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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

B
...
(I sem), 2020-21
MATHEMATICS-I (KAS-103T)
MODULE 1(MATRICES)
SYLLABUS: Types of Matrices: Symmetric, Skew-symmetric and Orthogonal Matrices;
Complex Matrices, Inverse and Rank of matrix using elementary transformations, Rank-Nullity
theorem; System of linear equations, Characteristic equation, Cayley-Hamilton Theorem and its
application, Eigen values and Eigen vectors; Diagonalisation of a Matrix
...
No
1
...
2
1
...
4
1
...
6
1
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8

MATRICES

1
...
14
1
...
16

Introduction
Types of Matrices
Basic Operations on Matrices
Transpose of a Matrix
Symmetric Matrix
Skew-Symmetric Matrix
Complex Matrices
Hermitian and Skew-Hermitian matrix
Unitary Matrix
Elementary Transformations (or Operations)
Inverse of a Matrix by E-Operations (Gauss-Jordan Method)
Rank of a Matrix
Nullity of a Matrix
Methods of Finding Rank of Matrix
Solution of System of Linear Equations
Linear Dependence and Independence of Vectors

1
...
18
1
...
20

Cayley-Hamilton Theorem
Similarity Transformation
Diagonalization of a matrix

1
...
10
1
...
12

1

PAGE
NO
...
1 INTRODUCTION
The term ‘Matrix’ was given by J
...
Sylvester about1850, but was introduced first by Cayley in
1860
...
Matrices (plural of matrix) find
applications in solution of system of linear equations, probability, mathematical economics,
quantum mechanics, electrical networks, curve fitting, transportation problems etc
...

Matrix inverse can be used in Cryptography
...


Matrix
A rectangular array of m
...
It is
also called m n matrix
...

Elements of a matrix are located by the double subscript ij where i denotes the row and j the

 a11
a
 21
column
...


am1

a12
a22

...
a2 n 

...
amn  mn


...

Note: Matrix has no numerical value
...
2 TYPES OF MATRICES
(a) Square Matrix
In a matrix, if the number of rows = number of columns = n, then it is called a square matrix of
order n
...


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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

(b) Row Matrix
A matrix with only one row is called a row matrix
...
a1n 


...
So, it is a row

matrix of type 1  n
...
It is a row matrix with 4 columns
...


(c) Column Matrix
A matrix with only one column is called a column matrix
...

 an1 

It is a column matrix with n rows
...


1 
 2
 
3
A

Also, Let
 
 4
5 
It is a column matrix with 5 rows
...


(d) Diagonal Matrix
A square matrix in which all the non- diagonal elements are zero is called a diagonal matrix
...

 4

(e) Scalar Matrix
In a diagonal matrix if all the diagonal elements are equal to a non-zero scalar  , then it is called
a scalar matrix
...

 

(f) Unit Matrix or Identity matrix
In a diagonal matrix if all the diagonal elements are equal to 1, then it is called a unit matrix or
identity matrix
...
They are denoted by I1, I2, I3
...


(g) Zero Matrix or Null matrix
In a matrix (rectangular or square), if all the entries are equal to zero, then it is called a zero
matrix or null matrix
...


(h) Triangular Matrix
A square matrix A= [aij] is said to be upper triangular matrix if all the entries below the main
diagonal are zero
...

A square matrix A= [aij] is said to be lower triangular matrix if all the entries above the main
diagonal are zero
...

Examples:

 1 2

1
The matrices A   0
 0 0

4
3 
0
4, 
0
5 
0

1

0

2

3

0

0

0

0

2 
1 
 2 are upper triangular matrices
...

2 1

(i) Singular and Non-singular Matrices
A square matrix A is said to be singular if A  0 and non-singular if A  0
...

 1 1 

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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

(j) Orthogonal Matrix
A square matrix A is called an orthogonal matrix if AA  AA  I
...


 cos
 sin 

Example: A  

sin  
cos 

(k) Nilpotent matrix
A square matrix A is said to be nilpotent if A2  0
...


2
3
1
0 1 

A 1
2
3 
Examples: A  

0 0  ,
 1  2  3
(l) Idempotent matrix
A square matrix A is idempotent if A  A
...

2
1 
 1
6
6
6

1
0



2
4

2
Examples: A  
, A 6
6
6 
0 1 
 1
 6  2 6 1 6 

(m) Involutary matrix
A square matrix A is said to be involuntary if A  I
...

2

 5  8 0 
Example: A   3
5
0 
...


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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

1 0 0
Example: Let I  0 1 0


0 0 1

1 0 0 
Operating R23 or C23 on I, we get the elementary matrix 0 0 1 
...



0 1 0

1
...
If A and B are two matrices of the same order , then their sum A+B is a matrix each
element of which is obtained by adding the corresponding elements of A and B
...
Then A+B=[cij] where
cij=aij+bij for all i, j and A  B is of type m n
...

Let A= [aij] and B = [bij] be two matrices of the same type m n
...


 1 2 3 1 2 3 
A
, 
 then
 0 1 5 1 0  2
  1  1 2  2 3  3  0 4 6 
A B  
  1 1 3
0

1
1

0
5

2

 


Example :If

We see that A and B are of type 2  3 and A  B is also of the type 2  3

 1  1 2  2
A B  
 0 1 1  0

3  3   2

5  2   1

0
1

0
7 

(b) Scalar Multiplication of Matrices
Let A= [aij] be an m n matrix and k be a scalar
...

Example:

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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

If

a
A   11
a21

a12
a22

Hence if k = -1, then

a13 
 ka11
then
kA

ka
a23 
 21

 a
 A   11
 a21

 a12
 a22

ka12
ka22

ka13 

...

 a23 

( c) Multiplication of Matrices
If A and B are two matrices such that the number of columns of A is equal to the number of rows
of B, then the product AB is defined
...
In the product AB, A is known as pre-factor and B is known as post-factor
...


That is cij is the sum of the products of the corresponding elements of the i th row of A and the

j th column of B
...


1
AB  0
2

1
1
2

2 1
3 3
1 2

2 1
...
3  2
...
2  1
...
1  8
1  0
...
3  3
...
2  1
...
1   9
1 2
...
3  1
...
2  2
...
1 10

5
4
7

Note: If A and B are square matrices of order n, then both AB and BA are defined,but not
necessarily equal
...
So, matrix multiplication is not commutative
...
If A, B, C are matrices of the same type, then
i
...

2
...
A  A
2

3

2

n

A

1
...

t

Clearly, (i) If the order of A is m n then order of A is n  m
...

th

th

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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

1
A

2
Example: If


0

2

1 3

1
0
5

, then A  
2
7 

5

2
 1
3

7

Properties of Transpose of a Matrix
If Aand B  denote the transposes of A and B respectively, then
(a)
(b)

 A  A i
...
, the transpose of the transpose of a matrix is the matrix itself
...
e
...


(c)

 AB  BA i
...
, the transpose of the product of two matrices is equal to the
product of their transposes taken in the reverse order
...
5: Symmetric Matrix

 

A square matrix A  a ij is said to be symmetric if A  A i
...
, if the transpose of the matrix
is equal to the matrix itself
...


 

  1 2 3  a
Example :  2 5 6,  h

 
 3 6 4  g

h
b
f

g
f  are symmetric matrices
...
6: Skew-Symmetric Matrix

 

A square matrix A  a ij is said to be symmetric if A   A i
...
, if the transpose of the matrix
is equal to the negative of the matrix
...


 

Putting j=i, aii  aii  2aii  0 or aii  0 for all i
...


2  3  0
0

Example :  2 0
1 ,  h

 3  1 0   g

h
0
f

 g
f  are skew-symmetric matrices
...

Proof : Let A be any square matrix
...

2
2

1
 A  A  1  A  A  A
2
2

1
 A  A  1  A   A   1  A  A  1  A  A  B
 2
2
2
2

 B is a symmetric matrix
...

Hence every square matrix A can be expressed as A= B+C, B 
and C 

1
 A  A is a symmetric matrix
2

1
 A  A is a skew-symmetric matrix
...


(1)

is another such representation in which P is a symmetric matrix, and Q is a skew-symmetric
matrix
...
e
...
(2)
From (1) and (2) , A  A  2P and A  A  2Q
...
so that P+Q =B+C
2
2

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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

Hence the result
...
7 Complex Matrices
A matrix having complex numbers as its elements is called a complex matrix
...


2  3i 4  6i 3 
2  3i 4  6i 3 

...

Clearly the transpose of the conjugate and conjugate of the transpose of matrix are equal i
...
,

A    A
...


3 
3 
5  6i
 4i

Thus if A  

Some Properties of Transposed Conjugate Matrix

 

(i) A



 A
...


(iii)  A  B   A  B
...



1
...

In a Hermitian matrix, the diagonal elements are all real, while every other element is the
conjugate complex of the element in the transposed position
...


 3i 1  i
0 
A square matrix A is called Skew- Hermitian if A   A
...
Every other element is the negative of the conjugate complex of the
element in the transposed position
...


  7 2  i
 i 

1
...

The determinant of a unitary matrix is of unit modulus
...

Example : A 

1  1 1  i


3 1  i  1  is a Unitary matrix
...


5 
 2i

i 
Solution: A 
 3
 1  3i 4  2i 
5 
 2i

A   A   3
 i 
 1  3i 4  2i 


5 
 2i
2  i 3  1  3i 

A A 3
 i  
 5 i 4  2i 

 1  3i 4  2i 


6  8i  19  17i 
 30

  6  8i
10
 5  5i   B(say)
 19  17i  5  5i
30 

6  8i  19  17i 
 30

Now B   6  8i
10
 5  5i 
 19  17i  5  5i
30 
 30
6  8i  19  17i 


B  B   6  8i
10
 5  5i   B
 19  17i  5  5i
30 

Hence B  A A is a Hermitian matrix
...

Solution: A and B are Hermitian  A  A and B  B
Now  AB  BA   AB   BA






B A  A B  BA  AB   AB  BA
 AB  BA is skew-Hermitian
...

Solution: A is a skew- Hermitian matrix  A   A
Now iA  i A   i  A  iA


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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

 iA is a Hermitian matrix
...

Solution: A be any square matrix
...


Now, A 









1
1
A  A  A  A  P  Q (say)
2
2

where P is Hermitian and Q is skew-Hermitian
...


Example 5 : Express the matrix

4 5  5i 
 2  2i

A   4i
3  i 4  2i  as the sum of a Hermitian
 3  3i  4
9 

matrix and Skew-Hermitian matrix
...

2
2
From (1) and (2), we have
1
1
A  A  A  A  A
2
2

Clearly



 



15

(1)

(2)

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

2  2i 1  4i   2i
2  2i 4  i 
 2



A  2  2i
3
i    2  2i
i
4  i 
1  4i
i
9    4  i  4  i
0 

Example 6: Show that the matrix A 
Solution: Given A 

1  1 1  i

 is unitary
...

Now AA  
3 1  i  1  1  i  1  3 0 3 0 1
 A   A 

A A  I
...

Hence A is unitary
...

Solution: Given

1  2i 
 0
N
0 
 1  2i
Also

1 0 
I

...

1 
IN
6 1  2i

Thus I - N I  N  
1



 1  2i   1
 1  2i 
1 1

...


2  4i 
1  4

6  2  4i  4 
2  4i    4  2  4i 
1  4
Pθ P 

 4 
36  2  4i  4  2  4i

θ
We have P  P 

 Pθ P 

1 36 0  1 0

 I
...
e
...


Practice Question:
  i
   i

1) Show that the matrix 

   i 
is unitary if  2   2   2   2  1
...

7 - 4i - 2  5i 
 2

-2
3  i  is a Hermitian matrix
...

4) Show that A   3  2i

 2 - i
- 3 - 4i - 2i 

 - 1 2  i 5 - 3i 

7
5i , Show that A is a Hermitian matrix and iA is a skew5) If A  2 - i

5  3i - 5i
2 
Hermitian matrix
...

1  i 1  i 

Prove that A  

i 0

7) Show that A  0 0

0 i
1  i
8) Prove that A  
1  i

0
i  is skew-Hemitian matrix and also unitary
...

1  i 

9) If A and B are two unitary matrices, show that AB is a unitary matrix
...


18

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

1
...

(i)
Interchange of two rows or two columns
...

The interchange of ith and jth columns is denoted by Cij
...

The multiplication of ith row by k is denoted by Ri(k)
...


(iii)

Addition of k times the elements of a row (or column) to the corresponding
elements of another row(or column), k≠0
...

The addition of k times the jth column to the ith column is denoted by Cij(k)
...
Two equivalent matrices A and B are written as A~B
...
11 Inverse of a matrix by E-Operations (Gauss-Jordan Method)
The elementary row transformations which reduce a square matrix A to the unit matrix, when
applied to the unit matrix, gives the inverse matrix A-1
...
Inverse of a n-square matrix A is denoted by A-1 and is
defined such that
A A-1 = A-1A=I where I is n x n unit matrix
...

(1) If we are to find out the inverse of a non-singular square matrix A, we first write A as
equivalent to I, a unit matrix of the same order
...
The objective is to reduce A to I
...

I~A-1
This is an elegant way of determining the inverse or reciprocal of a matrix A
...
e
...

(2) Inverse of a matrix is unique
...
e
...
e
...


Example1: Find the inverse of A by Gauss-Jordan method where
1 2 3
A  2 4 5
3 5 6
Solution : Writing A=IA i
...
,

1 2 3 1 0 0
 2 4 5   0 1 0  A

 

3 5 6 0 0 1
By R21(-2),R31(-3),we get

3   1 0 0
1 2
0 0  1   2 1 0 A

 

0  1  3   3 0 1

3   1 0 0
1 2

By R23 0  1  3    3 0 1 A

 

0 0  1  2 1 0
1 2 0    5 3 0
By R13(3) ,R23(-3) 0  1 0    3  3 1 A

 

0 0  1  2 1 0
1 2 0  5 3 0 
By R2(-1) ,R3(-1) 0 1 0   3 3  1 A

 

0 0 1  2  1 0 

20

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

By R12(-2)

Hence

1 0 0  1  3 2 
0 1 0   3 3  1 A

 

0 0 1  2  1 0 
 1 3 2 
A   3 3  1
...
e
...



 2 2  1
1

Example 3: Find the inverse of the following matrix by employing elementary
transformations:

 3  3 4
A  2  3 4
0  1 1 
Solution : Writing A=IA i
...
,

3  3 4 1 0 0
 2  3 4   0 1 0  A

 

0  1 1 0 0 1

1 0 0 1  1 0
By R12(-1) 2  3 4  0 1 0 A

 

0  1 1 0 0 1
1 0 0  1  1 0
By R21(-2) 0  3 4   2 3 0 A

 

0  1 1  0
0 1
1 0 0  1  1 0 
By R23(-4) 0 1 0   2 3  4 A

 

0  1 1  0
0
1 
1 0 0  1  1 0 
By R32(1) 0 1 0   2 3  4 A

 

0 0 1  2 3  3
 1 1 0 
Hence A   2 3  4
...
Writing A=IA i
...
,

0 1 2 1 0 0
1 2 3  0 1 0 A

 

3 1 1 0 0 1
1 2 3 0 1 0
By R11 0 1 2  1 0 0 A

 

3 1 1 0 0 1

3  0 1 0 
1 2

By R31(-3) 0 1
2   1 0 0 A

0  5  8 0  3 1

1 0  1  2 1 0
By R12(-2),R32(5) 0 1 2    1
0 0 A

 
0 0 2   5  3 1
1
0 
1 0  1   2



By R3(1/2) 0 1 2  1
0
0  A

 
0 0 1  5 / 2  3 / 2 1 / 2

1 0 0 1 / 2  1 / 2 1 / 2 
By R13(1),R23(-2) 0 1 0    4
3
 1  A

 
0 0 1 5 / 2  3 / 2  1 / 2
1 / 2  1 / 2 1 / 2 
Hence A 
...
Writing A=IA i
...
,

1
1
1  1
1
  2 1  1  2  0


  4  2  3 1  0

 
1
1
0  0
1
1
0
By R21 (2),R31(4),R41(-1) 
0

0

1
0
By R23(-1) 
0

0

1
0
By R13(-1) 
0

0

1 1
3 1 0   2

2 1 5 4
 
0 0  1  1

1 1

1 1
1 0  5  2

2 1 5 4
 
0 0  1   1
1 1

1
0
By R12(-2),R32(-2) 
0

0

0 0 0
1 0 0
A
0 1 0

0 0 1
0 0 0
1 0 0
A
0 1 0

0 0 1

0
1  1 0
A
0 1 0

0 0 1
0

0

6   3 1 1
1 0  5  2 1  1

0 1 15   8  2 3
 
0 0  1   1 0
0
0 1

0
0
A
...

0 1 15   8  2 3 0
 

0 0  1   1 0
0 1

1
0
By R14(-9), R24(-5), R34(15) 
0

0

0  4
1  2  9


1 0 0  3
1  1  5

A
...

0 1 0  7  2 3 15 
 

0 0 1  1
0
0  1

1  2  9
4
3
1  1  5
Hence A 1 
...
A  2 5 5 


2 5 11

 5 2 0 
1
Ans : A   2 3  1
...
A  2  3 4


0  1 1

 1  7 24 
1
Ans : A   2 3  4 
...
A  7  1 8


3 4 2

62 
 34 10
1 
Ans : A    10
0  30
...
A  2 8 4


1 2 8

2
3
5
...


392
  4  12 56 
-1

 23 29  64 / 5  18 / 5
 10  12 26 / 5
7 / 5 
Ans : A -1  
 1
2
6/5
2/5 


2
3/ 5
1/ 5 
 2

25

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

1
1
6
...
A  
1

 1

2 3 1
3 3 2
4 3 3

1 1 1

0
 1 2 1
 1  2 2  3

Ans : A -1  
0
1 1 1 


 2 3  2 3 
 17  3 15 7 

1  5  4
1 9
-1
Ans : A 
5  10  5 10 5 


0
1
 1 1

3
1  1 2
2 0 1

1 2 6
2

1

2 0  1
8
...
A  0 2 1 


5 2  3

 8  1  3
Ans : A   5 1
2 
 10  1  4

2  2
1

10
...
12 Rank of a Matrix
Let A be any m n matrix
...
The determinants of
these square sub-matrices are called minors of A
...

All the minors of order (r+1) or higher than r are zero
...

Note: (1) If A is a null matrix, then rank of A= 0
...

(3)If A is a non-singular n  n matrix , then rank of A  n
...
From this matrix A, delete all columns
and rows leaving a certain p columns and p rows
...
The determinant of this square matrix
is called a minor of A of order p
...
13 Nullity of a matrix
Let A be a square matrix of order n and if the rank of A is r, then n-r is called the nullity of the
matrix A and is usually denoted by N (A)
...
e
...

Example 2: Find the rank and nullity of the following matrices:

1 2 3
A  1 3 5
2 5 8
1 2 3


Solution: A  1 3 5
2 5 8

27

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

1 2 3 1 2 3
 A  1 3 5  1 3 5  0
...

Nullity N(A) = 3 – 2 =1
...
applying R 3  R 3 - 2R 2 & R 2  R 2 - 2R 1
4 8 12 0 0 0



So, the rank of A is less than 3
...
Therefore, rank of A is less than 2
...

Hence , rank of A=1
...


1
...
Then the number of non-zero rows is the
rank of A
...

 All the non-zero rows, if any, precede the zero rows
...
ie number of non-zero rows is equal to the rank of the
matrix
...


forms)
...

 Interchange rows (or columns) to obtain a non-zero element in I row and I column of
given matrix
...

 Obtain zeros in the remainder of I row and I column
...

 Continue the process down the main diagonal either until the end of diagonal is reached
or all the remaining elements of matrix are zero
...

 Reduce the matrix on L
...
S
...

 Every elementary row transformations on A must be accompanied by the same
transformation on the pre-factor on R
...
S
...
H
...


29

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

Determine the rank of the following matrices by reducing to Echelon form:
Example 1 : Determine the rank of the following matrices by reducing to Echelon form:

2
3 
4
A   8
4
6 
 2  1  1
...
So the rank of A is one
...


Solution:

1 5 4 
A  0 3 2 
2 13 10
1 5 4 
By R31(-2) ~ 0 3 2


0 3 2

1 5 4 
By R32(-1) ~ 0 3 2


0 0 0
Rank of A is 2 since the number of non-zero rows is 2
...


2 6 5 
A  B  2 5 4 
5 16 13
5 
2 6

By R21(-2), R31(-5/2), ~ 0  7  6 


0  1 1 / 2
Rank of A+B is 3 since the number of non-zero rows is 3
...


3 21 16 
BA  6 42 32
9 63 48

1 1 1
By R21(-2), R31(-3), ~ 0 0 0,


0 0 0
Rank of BA is 1 since the number of non-zero rows is 1
...

Example 4: Determine the values of b such that the rank of A is 3
...


(ii)If b = -6, number of non-zero rows is 3,rank of A=3
...


1 3  1 2 
0 11  5 3

A
2  5 3 1 


1 5
4 1
Solution: Applying elementary row operations on A

1 2 
1 3
0 11  5 3 

R31  2, R 41  4 ~ 
0  11 5  3


0  11 5  3

33

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

1 3  1 2
0 11  5 3

R32 1, R 42 1 ~ 
0 0 0 0 


0 0 0 0 

1

 1  0
R2   ~
 11  0

0

1

R12  3 ~ 0

0
0


3
1
0
0
0
1
0
0

1 2 
5 3 
11 11
0
0

0
0 
4 13 
11 11 
5 3 

11 11 
0
0
0
0 

This is the Echelon form
...


Determine the rank of the following matrices by reducing to Normal form:
Example 1: Reduce the matrix A to normal form and hence find the rank

1 2 3 4 
A  2 4 4 3 
3 0 5  10

1 2 3 4 
Solution: A  2 4 4
3 

3 0 5  10
3
4 
1 2

R21  2, R31  3 ~ 0 0 - 2 - 5 
0  6  4  22
1 0 0 0 
C21  2, C31  3, C41  4 ~ 0 0 2 5 
0 6 4 22

34

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

1 0 0 0 
1 
C2   ~ 0 0 2 5 
6
0 1 4 22
1 0 0 0
C32  4, C42  22 ~ 0 0 2 5
0 1 0 0
1 0 0 0
1
C3  , C43  5 ~ 0 0 1 0
2
0 1 0 0
1 0 0 0
R23 ~ 0 1 0 0  I 3
0 0 1 0

0

Hence, Rank of A =3
Example 2: Reduce the matrix A to normal form and hence find the rank:

2
0
A
2

2

4
3 4 1
3 7 5

5 11 6

1

3

2
0
Solution: A  
2

2

4
3 4 1
3 7 5

5 11 6

1

3

2
0
R31  1, R41  1 ~ 
0

0
2
0
R43  2 ~ 
0

0

1 3 4
3 4 1
2 4 1

4 8 2

1 3 4
3 4 1
2 4 1

0 0 0

35

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

2
0
R23  1 ~ 
0

0

1 3 4
1 0 0
2 4 1

0 0 0

2
0
R32  2, R12  1 ~ 
0

0

0 3 4
1 0 0
0 4 1

0 0 0

1
0
1
C1  , C31  3, C41  4 ~ 
0
2

0
1
0
1
C3  , C43  1, ~ 
0
4

0

0 0 0
1 0 0
0 4 1

0 0 0

0 0 0
1 0 0 I 3

0 1 0  0

0 0 0

0
, rank of A  3
...

0

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

Example 5: Reduce the matrix A to normal form and hence find the rank

 2 1 - 3 - 6
A  3 - 3 1 2 
1 1 1 2 

 2 1 - 3 - 6
Solution: A  3 - 3 1
2 

1 1 1 2 
1
2 
1 1

R21  3, R31  2 ~ 0 - 6 - 2 - 4 
0  1  5  10

0
0 
1 0

C21  1, C31  1, C41  2 ~ 0 - 6 - 2 - 4 
0  1  5  10
1 0 0 0 
R2  1, R3  1 ~ 0 6 2 4 
0 1 5 10
1 0 0 0 
R23 ~ 0 1 5 10
0 6 2 4 

0
0 
1 0
R32  6 ~ 0 1
5
10 
0 0  28  56
0
0 
1 0

C32  5, C42  10  ~ 0 1
0
0 
0 0  28  56
1 0 0 0
 1  
R3    ~ 0 1 0 0
 28 
0 0 1 2

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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

1 0 0 0
C43  2 ~ 0 1 0 0  I 3
0 0 1 0

0, rank of A  3
...

0 0
1
0
 1  2 1


0
1
0 0
Example 2: Find the non-singular matrices P and Q such that the normal form of A is PAQ where

1  1  1
A  1 1 1 
3 1 1 


...
H
...
Hence Rank of A  2
0 

Thus I2=PAQ where


1
 1
P   2
 -1
 4

0
1
2
1
2


0
1 1 0 

0  , Q  0 1 - 1 , Rank of A  2
...
Hence find the rank
...
H
...
Hence Rank of A  2
0 

Thus I2=PAQ where

 1 0 0
1 - 1 - 1


P  - 1 1 0 , Q  0 1 - 1 , Rank of A  2
...
Hence find the rank
...


Solution: Consider A3X3 = I 3X3 A 3X3 I3X3

 3  3 4
 2  3 4


0  1 1

1 0 0 1 0 0
 0 1 0 A 0 1 0
0 0 1 0 0 1

1 0 0 1 - 1 0 1 0 0
R12  1, pre2  3 4  0 1 0 A 0 1 0
0  1 1 0 0 1 0 0 1

1 0 0  1 - 1 0 1 0 0
R21  2, pre0  3 4  - 2 3 0 A 0 1 0
0  1 1  0 0 1 0 0 1

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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

1 0 0  1 - 1 0  1 0 0
R23  4, pre0 1 0  - 2 3 - 4 A 0 1 0
0  1 1  0 0 1  0 0 1
1 0 0  1 - 1 0  1 0 0
C23 1, post 0 1 0  - 2 3 - 4 A 0 1 0
0 0 1  0 0 1  0 1 1
Thus the L
...
S is in the normal form I 3




...


Thus I3=PAQ where

 1 -1 0 
1 0 0


P  - 2 3 - 4 , Q  0 1 0 , Rank of A  3
...


 
 

0 1 1  0 0 1  - 2 3  3
1

Practice Questions:
(1)Find the rank of the following matrices by reducing it to Echelon form:

(i)

1  3 1 2 
A  0 1 2 3  Ans : 3
3 4 1  2

44

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

(ii)

(iii)

(iv)

1 2 3 4 5 
A  2 3 4 5 1 Ans : 2
5 8 11 14 7
4 1 1
3
2
4 3 6 
A
Ans : 4
 1  2 6 4 


 1  1 2  3
1 2 3 1 
A  2 4 6 2 Ans : 2
1 2 3 2

(2) Find the rank of the following matrices by reducing it to normal
form(or canonical form) :
1 2  1
4 1 2
(i) A  
3  1 1

1 2 0

3
1
Ans : 3
2

1

1 2 3 2 
(ii) A  2 3 5 1  Ans : 2


1 3 4 5
3
2 1

(iii) A  4 7 13  Ans : 2


4  3  1

0 1 2  2
(iv) A  4 0 2 6  Ans : 2


2 1 3 1 
(3)Determine the non-singular matrices P and Q such that PAQ is in the normal

form for A
...

1 1 2 
A  1 2 3 
1
0  1  1

 1 0 0
1  1  1


Ans : P   1 1 0Q  0 1  1 ; rankA  2
 1 1 1
0 0 1 

45

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

2

3

4

1  1  1 2 
A  4 2 2  1
2 2 0  2


 1
 2
Ans : P  
 3
 1
 3

0
1
6
1
3



1
0

 1
 Q  0
2

 1
0
0
2 


 1
2
3
 ; rankA  3
1 1
2
0 0
1
0 1
0 

1

0

3 2  1 5 
A  5 1 4  2 
1  4 11  19

9 
1 4 9
7 119 217 
0


0
1


0 1 7  1 7  1 7 

5
1
Ans : P   0
Q  
 ; rankA  2
0 
 1  13 1 3 
0 0  117
 2

3
6 
1 
0
0 0
31 

1  1 2  1
A  4 2  1 2 
2 2  2 0 

 1

Ans : P   2
 1 3
 3

 2 1  3  6
A  3  3 1
2 
5
1 1 1
2 

1
0
0 


1
 1  Q  0

6
2
0
1  1 
3
2
0

1 
2
3
1 1
2  ; rankA  3
0 0
1 
0 1
0 

1

0

0
1  1 4
 0
0
1 



0 1  5 0 

Ans : P    1
0
2 Q 
; rankA  3
0 0 1  2 
 3

9
1


 14 28 28
0 0 0 1 

1
...

 
 d 3 
HOMOGENEOUS LINEAR EQUATIONS
If all constants d1  d 2  d3  0 ,then B  0 and the matrix equation AX  B reduces to

AX  0
...


NON-HOMOGENEOUS LINEAR EQUATIONS
If at least one of constants d1 , d 2 , d 3 is non-zero, then B  0
...


Augmented matrix
The matrix A : B  in which the elements of A (coefficient matrix) and B (Column matrix of
constant) are written side by side is called the augmented matrix
...
e
...
Then x   , y   , z  
...
It may have no solution or a
unique solution or an infinite number of solutions
...

(4) A system of equations having one or more solution is called a consistent system of
equations
...

(1)

X= 0 is always a solution
...
Thus a homogeneous system is
always consistent
...


(2)

If rank (A) = number of unknowns,the system has only the trivial solution
...


Example1: Examine the following equations for consistency and if consistent, find the complete
Solution
...

Solution: Here

x  y  z  3
3x  y  2 z  2

48

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

2 x  4 y  7 z  7
...
⸫ Rank of [A:B] =3
...

Hence the given systems are inconsistent
...


49

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

Solution: The matrix equation AX=B is given by

 1 3 2 
A   3 3  1
 2  1 0 
1 :
1 2
3 1  2 :
A : B  
4  3  1 :

2 :
2 4
1

2
1
3

4
1
0
R21  3, R31  4, R41  2 ~ 
0

0

1
0
R32  1 ~ 
0

0

(1)

2
 5  5 :  5
 11  5 :  5

0
0 : 0
2

1

:

2
 5  5 :  5
6 0 : 0 

0
0 : 0
1 2 1 : 2


  1    1  0 1 1 : 1 
R2 
, R3 
~
 5   6  0 1 0 : 0 


0 0 0 : 0 
2

1

:

Which is in Echelon form
...

The given systems of equations are consistent and have a unique solution
...

Solving x  1, y  0, z  1
...


x  2 y  z  3,3x  y  2 z  1,2 x  2 y  3z  2
...

5
0 0 1 : 4

1
R32  6  ~ 0
0

⸫ Rank of [A:B] =3=Rank of A=Number of unknowns
...
Hence equation (1) can
be written as

1
0

0

2
1
0

 1  x  3
0   y   4
1   z  4

 x  2 y  z  3, y  4, z  4
...


51

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

Example 4: Prove that the following equations are consistent and solve

2x  4 y  z  9,3x  y  5z  5,8x  2 y  9z  19
...
⸫ Rank of [A:B] =2=Rank of A < Number of unknowns
...
The equations (1)
becomes

1
0

0

5
1
0

6  x   4 
 13   y   17 
14    14
0   z   0 

52

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD





 x  5 y  6 z  4, y  13 z  17
...

Example 5:

Prove that what values of ,  the equations
x  y  z  6, x  2 y  3z  10, x  2 y  z  
Have (i) no solution (ii) a unique solution (iii) infinite many solutions
...
If   3 ,then Rank of A = Rank of [A:B]=3=Number of unknowns
...

Case II
...

Hence in this case the equations are consistent and will have infinite many solutions
...
If   3,   10 ,then Rank of A = 2,Rank of [A:B]= 3
...

Hence in this case the equations are inconsistent and have no solution
...

Solution: The matrix equation AX=B is given by

1 1 1   x   1 
1 2 4   y     

   
1 4 10  z  2 
1 1 1 : 1 
 A : B  1 2 4 :  
1 4 10 : 2 

(1)

1 1 1 : 1 
R21  1, R31  1 ~ 0 1 3 :   1 
0 3 9 : 2  1
1
R32  3 ~ 0
0

1
1
0

1
:1

3 :   1 
0 : 2  3  2

Deleting the last column of matrix [A: B] clearly rank of A=2,Therefore the equations are
consistent if the rank of the augmented matrix [A:B] is also 2
...

Case I
...

Case II
...


Example 7: Solve the following system of equations :

3x  y  z  0,15x  6 y  5z  0,5x  2 y  2 z  0
...


1 1   x 
 3

AX   15 6  5  y   0
...


Hence the system of equations have Trivial solutions which is x  y  z  0
...


55

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD

Solution: The given system of equations is homogeneous and hence it is consistent

---