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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.
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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
...
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
...
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
...
y
4
z
1 1
0
4
0 7
1 1
0 1
0 1
⸫ Rank of A=2 < Number of unknowns,(n=4)
...
Here n-rank of A = 4-2=2
...
2
Now equation (1) becomes AX 0
0
3
3
0
w
1 1
x
0 1 0
...
3
2
Example 9: Show that the equations :
2 x y z a, x 2 y z b, x y 2 z c
...
Find
the solutions when a 1, b 1, c 2
...
Therefore ,if the
given, system of equations have a solution then the rank of the augmented matrix must also be
2,which is possible only if a+b+c=0
...
When a+b+c=0, the matrix equation (1) reduces to
1
0
0
1
1
0
2 x
c
c
b
1 y
3
0 z
0
x y 2 z c, y z c b
3
when a 1, b 1, c 2, we have x y 2z 2 and y z 1
Taking z k, we have y k -1, x -2 - k 1 2k i
...
x k -1, y k -1, z k
...
57
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
Example 10: Solve
x 2 y 3z 14,3x y 2z 11,2x 3 y z 11
Solution:
The matrix equation AX=B is given by
1
3
2
2
1
3
3 x 14
2 y 11
1 z 11
1
A : B 3
2
(1)
3 : 14
2 : 11
1 : 11
1 2
R21 3, R31 2 ~ 0 5
0 1
2
1
3
1
R2 1, R3 1 ~ 0
0
1
R23 ~ 0
0
2
5
1
3 : 14
7 : 31
5 : 17
3 : 14
7 : 31
5 : 17
3 : 14
5 : 17
7 : 31
3
: 14
1 2
R32 5 ~ 0 1
5
: 17
0 0 18 : 54
1 2 3 : 14
R3 1
~ 0 1 5 : 17 which is in Echelon form
...
The give equations are consistent and have a unique solution
...
Practice Questions:
1
What will be the value of
...
are consistent
...
find the values of x,y,z
...
Ans (i) a=3,b≠9(ii) a≠3, b any value (iii) a=3,b=9
...
Ans:x=(k1+k2)/2, y =k2,z=-k1,w=k1,where k1 and k2 are two are arbitrary constants
...
Ans: x=-10/7k, y=8/7k, z=k, where k is arbitrary
...
for which the equations
11-λ x-4 y-7 z 0
7 x- λ 2 y-5 z 0
10 x- 4 y-6 λ z 0
possess a non-trivial solution
...
,find the solution also
...
Ans : a 5 , 9 (b) 5 , arbitrary (c) 5 , 9
8
Verify that the following set of equations have a non-trivial solution:
9
Verify that the following system of equations has always a solution:
x 3 y 2z 0,2x y 4z 0, x 11y 14 z 0
4x 3 y z 0,3x 4 y z 0, x y 2z 0,5x y 4z 0
...
Ans: No solution, system inconsistent
1
...
Xn } be a set of vectors (row matrices or column matrices)
...
Xn are said to be linearly dependent (L
...
) if
(i)
X1,X2,……
...
(1)
If a1≠ 0, then transposing a1X1 to the other side and dividing by (-a1), the relation (1)may
be written as
X = b2X2+ b3X3+……
...
(2)
The relation (2) shows that the vector X1 is a linear combination of the vectors X2,X3,…
...
(B) The vectors X1,X2,…………
...
I) if every relation of
the form
a1 X1+ a2X2+……+ an Xn= 0
...
an=0 i
...
, all the n scalars a1, a2……………an vanish
...
Find the relation between them
...
Putting the values of a1 , a 2 , a3 , a 4 in (1), we get
X 1 X 2 2X 3 3X 4 0
X 1 X 2 2 X 3 3 X 4 0 which is the required solution
...
Solution: Consider the vector equation
a1 X 1 a2 X 2 a3 X 3 0
a1 2,1,3,0 a2 1,2,1,5 a3 1,3,2,7 0,0,0,0
(1)
2a1 a2 a3 0,-a1 2a2 3a3 0,3a1 a2 2a3 0,0a1 5a2 7a3 0
(2)
The system (2) of equations is homogeneous and so it is consistent
...
Rank of A 3 number of unknowns
...
e
...
Hence the given vectors are linearly independent
...
Solution: The given vectors will be linearly dependent if one vector can be expressed as a linear
combination of the remaining vectors
...
7
Putting for a1 and a2 in (4), we get
5
7
3
5
...
Practice Question
Test the following system of vectors for linear dependence
...
Ans: Linearly independent
...
Ans: Linearly independent
...
Ans: Dependent, 2X1-6X2+X3-2X4=0
...
Ans: Dependent, 2X1+X2-X3=0
5)
X 1 1,3,2, X 2 (1,7 8), X 3 (2,1,1)
...
1
...
63
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
Then geometrically each vector on the line through the origin determined by X gets mapped back
onto the same line under multiplication by A
...
Thus the finding of non-zero vectors that get mapped into scalar multiples of themselves under a
linear operator are most important in the study of vibrations of beams, probability (Markov
process), Economics (Leontief model),genetics, quantum mechanics, population dynamics and
geometry
...
Eigen values
If A is a square matrix of order n, we can form the matrix A - I
...
The determinant of this matrix equated to zero,i
...
,
a11
A - I
a 21
a12
...
a 2n
...
a n1
a n2
...
0
is called the characteristic equation of A
...
a nn
On expanding the determinant, the characteristic equation can be written as a polynomial
equation of degree n in
of the form -1n n k1n1 k2n2
...
The roots of this equation are called characteristic roots or latent roots or eigenvalues of A
...
In practice, we are often required
to find those vectors X which transform into scalar multiples of themselves
...
Then Y X
(2)
From (1) and (2), AX X AX IX 0 A I X 0
This matrix equation gives n homogeneous linear equations
(3)
a11 x1 a12 x2
...
a2 n xn 0
...
ann xn 0
These equations will have a non-trivial solution only if the co-efficient matrix
64
(4)
(5)
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
A I 0
This equation is called Characteristic equation of the matrix A and has n roots
which are the Eigen values of A
...
...
If A is of nth order, the
number of Eigen values arte n or less than n(with repeated real roots or complex
conjugate pairs
...
(3)If all the n Eigen values of A are distinct, then there corresponds n distinct linearly
independent Eigenvectors
...
Properties of Eigen values and Eigen Vectors
1
...
2
...
3
...
4
...
5
...
6
...
7
...
65
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
8
...
9
...
10
...
11
...
12
...
Example 1:Calculate the eigenvalues and eigenvector for the matrix
5 4
A
...
⸫ Eigen values of A are 1,6
...
1
The eigen vector corresponding to eigen value 6 is given by
A I X 2 0 A 6I X 2 0
4 x1
5 6
1 4 x1
0
1
0
2 6 x2
1 4 x2
1 4 x1
R21 1 ~
0
0 0 x2
x1 x2
This gives equation x1 4 x2 0 or 4 1
4
X 2 is the eigen vector of A for the eigenvalue 6
...
1 2
The eigen vector corresponding to eigen value 3 is given by
A I X 2 0 A 3I X 2 0
2 x1
8 3 6
5 6 2 x1
6 7 3 4 x2 0 or 6 4 4 x2 0
2
2 4 0 x3
4 3 3 x3
2 x1
1 2
R13 2, R23 3 ~ 0 8 4 x2 0
2 4 0 x3
2 x1
1 2
R31 2 ~ 0 8 4 x2 0
0 8 4 x3
2 x1
1 2
R32 1 ~ 0 8 4 x2 0
0 0
0 x3
x1 2 x2 2 x3 0,8 x2 4 x3 0
x2 x3
2
x
x
k say x2 k , x3 2k
...
2
The eigen vector corresponding to eigen value 15 is given by
68
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
A I X 3 0 A 15I X 3 0
2 x1
8 15 6
6 7 15 4 x2 0
2
4 3 15 x3
7 6 2 x1
6 8 4 x2 0
2 4 12 x3
1 22 46 x1
R13 4, R23 30 20 40 x2 0
2 4 12 x3
1 22 46 x1
R31` 2 ~ 0 20 40 x2 0
0 40 80 x3
1 22 46 x1
R32` 2 ~ 0 20 40 x2 0
0
0
0 x3
x1 22 x2 46 x3 0,20 x2 40 x3 0
...
2 1
x1 44k 46k 0 x1 2k
...
1
Example 3: Calculate the eigenvalues and eigenvector for the matrix
3 1 4
A 0 2 6
...
x2 x1
1
X 1 1, is the eigen vector for 2
...
Solving these equations, we have
x3 0, x2 0, x1 1say
1
X 2 0, is the eigen vector for 3
...
e
...
e
...
1
Example 4: Calculate the eigenvalues and eigenvector for the matrix
1 2 2
A 0 2 1
1 2 2
Solution: Characteristic equation is given by A I 0
1
2
2
2
0
2
1 0 1 2 0 1,2,2
1
2
2
The eigen vector corresponding to eigen value 1 is given by
A I X 1 0 A 1I X 1 0
2 x1
1 1 2
0 2 2 x1
0 2 1 1 x2 0 0 1 1 x2 0
1
1 2 1 x3
2
2 1 x3
0 0 0 x1
R12 2, R32 1 ~ 0 1 1 x2 0
1 1 0 x3
x2 x3 0 and x1 x2 0
x1 x2 x3
1
1 1
1
Hence X1 1 , is the eigen vector for 1
...
Hence arbitrary values will be
given to one variable i
...
, equation has only one linearly independent solution
...
2
The the only eigen vector for 2 is X 2 1
...
k1
1
0
The eigen vector corresponding to eigen value 5 is given by
A I X 0 A 5I X 0
7 2 3 x1
2 4 6 x2 0
1 2 5 x3
On solving, we get
x1 k , x2 2k , x 3 k3
1
Hence , eigen vector for 5 is X 2 2
...
Find the eigen values and eigenvector of the matrix
Ans:4,6,
8 4
2 2
2
...
Find the eigen values and eigenvector of the matrix
2 1 1
1 2 1
0 0 1
Ans:1,1,3
4
...
For the matrix A= 0 3 2 ,find the eigenvalues of
0 0 2
3A3+5A2-6A+2I
...
Find the eigen values and eigenvector of the matrix
1 1 3
1 5 1
3 1 1
7
...
Find the eigen values and eigenvector of the matrix
3 10 5
2 3 4
3
5
7
9
...
Find the eigen values and eigenvector of the matrix
74
1 1 1
1, 0 , 1
0 1 0
2 2 1
2, 1, 2
1 2 2
Ans:4,110,10
Ans:-2,3
...
18 Cayley-Hamilton Theorem
Statement: Every square matrix satisfies its own characteristic equation
...
This method of finding inverse of a matrix A is most suitable for large
order matrix
...
etc
...
2 −1
|𝐴 − 𝜆𝐼| = 0
...
𝐴 = [
2
2 1 2
5
][
]= [
−1 2 −1
0
From equation (1), 𝐴2 − 5𝐼 = [
0
]
5
1
5 0
] − 5[
0
0 5
0
0 0
]=[
]=𝑂
1
0 0
Hence, Cayley- Hamilton theorem is verified
...
Solution: The characteristic matrix of A is
2−𝜆
| −1
1
i
...
Hence
−1 2
|𝐴 − 𝜆𝐼| = 0
...
𝐴 = [−1 2 −1] [−1 2 −1] = [−5 6 −5]
1 −1 2
1 −1 2
5 −6 6
2 −1 1
22 −22 −21
6 −5 5
𝐴3 = 𝐴2
...
0
Now for 𝐴−1 , Pre multiplying equation (1) by 𝐴−1 ,
We have,𝐴−1 (𝐴3 − 6𝐴2 + 9𝐴 − 4𝐼) = 𝐴−1 𝑂 = 𝑂
(𝐴2 − 6𝐴 + 9𝐼 − 4𝐼𝐴−1 ) = 𝑂
𝐴2 − 6𝐴 + 9𝐼 = 4𝐴−1
2 −1 1
1 0
6 −5 5
⇒ 4𝐴−1 = [−5 6 −5] − 6 [−1 2 −1] + 9 [0 1
1 −1 2
0 0
5 −5 6
1 3 1
𝐴−1 = [ 1 3
4
−1 1
76
−1
1]
3
0
3 1 −1
=
]
[
0
1 3 1]
1
−1 1 3
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
3 0
1
Example-3:Verify Cayley-Hamilton theorem for the matrix 𝐴 = [0 2 0]
...
Solution: The characteristic matrix of A is
3−𝜆
| 0
0
i
...
𝐴 = [0 2
0 0
9 0
and𝐴3 = 𝐴2
...
0 0 0
1
2
Example-4:Show that the matrix 𝐴 = [2 −1
0
0
0
0 ] satisfies its own characteristic equation
−1
and hence
or otherwise obtain 𝐴−2
...
1−𝜆
⇒| 2
0
2
−1 − 𝜆
0
0
0 |=0
−1 − 𝜆
On simplification, 𝜆3 + 𝜆2 − 5𝜆 − 5 = 0
According to Cayley- Hamilton theorem we have, 𝐴3 + 𝐴2 − 5𝐴 − 5𝐼 = 𝑂
1
Now, 𝐴2 = 𝐴
...
𝐴=[0 5
0 0
0
0 1 2
5
]
[
]
=
[
0 2 −1 0
10
1 0 0 −1
0
10
0
−5
0]
0 −1
From equation (1),
5
𝐴3 + 𝐴2 − 5𝐴 − 5𝐼 = [10
0
1 0 0
1
2
0
10
0
5 0 0
0 ] − 5 [ 0 0 0]
−5
0] + [0 5 0] − 5 [2 −1
0 0 1
0
0 −1
0 −1
0 0 1
0
= [0
0
0 0
0 0] = 𝑂
0 0
For 𝐴−2 ,now pre multiplying equation (1) by 𝐴−1
𝐴−1 (𝐴3 + 𝐴2 − 5𝐴 − 5𝐼) = 𝐴−1 𝑂 = 𝑂
⇒ 𝐴2 + 𝐴 − 5𝐼 − 5𝐴−1 = 𝑂
(2)
Again pre multiplying equation (2) by 𝐴−1 ,𝐴−1 (𝐴2 + 𝐴 − 5𝐼 − 5𝐴−1 ) = 𝐴−1 𝑂 = 𝑂
⇒ 5𝐴−2 = 𝐴 − 5𝐴−1 + 𝐼
⇒ 5𝐴−2 = 𝐴 − (𝐴2 + 𝐴 − 5𝐼) + 𝐼 = -𝐴2 + 6𝐼
1 0 0
5 0 0
⇒ 5𝐴−2 = − [0 5 0] + 6 [0 1 0]
0 0 1
0 0 1
6 0
−5 0
0
⇒ 5𝐴−2 = [ 0 −5 0 ] + [0 6
0 0
0
0 −1
−2
⇒𝐴
1 1 0
= [0 1
5
0 0
0
1 0
0] = [0 1
6
0 0
0
0]
5
0
0]
5
2 1 1
Example-5:Find the characteristic equation of the matrix 𝐴 = [0 1 0] and hence compute
1 1 2
𝐴
...
−1
8
2
7
6
1 1
Solution:Given matrix is 𝐴 = [0 1 0]
1
1 2
The Characteristic equation of A is |𝐴 − 𝜆𝐼| = 0
78
5
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
2−𝜆
⇒| 0
1
1
1−𝜆
1
1
0 |=0
2−𝜆
On simplification,𝜆3 − 5𝜆2 + 7𝜆 − 3 = 0
Now, to verify Cayley-Hamilton theorem we have to show that
𝐴3 − 5𝐴2 + 7𝐴 − 3𝐼 = 𝑂
(1)
For 𝐴−1 , Pre multiplying equation 1) by 𝐴−1
𝐴−1 (𝐴3 − 5𝐴2 + 7𝐴 − 3𝐼) = 𝐴−1 𝑂
⇒ 𝐴2 − 5𝐴 + 7𝐼 − 3𝐴−1 = 𝑂
⇒ 3𝐴−1 = 𝐴2 − 5𝐴 + 7𝐼
2 1
Now, 𝐴2 = 𝐴
...
using equation (1)
= 𝐴4 − 5𝐴3 + 8𝐴2 − 2𝐴+I
= 𝐴4 − 5𝐴3 + 7𝐴2 − 3𝐴 + 𝐴2 + 𝐴+I
= 𝐴(𝐴3 − 5𝐴2 + 7𝐴 − 3𝐼)+𝐴2 + 𝐴+I
= 𝐴(𝑂)+𝐴2 + 𝐴+I , using equation (*1)
=𝐴2 + 𝐴+I
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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
5 4 4 2 1
= [0 1 0]+[0 1
4 4 5 1 1
1
1 0
0] + [ 0 1
2
0 0
0
0]
0
8
= [0
5
5 5
3 0]
5 8
Practice Questions:
1
...
Using Cayley- Hamilton theorem, find the inverse of 𝐴 = [6
2
4
20
1
Ans:𝐴−1 = 130 [−42
50
20 −30
3
4 ]
8
9
−3 ]
20
4
3
1
3
...
If𝐴 = [ 1
2
], Use Cayley-Hamilton theorem to express 𝐴6 − 4𝐴5 + 8𝐴4 − 12𝐴3 + 14𝐴2 as
−1 3
a linear polynomial in A
...
Given 𝐴 = [0
3
2
1
−1
−1
−1 ], find 𝑎𝑑𝑗𝐴, by using Cayley-Hamilton theorem
...
19 Similarity Transformation:
80
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
Let A and B be any two square matrices of order ‘ n ‘
...
Note: Similar matrices have the same set of Eigen values
...
1
...
The process of reduction of matrix A to a diagonal form
D is known as Diagonalization of matrix A
...
where D is known as the
spectral matrix of A
...
Note: The matrix B which diagonalizes A is called the modal matrix of A and which can be
obtained by grouping the Eigen vectors (suitably by choosing arbitrary constants as 1) of A into a
square matrix
...
Solution: Given matrix is 𝐴 = [4 1]
2 3
The characteristic equation of A is |𝐴 − 𝜆𝐼| = 0
|
4−𝜆
2
1
|=0
3
On expanding the determinant,(4 − 𝜆)(3 − 𝜆) − 2=0
⇒ 𝜆2 − 7𝜆+10= 0 ⇒ 𝜆 = 2, 5
Now, we have to find the eigen vectors corresponding to the eigen values 𝜆1 = 2, 𝜆2 = 5
For 𝜆1 = 2 eigen vector 𝑋1 is given by 𝐴𝑋1 = 𝜆1 𝑋1 ⇒ (𝐴 − 𝜆1 𝐼)𝑋1 = 𝑂
81
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
[
⟹[
Where,
𝐵=[
2 1
]
2 1
𝑥1
4−2
1
0
] [𝑥 ] = [ ]
2
3−2 2
0
1 𝑥1
0
] [𝑥 ] = [ ] ⟹ 𝐵𝑋1 = 𝑂
1
0
2
2
2
𝑅1
2
, 𝑅1 →
1
2]
1
∼ [1
2
, 𝑅2 → 𝑅2 − 2𝑅1
∼ [1
0
1
′
2] ∼ 𝐵
0
The equivalent system of equations is 𝐵′ 𝑋1 = 𝑂
[
⟹ 𝑥1 +
1 1/2 𝑥1
0
][ ] = [ ]
0
0
0 𝑥2
𝑥2
=0
2
(1)
Let 𝑥2 = 𝑘1 , from equation (1) we have 𝑥2 = −𝑘1 /2
−𝑘1 1
𝑥1
−𝑘 /2
1
∴ 𝑋1 = [𝑥 ] = [ 1 ] =
[ ] = 𝑘′ [ ]
𝑘1
−2
2
2 −2
For 𝜆2 = 5 eigen vector 𝑋2 is given by 𝐴𝑋2 = 𝜆2 𝑋2 ⇒ (𝐴 − 𝜆2 𝐼)𝑋2 = 𝑂
[
𝑥1
0
4−5
1
] [𝑥 ] = [ ]
0
2
3−5 2
𝑥1
1
0
] [𝑥 ] = [ ] ⟹ 𝐶𝑋2 = 𝑂
−2
0
2
−1
2
⟹[
Where,
𝐶=[
−1 1
]
2 −2
, 𝑅1 → −𝑅1
1
∼[
2
−1
]
−2
∼[
, 𝑅2 → 𝑅2 − 2𝑅1
1 −1
] ∼ 𝐶′
0 0
The equivalent system of equations is 𝐶 ′ 𝑋2 = 𝑂
1
[
0
−1 𝑥1
0
][ ] = [ ]
0 𝑥2
0
⟹ 𝑥1 − 𝑥2 = 0
(2)
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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
Let 𝑥2 = 𝑘2 , from equation (1) we have 𝑥1 = 𝑘2
𝑥1
𝑘
1
∴ 𝑋2 = [𝑥 ] = [ 2 ] = 𝑘2 [ ]
𝑘
1
2
2
On choosing 𝑘′ = 𝑘2 = 1
𝑋1 = [
1
1
] , 𝑋2 = [ ]
−2
1
Now the Modal matrix of A is
1 1
𝑀= [
]
−2 1
And𝑀−1 =
1 1
[
3 2
−1
]
1
𝑀−1 𝐴𝑀 =
1 1
[
3 2
1 1 −1
−1 4 1 1 1
2 0
2 5
][
][
] = 3[
][
]= [
]=𝐷
1 2 3 −2 1
2 1 −4 5
0 5
Ans
...
Now,
𝑀−1 𝐴𝑀 = 𝐷
Pre multiplying equation (α)by𝑀 and post multiplying by 𝑀−1
𝑀(𝑀−1 𝐴𝑀)𝑀−1 = 𝑀𝐷𝑀−1
⇒ (𝑀𝑀−1 )𝐴(𝑀𝑀−1 ) = 𝑀𝐷𝑀−1
⇒ 𝐴 = 𝑀𝐷𝑀−1
Now,𝐴2 = 𝐴
...
Example-3:The eigen vectors of a 3 × 3 matrix a corresponding to the eigen values 1,1,3 are
[1,0
...
−1]′ and[1,1
...
Find the matrix A
...
−1]′, [0, 1
...
0]′
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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
1
0 1
The modal matrix of A is 𝑀 = [ 0
1 1]
−1 −1 0
1 0
And 𝐷 = [0 1
0 0
0
0]
3
Now we have to find 𝑀−1 :
𝑀~𝐼
𝟏
0 1
1 0
[0
1 1 ] ~ [0 1
−1 −1 0
0 0
0
0]
1
, 𝑅3 → 𝑅3 + 𝑅1
1 0 1
1 0
⇒ [0 𝟏 1] ~ [0 1
0 −1 1
1 0
0
0]
1
, 𝑅3 → 𝑅3 + 𝑅2
1 0
⇒ [0 1
0 0
1
⇒ [0
0
1
0 1
0
1 1] ~ [1
0 𝟏
2
1
⇒ [0
0
⇒ 𝑀−1
1
1 0 0
1] ~ [0 1 0]
2
1 1 1
0
1
1
2
0
0
1]
2
, 𝑅3 →
𝑅3
2
, 𝑅1 → 𝑅1 − 𝑅3 , 𝑅2 → 𝑅2 − 𝑅3
1 −1 −1
2
2
2
0 0
−1 1 −1
1 0] ~
2
2
2
0 𝟏
1
1
1
[2
2
2]
1 −1 −1
2
2
2
1 1 −1 −1
−1 1 −1
=
= [−1 1 −1]
2
2
2
2
1
1
1
1
1
1
[2
2
2]
We know that, 𝑀−1 𝐴𝑀 = 𝐷
Pre multiplying equation (α) by 𝑀 and post multiplying by 𝑀−1
𝑀(𝑀−1 𝐴𝑀)𝑀−1 = 𝑀𝐷𝑀−1
87
KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
⇒ (𝑀𝑀−1 )𝐴(𝑀𝑀−1 ) = 𝑀𝐷𝑀−1
⇒ 𝐴 = 𝑀𝐷𝑀−1
1
1
⇒ 𝐴 = 2[ 0
−1
0
1
−1
1
𝐴=[0
−1
0
1
−1
1 1
1] [ 0
0 0
0
1
0
1 1
1] [−1
0 3
−1
1
3
−1
2
−1] = [1
3
0
0 1 1
0] [−1
2
3
1
1
2
0
1
1]
1
−1
1
1
−1
−1]
1
Ans,
Practice Questions:
9
−1 −9
−1 3 ] by, means of similarity transformation
...
Diagonalize the matrix 𝐴 = [ 3
1 −6 −4
4
2 ]by ,means of similarity transformation
...
Diagonalize the matrix 𝐴 = [0
3
1
3
...
Hence, find P such that 𝑃−1 𝐴𝑃 is a
2
diagonal matrix
...
Diagonalize the matrix𝐴 = [
]
7 31
1
2 −2
2
1 ], find the modal matrix P and resulting diagonal matrix D of A
...
If 𝐴 = [ 1
2
2 −2
2
Ans
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.