Search for notes by fellow students, in your own course and all over the country.
Browse our notes for titles which look like what you need, you can preview any of the notes via a sample of the contents. After you're happy these are the notes you're after simply pop them into your shopping cart.
Title: Matrix for Engineering mathematics
Description: this is matrix chaptre notes . you can learn easily beacause everthing is easily understandable you don't need to take other person help.
Description: this is matrix chaptre notes . you can learn easily beacause everthing is easily understandable you don't need to take other person help.
Document Preview
Extracts from the notes are below, to see the PDF you'll receive please use the links above
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
...
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 mn
...
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
...
2
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
5
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
(j) Orthogonal Matrix
A square matrix A is called an orthogonal matrix if AA AA 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
...
6
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:
7
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
9
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 Aand 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 BA 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
11
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
4i
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
2i
i
Solution: A
3
1 3i 4 2i
5
2i
A A 3
i
1 3i 4 2i
5
2i
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
14
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
IN
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
39
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, pre2 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, pre0 3 4 - 2 3 0 A 0 1 0
0 1 1 0 0 1 0 0 1
43
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
1 0 0 1 - 1 0 1 0 0
R23 4, pre0 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 0Q 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
...
x2 x3
k x2 2k , x3 k
...
x2 2 x3
x1 x2 x3
2 2 1
2
X 3 2, is the eigen vector for 15
...
0 0 5
Solution: Characteristic equation is given by A I 0
69
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
3
1
4
0
2
6 0 3 2 5 0 2,3,5
0
0
5
The eigen vector corresponding to eigen value 2 is given by
A I X 1 0 A 2I X 1 0
1
4 x1
3 2
1 1 4 x1
0
22
6 x2 0 or 0 0 6 x2 0
0
0 0 3 x3
0
5 2 x3
1 1 4 x1
R23 2 ~ 0 0 0 x2 0
0 0 3 x3
1 1 4 x1
R23 ~ 0 0 3 x2 0
0 0 0 x3
x1 x2 4 x3 0,3x3 0 or x3 0
...
0
The eigen vector corresponding to eigen value 3 is given by
A I X 2 0 A 3I X 2 0
1
4 x1
3 3
0 1 4 x1
0
23
6 x2 0 or 0 1 6 x2 0
0
0 0 2 x3
0
5 3 x3
x2 4 x3 0, x2 6 x3 0 , 2x 3 0
...
0
The eigen vector corresponding to eigen value 5 is given by
A I X 3 0 A 5 I X 3 0
1
4 x1
3 5
2 1 4 x1
0
25
6 x2 0 or 0 3 6 x2 0
0
0
0
5 5 x3
0 0 x3
70
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
2 x1 x2 4 x3 0,3x2 6 x3 0
Solving these equations, we have
x
x
x2 2 x3i
...
, 2 3 k say
2
1
x2 2k , x3 k and 2x1 x2 4 x3 2k 4k 6k i
...
x1 3k
3
X 3 2, is the eigen vector for 5
...
1
71
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
The eigen vector corresponding to eigen value 2 is given by
A I X 2 0 A 2I X 2 0
2
2 x1
1 2
1 2 2 x1
0
22
1 x2 0 0 0 1 x2 0
1
1 2 0 x3
2
2 2 x3
0 0 0 x1
R13 1, R12 2 ~ 0 0 1 x2 0
1 2 0 x3
The rank of coefficient matrix in above equation is 2
...
e
...
Now above equation
x3 0 and x1 2 x2 0
(1)
If x 2 1, then (1) gives x1 2, x3 0
...
0
Example 5: Calculate the eigenvalues and eigenvector for the matrix
2 2 3
A 2
1 6
1 2 0
Solution: Characteristic equation is given by A I 0
2
2
3
2
1 6 0 3 2 21 45 0 3,3,5
1
2
The eigen vector corresponding to eigen value 3 is given by
A I X 1 0 A 3I X 1 0
72
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
1 2 3 x1
2
4 6 x2 0
1 2 3 x3
On solving, we get only one independent solution
x1 2 x2 3x3 0
Let us suppose
x3 k1 , x2 k 2 then x1 3k1 2k 2
3k1 2k 2
3
2
Hence X1 k 2 k1 0 k 2 1 , are the eigen vector for -3
...
1
Practice Question:
1
...
Find the eigen values and eigenvector of the matrix
1 0 1
1 2 1
2 2 3
1 1
1 , 2
Ans:1,2,3
73
1 2 1
1, 1, 1
0 2 2
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
3
...
Find the eigen values and eigenvector of the matrix
5 2 0
2 6 2
0
2 7
Ans:3,6,9
1 2 3
5
...
6
...
Find the eigen values and eigenvector of the matrix
3 7 5
2
4
3
1
2
2
8
...
Find the eigen values and eigenvector of the matrix
6 2 2
2 3 1
2 1 3
10
...
6
1 1 1
0 , 1, 2
1 1 1
Ans:1,1,1
3
1
1
Ans:2,2,3
5 1
2 , 1
3 2
Ans:-2,2,8
1 1 2
0 , 2, 1
2 0 1
Ans:2,2,2
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
2 1 0
0 2 1
0 0 2
1
0
0
1
...
Note: 1) Cayley-Hamilton theorem gives another method for computing the inverse of a
square matrix
...
2)By using the Cayley- Hamilton theorem we can find 𝐴−2 provided 𝐴−1 is
known and hence 𝐴−3 ,𝐴−4 ,………………
...
Examples based on Cayley-Hamilton theorem:
Example-1:Verify Cayley-Hamilton theorem for the matrix [1
2
]
...
Solution: The characteristic matrix of A is
1−𝜆
|
2
2
|=0
−1 − 𝜆
⇒ 𝜆2 + 5 = 0
On simplification, (1 − 𝜆)(−1 − 𝜆) − 4 = 0
Now, To verify Cayley-Hamilton theorem we have to show that 𝐴2 + 5𝐼 = 𝑂
1
Now 𝐴2 = 𝐴
...
75
(1)
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
2
Example-2:Verify Cayley-Hamilton theorem for the matrix 𝐴 = [−1
1
compute 𝐴−1
...
e,
−1
2−𝜆
−1
−1 1
2 −1]
...
1
−1 | = 0
2−𝜆
on simplification, 𝜆3 − 6𝜆2 + 9𝜆 − 4 = 0
To verify Cayley-Hamilton Theorem, We have to show that
𝐴3 − 6𝐴2 + 9𝐴 − 4𝐼 = 𝑂
(1)
2 −1 1
2 −1 1
6 −5 5
Now, 𝐴2 = 𝐴
...
𝐴 = [−5 6 −5] [−1 2 −1] = [−21 22 −21]
1 −1 2
21 −21 22
5 −6 6
∴ 𝐴3 − 6𝐴2 + 9𝐴 − 4 𝐼
22 −22 −21
2 −1 1
6 −5 5
= [−21 22 −21] − 6 [−5 6 −5] + 9 [−1 2 −1]
21 −21 22
1 −1 2
5 −6 6
1 0 0
− 4 [ 0 1 0]
0 0 1
0 0
⇒ 𝐴 − 6𝐴 + 9𝐴 − 4𝐼 = [0 0
0 0
3
2
0
0] = 𝑂, Hence Cayley-Hamilton Theorem is verified
...
0 0
|𝐴 − 𝜆𝐼| = 0
...
e,
0
2−𝜆
0
1
1
0 |=0
1−𝜆
on simplification, 𝜆3 − 6𝜆2 + 11𝜆 − 6 = 0
verify Cayley-Hamilton Theorem, We have to show that 𝐴3 − 6𝐴2 + 11𝐴 − 6𝐼 =
𝑂
(1)
3 0
Now, 𝐴2 = 𝐴
...
𝐴 = [0 4
0 0
1 3 0
0] [ 0 2
1 0 0
4 3 0
0] [ 0 2
1 0 0
1
9 0
0] = [ 0 4
1
0 0
1
27 0
0] = [ 0 8
1
0 0
27 0
∴ 𝐴 − 6𝐴 + 11𝐴 − 6𝐼 = [ 0 8
0 0
3
4
0]
1
13
0]
1
13
9 0
0 ] − 6 [0 4
1
0 0
2
4
3
0] + 11 [0
1
0
0 1
1
2 0] − 6 [0
0 1
0
0 0
1 0]
0 1
0 0 0
⇒ 𝐴3 − 6𝐴2 + 11𝐴 − 6𝐼 = [0 0 0] = 𝑂, Hence Cayley-Hamilton Theorem is verified
...
Solution:The characteristic equation of the matrix A is |𝐴 − 𝜆𝐼| = 0
Title: Matrix for Engineering mathematics
Description: this is matrix chaptre notes . you can learn easily beacause everthing is easily understandable you don't need to take other person help.
Description: this is matrix chaptre notes . you can learn easily beacause everthing is easily understandable you don't need to take other person help.