for all terms
below a pivot in a pivot
column.
←
1
2
𝑅2
Creating a pivot in second
column. Our consideration is on
converting 2 to 1.
← 𝑅3 − 3𝑅2
Making zeros for all terms
below a pivot in a pivot
column. Start with terms below
and lastly finish with entries in
row 1.

1 −3 4 −3 2 5
0 1 −2 2 1 −3
0 0 0 0 1 4
1 −3 4 −3 2 5
0 1 −2 2 0 −7
0 0 0 0 1 4
1 −3 4 −3 0 −8
0 1 −2 2 0 −7
0 0 0 0 1 4
← 𝑅2 − 𝑅3
We have a new pivot in row 3.
we need to change everything in
this column to 0’s above 1.
← 𝑅1 − 2𝑅3
We have now changed all
terms in pivot columns in
lower rows. Let’s do for row 1
← 𝑅1 + 3𝑅3
Make sure that all pivot
columns have been treated
such that all terms that are not
pivots are 0’s.

‣ Hence, we finally get to the final row reduced echelon form of the
given matrix. And we have;
𝐴∗ =
1 0 −2 3 0 −24
0 1 −2 2 0 −7
0 0 0 0 1 4
‣ We may also note that every row operation targets at least one entry
changing it to either 1 or 0. However, each operation affects the entire
row.
Pivots

‣ This algorithm provides a method for using row operations to take a matrix to
its echelon form. We begin with the matrix in its original form.
1. Starting from the left, find the first non-zero column. This is the first
pivot column, and the position at the top of this column will be the
position of the first pivot entry. Switch rows if necessary to place a non-
zero number in the first pivot position.
2. Use row operations to make the entries below the first pivot entry (in the
first pivot column) equal to zero.
3. Ignoring the row containing the first pivot entry, repeat steps 1 and 2 with
the remaining rows. Repeat the process until there are no more non-zero
rows left. Back to slide 29

Example 3: Find the reduced echelon form of 𝐴 =
1 2 1 3 2
1 3 6 0 2
3 7 8 6 6
1 2 1 3 2
1 3 6 0 2
3 7 8 6 6
𝑅2: (𝑅2 − 𝑅1)
1 2 1 3 2
0 1 5 −3 0
3 7 8 6 6
𝑅3: (𝑅3 − 3𝑅1)
1 2 1 3 2
0 1 5 −3 0
0 1 5 −3 0
𝑅3: (𝑅3 − 𝑅2)
1 2 1 3 2
0 1 5 −3 0
0 0 0 0 0

1 2 1 3 2
0 1 5 −3 0
0 0 0 0 0
𝑅1: (𝑅1 − 2𝑅2)
1 0 −9 9 2
0 1 5 −3 0
0 0 0 0 0
‣ In this question, the first column is a non-zero column, hence, it is our
first pivot column. Since the first entry was already a 1, nothing can be
done here than making everything in this column equal 0.
‣ In column 2 we already get another pivot in column 2. The process
continues until when all the pivot columns have 1 (pivot) and zeros
everywhere. In addition, all entries in a row on the left of a 1 (pivot) are
zeros.

A. Determine if the given matrices are in reduced echelon form or not.
1.
1 0 0 4
0 1 0 7
0 0 1 −1
2.
0 0 0
0 0 1
0 1 0
1 0 0
3.
1 3 −3 3
0 0 0 0
4.
1 0 2
0 1 −1
0 0
1
2
NOTE: Remember to describe the
properties in your critics.
Understand the reasons why a given
matrix is in row-reduced echelon
form or not.

B. Find the echelon and reduced echelon forms of the following matrices
1.
1 2 3 4
4 5 6 7
6 7 8 9
2.
−2 −3 −2
3 −2 −2
3 −2 −1
−1 −1 −2
3.
1 3 −3 −3
0 0 −2 2
4.
1 2 0
1 3 3
−1 0 −1
−3 0 0
5.
0 0 −2 0 7 12
2 4 −10 6 12 28
2 4 −5 6 −5 −1

MAT222 - INTRODUCTION TO LINEAR ALGEBRA WITH APPLICATIONS | ROW-REDUCED ECHELON FORM AND VECTOR SPACES.pdf

‣ A system of linear equations consists of a series of linear equations which
is converted and solved using matrix mathematics.
‣ The general form of these equations is Aത
x = ത
b where A is the matrix of
coefficients in the given equations, ത
x is the column vector of all the
variables available in the equations and ത
b is the column vector of the
solution associated with given equations.
‣ We have two types of equations based on the nature of the solutions in
the equations as well as the nature of the values of the variables. These
include the non-homogenous equations and homogenous equations.
LINEAR SYSTEM OF EQUATIONS

‣ As a matter of definition; A linear system of equations Aത
x = ത
b is called
homogeneous if ത
b = 0, and non-homogeneous if b ≠ 0.
‣ Notice that ത
x = 0 is always solution of the homogeneous equations. That
is, 𝑥1 = 0, 𝑥2 = 0, … 𝑥𝑛 = 0. This solution is called the trivial solution.
‣ If there are other solutions, they are called nontrivial solutions. Because a
homogeneous linear system always has the trivial solution, there are only
two possibilities for its solutions:
• The system has only the trivial solution.
• The system has infinitely many solutions in addition to the trivial
solution

‣ In nontrivial cases, we tend to find a non-zero vector that satisfies the
equation Aത
x = 0. The homogeneous equation Aത
x = 0 has a nontrivial
solution if and only if the equation has at least one free variable.
‣ For non-homogenous equations Aത
x = ത
b, we have various forms and their
own solutions. The other methods for solving the equations Aത
x = ത
b like
the inverse method and Cramer's rule are mostly applicable if and only if
the coefficient matrix A is a non-singular square matrix (det A ≠ 0).
‣ If the coefficient matrix is not a square matrix there is a slight difference
in the nature of the solutions.

‣ Consider the general for a system of linear equations and their respective
augmented matrix when using row-reduced echelon form.
Given the following set of linear equations
𝑎11𝑥1 + 𝑎12𝑥2 + ⋯ + 𝑎1𝑛𝑥𝑛 = 𝑏1
⋮
𝑎𝑚1𝑥1 + 𝑎𝑚2𝑥2 + ⋯ + 𝑎𝑚𝑛𝑥𝑛 = 𝑏𝑚
𝑎11 ⋯ 𝑎1𝑛
⋮ ⋱ ⋮
𝑎𝑚1 ⋯ 𝑎𝑚𝑛
𝑏1
⋮
𝑏𝑚
Definition: Augmented matrix of a system of linear equations

Example 4: Solve the linear equations: 𝑥1 + 4𝑥2 + 3𝑥3 = 11
2𝑥1 + 10𝑥2 + 7𝑥3 = 27
𝑥1 + 𝑥2 + 2𝑥3 = 5
‣ The augmented matrix is
1 4 3
2 10 7
1 1 2
11
27
5
‣ To carry the augmented matrix to reduced echelon form, we will use the
method discussed earlier (slide 19). Notice that the first column is non-zero,
so this is our first pivot column. The first entry in the first row, 1, is the first
pivot entry and we will proceed with the rest.

1 4 3
2 10 7
1 1 2
11
27
5
1 4 3
0 2 1
1 1 2
11
5
5
1 4 3
0 2 1
0 −3 −1
11
5
−6
‣ Changing the last row to zeros except the last entry which changes to 1.
1 4 3
0 2 1
0 −3 −1
11
5
−6
1 4 3
0 2 1
0 −6 −2
11
5
−12
1 4 3
0 2 1
0 0 1
11
5
3
‣ Changing the remaining pivot columns to zeros in row 2.
1 4 3
0 2 1
0 0 1
11
5
3
1 4 3
0 2 0
0 0 1
11
2
3
1 4 3
0 1 0
0 0 1
11
1
3
← 𝑅2 − 2𝑅1
← 𝑅3 − 𝑅1
← 2𝑅3
← 𝑅3 + 3𝑅2
← 𝑅2 − 𝑅3 ←
1
2
𝑅2

‣ Let’s finish with row 1, make 0’s on all pivot columns
1 4 3
0 1 0
0 0 1
11
1
3
1 4 0
0 1 0
0 0 1
2
1
3
1 0 0
0 1 0
0 0 1
−2
1
3
‣ Since all the rows have been entirely reduced, each pivot holds a value
of one variable.
‣ Hence, ത
x =
𝑥1
𝑥2
𝑥3
=
−2
1
3
; in other words 𝑥1 = −2, 𝑥2 = 1 and 𝑥3 = 3.
← 𝑅1 − 3𝑅3
← 𝑅1 − 4𝑅2

Example 5: Solve the following equations: 3𝑥1 − 𝑥2 + 5𝑥3 = 8
𝑥2 − 10𝑥3 = 1
6𝑥1 − 𝑥2 = 17
3 −1 5
0 1 −10
6 −1 0
8
1
17
3 −1 5
0 1 −10
0 1 −10
8
1
1
3 −1 5
0 1 −10
0 0 0
8
1
0
‣ This is the echelon form not the reduced echelon form. Converting this to
reduced echelon form will consists of multiplying row 1 by
1
3
hence, we will
have fractions. In this form, we can deduce equations that satisfy the given
equation solutions.
← 𝑅3 − 2𝑅1 ← 𝑅3 − 𝑅2

‣ In the echelon form,
𝑥1 𝑥2 𝑥3
3 −1 5
0 1 −10
0 0 0
𝑏
8
1
0
‣ Variables that are represented by pivot columns are called pivot
variables and the variables that aren’t represented by pivot columns are
called free variables.
‣ In this case, 𝑥1 and 𝑥2 are pivot variables and 𝑥3 is a free variable.
Pivots

‣ Thus, our solution to the equations include
3𝑥1 − 𝑥2 + 5𝑥3 = 8
𝑥2 − 10𝑥3 = 1
‣ We may notice that 𝑥3 is not constrained by any other equation, that is, we
can let 𝑥3 to be equal to any number. If we let 𝑥3 to be equal to 𝑡, 𝑡 will be
known as a parameter. Thus, 𝑥2 − 10𝑡 = 1; 𝑥2 = 1 + 10𝑡 and 3𝑥1 −
1 + 10𝑡 + 5𝑡 = 8; 𝑥1 = 9 +
5
3
𝑡 (infinite number of solutions)
‣ Therefore, 𝑥1 = 9 +
5
3
𝑡, 𝑥2 = 1 + 10𝑡, 𝑧 = 𝑡. If we let the value of 𝑡 to be 4,
we get the following values: 𝑥1 =
29
3
, 𝑥2 = 41 and 𝑥3 = 4.

‣ Consider the following equations;
𝑥 + 2𝑦 − 2𝑧 + 2𝑤 = 7
𝑥 + 2𝑦 − 𝑧 + 3𝑤 = 5
𝑥 + 2𝑦 − 3𝑧 + 𝑤 = 1
‣ The augmented matrix and its reduced echelon form are
1 2 −2 2
1 2 −1 3
1 2 −3 1
3
5
1
RREF
1 2 0 2
0 0 1 1
0 0 0 0
7
2
0
‣ Thus, ത
x =
𝑥
𝑦
𝑧
𝑤
=
7 − 2𝑦 − 2𝑤
𝑦
2 − 𝑤
𝑤
=
7 − 2𝑡 − 2𝑠
𝑡
2 − 𝑠
𝑠
for 𝑦 = 𝑡, 𝑤 = 𝑠 and ∀{𝑦, 𝑡} ∈ ℝ.

‣ The following are the properties of a linear system of equations.
1. No solution: If the echelon form has a row of the form
0 0 0 b , where b ≠ 0, then the system is inconsistent and has
no solution.
2. One solution: If every column of the coefficient matrix of the
echelon form is a pivot column, the system has exactly one solution.
3. Infinitely many solutions: For a consistent system of equations: If
not all columns of the coefficient matrix of the echelon form are
pivot columns, then the system has infinitely many solutions.

Example 6: For homogenous, solve the following system of equations.
2𝑥1 + 𝑥2 + 𝑥3 + 4𝑥4 = 0
𝑥1 + 2𝑥2 − 𝑥3 + 5𝑥4 = 0
2 1 1 4
1 2 −1 5
0
0
RREF
1 0 1 1
0 1 −1 2
0
0
‣ According to the reduced echelon form,
𝑥1 = −𝑥3 − 𝑥4 and 𝑥2 = 𝑥3 − 2𝑥4
That is,
𝑥1
𝑥2
𝑥3
𝑥4
=
−𝑥3 − 𝑥4
𝑥3 − 2𝑥4
𝑥3
𝑥4
= 𝑥3
−1
1
1
0
+ 𝑥4
−1
−2
0
1

‣ Notice that we have constructed a column from the coefficients of 𝑥3 in each
equation, and another column from the coefficients of 𝑥4.
‣ Consider what happens when we let the parameters to be 𝑥3 = 1 and 𝑥4 = 0;
𝑥1
𝑥2
𝑥3
𝑥4
= 1
−1
1
1
0
+ 0
−1
−2
0
1
=
−1
1
1
0
‣ And if we let 𝑥3 = 0 and 𝑥4 = 1;
𝑥1
𝑥2
𝑥3
𝑥4
= 0
−1
1
1
0
+ 1
−1
−2
0
1
=
−1
−2
0
1
These two vector
are called basic
solutions. We say
that the general
solution of the
homogeneous
system is a linear
combination of its
basic solutions

Question: Solve the following systems of linear equations.
1. 2𝑥1 + 3𝑥2 + 4𝑥3 = 0
𝑥1 − 3𝑥2 + 𝑥3 = 0
4𝑥1 − 𝑥2 + 6𝑥3 = 0
2. 𝑥 + y − z + 2w = 0
𝑥 + 3𝑦 + 𝑧 + 6𝑤 = 0
𝑥 + 2𝑦 + 4𝑤 = 0
3. −𝑥2 + 𝑥3 = 3
𝑥1 − 𝑥2 − 𝑥3 = 0
−𝑥1 − 𝑥2 = −3
4. 𝑥2 − 4𝑥2 = 8
2𝑥1 − 3𝑥2 + 2𝑥3 = 1
4𝑥1 − 8𝑥2 + 12𝑥3 = 1

‣ In this section we will study some important vector spaces that are associated
with matrices. Our work here will provide us with a deeper understanding of
the relationships between the solutions of a linear system and properties of its
coefficient matrix.
▪ Row space: This refers to the space spanned by the row vectors of a matrix.
In other words, it's the set of all possible linear combinations of the rows of
a matrix.
▪ Column space: It's the space spanned by the column vectors of a matrix.
Like the row space, it's the set of all possible linear combinations of the
columns.
▪ Null space: This is the set of all vectors that satisfy the homogenous
equations. In essence, it represents the solutions to the equation Aത
x = 0,
where A is the matrix and ത
x is a vector of appropriate dimension.
ROW SPACE, COLUMN SPACE AND NULL SPACE

‣ For an 𝑚 × 𝑛 matrix
A =
𝑎11 𝑎12 ⋯ 𝑎1𝑛
𝑎21 𝑎22 ⋯ 𝑎2𝑛
⋮ ⋮ ⋮
𝑎𝑚1 𝑎𝑚2 ⋯ 𝑎𝑚𝑛
‣ The vectors 𝑟1 = 𝑎11 𝑎12 ··· 𝑎1𝑛 , 𝑟2 = 𝑎21 𝑎22 ··· 𝑎2𝑛 and so on
up to 𝑟𝑚 = [𝑎𝑚1 𝑎𝑚2 ··· 𝑎𝑚𝑛] in 𝑅𝑛
that are formed from the rows of
A are called the row vectors of A.
‣ This implies that the vectors 𝑅𝑚 formed from the columns of A are called
the column vectors of A.

‣ From the given matrix A, below are the column vectors.
𝑐1 =
𝑎11
𝑎21
⋮
𝑎𝑚1
, 𝑐2 =
𝑎12
𝑎22
⋮
𝑎𝑚2
and so on up to 𝑐3 =
𝑎1𝑛
𝑎2𝑛
⋮
𝑎𝑚𝑛
Let A be an 𝑚 × 𝑛 matrix. The column space of A, written col(A), is the
span of the columns. The row space of A, written row(A), is the span of
the rows. The null space of A, written null(A), is the set
null(A) = {ത
x | Aത
x = 0}.
Definition: Column space, row space, null space

‣ Now, we may note that the product Aത
x, in the equation Aത
x = 0, can be
written as a linear combination of the row and column matrix consisting
of coefficients from ത
x. That is
Aത
x = 𝑥1𝑐1 + 𝑥2𝑐2 +··· + 𝑥𝑛𝑐𝑛
‣ Thus, a linear system, Aത
x = ത
b, of m equations in n unknowns can be
written as 𝑥1𝒄𝟏 + 𝑥2𝒄𝟐 +··· + 𝑥𝑛𝒄𝒏 = ത
b from which we conclude that
Aത
x = ത
b is consistent if and only if ത
b is expressible as a linear
combination of the column vectors of A.

Let Aത
x = ത
b be the linear system
−1 3 2
1 2 −3
2 1 −2
𝑥1
𝑥2
𝑥3
=
1
−9
−3
Show that ത
b is in the column space of A by expressing ത
b as a linear
combination of column vectors of A.
Example 7: A Vector ത
b in the Column Space of A

‣ Solving this system using Gaussian Elimination method gives 𝑥1 = 2,
𝑥2 = −1 and 𝑥3 = 3 (please verify this)
‣ Since 𝑐1 =
−1
1
2
, 𝑐2 =
3
2
1
, 𝑐3 =
2
−3
−2
and 𝑥1𝒄𝟏 + 𝑥2𝒄𝟐 + 𝑥3𝒄𝟑 = ത
b,
Thus,
𝑥1
−1
1
2
+ 𝑥2
3
2
1
+ 𝑥3
2
−3
−2
= 2
−1
1
2
−
3
2
1
+ 3
2
−3
−2
=
1
−9
−3

‣ We know that performing elementary row operations on the augmented matrix
[A |ത
b] does not change the solution set of that system. This is also true if the
system is homogeneous, where the augmented matrix is [A | 0].
Find the basis for null space of the matrix
1 3 −2 0 2 0
2 6 −5 −2 4 −3
0 0 5 10 0 15
2 6 0 8 4 18
0
0
0
0
Example 8: Finding a Basis for the Null Space of a Matrix

‣ After the row operations, we get the following matrix in RREF.
1 3 0 4 2 0
0 0 1 −2 0 0
0 0 0 0 0 1
0 0 0 0 0 0
0
0
0
0
‣ This results into the following equations;
𝑥1 = −3𝑥2 − 4𝑥4 − 2𝑥5
𝑥3 = −2𝑥4
𝑥6 = 0
‣ This means that 𝑥1, 𝑥3 and 𝑥6 are pivot variables while 𝑥2, 𝑥4 and 𝑥5 are free
variables.

‣ Therefore,
𝑥1
𝑥2
𝑥3
𝑥4
𝑥5
𝑥6
=
−3𝑥2 − 4𝑥4 − 2𝑥5
𝑥2 is free
−2𝑥4
𝑥4 is free
𝑥5 is free
0
and if we factorise;
𝑥1
𝑥2
𝑥3
𝑥4
𝑥5
𝑥6
= 𝑥2
−3
1
0
0
0
0
+ 𝑥4
−4
0
−2
0
0
0
+ 𝑥5
−2
0
0
0
1
0

‣ Therefore;
𝑢 =
−3
1
0
0
0
0
, 𝑣 =
−4
0
−2
0
0
0
and 𝑤 =
−2
0
0
0
1
0
‣ Observe that the basis vectors 𝑢,𝑣, and 𝑤 in this example are the
vectors that result by successively setting one of the parameters (free
variables) in the general solution equal to 1 and the others equal to 0.
Basis

Find a spanning set for the null space of the matrix
−3 6 −1 1 −7
1 −2 2 3 −1
2 −4 5 8 −4
Example 9: Finding the Null Space of a Matrix

‣ The first step is to find the general solution of Aത
x = 0 in terms of free
variables. The reduced echelon form of the augmented matrix is
1 −2 0 −1 3
0 0 1 2 −2
0 0 0 0 0
0
0
0
‣ At this point we need to write the pivot variables in terms of the free
variables.
𝑥1 = 2𝑥2 + 𝑥4 − 3𝑥5
𝑥3 = −2𝑥4 + 2𝑥5
‣ 𝑥1 and 𝑥3 are the only pivot variables. The rest are free.

‣ We now decompose the vector giving the general solution into a linear
combination of vectors.
𝑥1
𝑥2
𝑥3
𝑥4
𝑥5
=
2𝑥2 + 𝑥4 − 3𝑥5
𝑥2 is free
−2𝑥4 + 2𝑥5
𝑥4 is free
𝑥5 is free
= 𝑥2
2
1
0
0
0
+ 𝑥4
1
0
−2
1
0
+ 𝑥5
−3
0
2
0
1
‣ Thus, Null(A) = 𝑢, 𝑣, 𝑤 =
2
1
0
0
0
,
1
0
−2
1
0
,
−3
0
2
0
1
𝑢 𝑣 𝑤

Find a basis for the column space, row space of the matrix
1 2 1 3 2
1 3 6 0 2
3 7 8 6 6
Example 10: Basis of column space and row space

‣ The column space is the span of the columns of the matrix A, i.e.;
Col(A)=Span
1
1
3
,
2
3
7
,
1
6
8
,
3
0
6
,
2
2
6
‣ The row-reduced echelon for of A is
1 0 −9 9 2
0 1 5 −3 0
0 0 0 0 0
and the pivot
column in the RREF of A represent the basis pf Col(A)
Basis of Col(A) =
1
1
3
,
2
3
7

‣ Row space of A is the span of the rows, i.e.;
Row(A) = span 1 2 1 3 2 , 1 3 6 0 2 , 3 7 8 6 6
‣ Non-zero rows of the row-reduced echelon form of the given matrix
gives the basis for the row space of A.
‣ Therefore;
Basis of Row(A) = 1 0 −9 9 2 , 0 1 5 −3 0

Question: Find null(A) for the following matrices.
1. A =
2 3
4 6
2. A =
1 0 −1
−1 1 3
3 2 1
3. A =
2 −1 3 5
2 0 1 2
6 4 −5 −6
0 2 −4 −6
4. A = 1 3 5 0
0 1 4 2
5. A =
1 5 −4 −3 1
0 1 −2 1 0
0 0 0 0 0

Josophat Makawa
bsc-mat-14-21@unima.ac.mw
+265 999 978 828
+265 880 563 256

MAT222 - INTRODUCTION TO LINEAR ALGEBRA WITH APPLICATIONS | ROW-REDUCED ECHELON FORM AND VECTOR SPACES.pdf

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