Definition
A matrix A of order (or size) m n is a rectangular array of mn elements arranged in m rows and n columns. The element in the i th row and j th column is called the ( , ) -entry of the matrix and is denoted by a ij
.
A

[ a ij
]
 


 a a
11
21 a a
12
22

 a a
1 n
2 n
M M M M



 a m 1 a m 2
 a mn
An n n matrix is called a square matrix of order n . A matrix of order m

1 is called a column vector, and one of order 1
 n is called a row vector.
Here, consider matrices whose elements are, in general, complex numbers. Two matrices A and B are equal when their corresponding elements are equal i.e., a ij
 b ij
for all i , j (which implies that A and B are of the same order).
Addition: Matrices are added (or subtracted) element-wise:
[ a ij
  a ij
 b ij
] , which is of same order.
A matrix A

[ a ij
] can also be multiplied by a scalar (complex number) k as kA

[ ij
 k a ij
] .
An m n matrix A can be multiplied with an p q matrix B if and only if n
 p . Their product is defined as AB
 k n 
.

1


Example


2 3


5 7
11 13







1

1
4 6







2 1 3 4
5 1 7 4
  
   
5 ( 1) 7 6
   
 
14 16
 
33 37
63 67





Matrix multiplication is associative, ( )

( ) ; and distributive,
(

C )

AB

AC , but not commutative (in general), AB

BA . Indeed, the product BA may not even be defined, as in the case of the matrices in the above example.
The identity matrix I of order n is defined as the square matrix of order n with
( , ) -entry 1 if i
 j and 0 if i
 j for all i j

1, 2
 n .
I
 



1
0
0
1


0

0

M M M M
0 0

1



Multiplying any matrix by an identity matrix of appropriate order results in the same matrix, i.e., if A is any m n matrix, then I A

AI n

A . It is easy to show that the identity matrix I the only matrix with this property. For, suppose
K is another such matrix, then IK

K , because of the property of I , and
IK

I , because of the property of K , and thus, I

K .
The transpose of an m n matrix A

[ a ] is defined as A
T 
[ a ] .
Example
If A



1
0

2
3

1
1 4
7 
 9

2 5

1




, then A
T  



1
2

0
3

9

1 1 5
2 



.
4 7

1
Hermitian transpose
The conjugate transpose of A (also called the Hermitian transpose or transjugate) is matrix A * obtained by replacing every element of A
T
by its complex conjugate. Thus, if A

[ a ] , then A *

A
T a .

Example
If A




2
1
6

 i i i
4
4
1 3
 i i
  i


5 i




, then A *





2
 i
4
2 i
1

 i i 4

6
 i i

5
 i




.
Symmetric, Skew-symmetric, Hermitian matrices
A square matrix A is said to be
(i) Symmetric if A

A
T
(ii) Skew-symmetric if
(iii) Hermitian of A

A
A
*
 
A
T
Note: The principal or main diagonal of a square matrix A

[ a ij
] of order n consists of the entries a ii
for every i

1,2
 n .
Trace of a matrix
The trace of a square matrix is the sum of all the elements on the main diagonal, and is denoted by trace( ) or
 n 
. i

1 a ii
Example
If A





10

1 1
3

2 1

4 7 5




, then tr( )
 a
11
 a
22
 a
33

10 ( 2) 5 13 .
Properties
1.
( A

B )
T 
A
T 
B
T and ( A

B )*

A *

B * for all m n matrices A and B .
2.
  T  cA
T and
 
*

* * for all matrices A and scalars c .
3.
 
T
   
A for all matrices A .
4.
 
T  T
B A
T and
 
*

* * (provided the product AB is defined).
5.
tr( A

B )
 tr( )

B for all n n matrices A and B .
6.
tr( cA )
 c tr( ) for all square matrices A and scalars c .
7.
tr
 and tr
  
A for all square matrices A .
8.
tr( AB )
 tr( BA ) (provided the product AB is defined and is a square matrix).

Example
Let A
Then


AB
3

2 5
1

 and B


2 7
1 4


.


7 17
1 34

 and BA


8 37
11 19


.
We observe that tr( AB ) 7 34

27 and tr( BA )
  
27 . Thus, although the products AB and BA , and indeed, even the main diagonal entries of the products are different, tr( AB )
 tr( BA ) .
For every square matrix A , we can associate a unique real (or complex) number, called the determinant of A , and denoted by | A | A , having certain important properties.
Definition
1.
2.
For a 1 1
For an matrix n n
A

[ a
11
] , the determinant is defined to be | A |
 a
11
.
matrix A

[ a ij
] (where n

1 ), the determinant is defined as
| A |
 n
( 1) j


1 j

1 a
1 j
A (1| ) where ( | ) denotes the ( n 1) ( n 1) matrix obtained by deleting the i th row and j th column of A (the determinant A i j is itself called the minor and is denoted by M ij
.
Example
1.
Let A

1 2

.
 3 4 
Then | A |

1 2
3 4
   
 
2 .

2.
Let A




1
2
3
1

1
1 1 2
2




1 3 1
Then | A |
 
1 1 2
2 1

2
1 2

1 2

1 1
1

2 2

2 2 1
 
2 1
 
2 2
 
1 1 1 2

   
 
1 .
.
For a square matrix A of order n , another matrix B of the same order is said to be an inverse of A if AB

BA

I , the identity matrix of order n .
A square matrix that has no inverse is said to be singular, and one that has an inverse is said to be non-singular or invertible.
Theorem
Every invertible matrix has a unique inverse.
Proof
Let A be an invertible matrix, and suppose B
1
and B
2
are inverses of A .
Then B
2

B I (where I is the identity matrix of appropriate order)


(
2
)
1

IB
1
(
1
)
(as

B
1
(as
B
2
B
1
is an inverse of A )
(as matrix multiplication is associative)
is an inverse of A )
.

B
2

B
1
, i.e., the inverse of A is unique. ∎
The unique inverse of A is denoted by A

1
.

Example
If A




1 3 1
1 1 2
2 1

2




, then A

1  




4

7

5
2 4 3


3

5


.
For, AA

1 




1

2
3
1 1
1

1
2
2









4
3
2


4
7
5


3
5
4
 
 

1
0
0
0 1
0
0
0 1




.
Properties
1.
  
1

A for all invertible matrices A .
2.
  
1  1

1 for all invertible matrices A and B of the same order.
3.
   
T for all invertible matrices A .
A square matrix is non-singular if and only if its determinant is non-zero. The inverse of a non-singular matrix A can be calculated in terms of determinants of
A and its submatrices.
Definition
The cofactor C ij
of A i j -entry is defined as
C ij
M ij
, where M ij
is the minor.
The cofactor matrix C of A is the matrix with ( , ) -entry as the cofactor C ij
, for every i , j .
The adjoint (or adjugate) of A , denoted by adj A is the transpose of the cofactor matrix C .
The inverse of a non-singular matrix A is given by the formula
A

1 
1
| A |
 adj A

Example
Let A
 



1 0 2

2 3 1

1 2 1


.

The cofactor matrix is C






1
4


1
1

1
2


6 3 3


and the determinant is | A | 3 .
Thus, adj A

C
T

 

1 4

6
1
1


1
2
3
3




.

A

1  adj A

1
| A | 3




1 4

6

1
1


1
2
3
3




.
A matrix A

[ a ij
] be an n n (real or complex) matrix is said to be
1.
diagonal if a ij

0 for all i
 j .
2.
tridiagonal if a ij

0 for all i , j with | i j | 1 , i.e., all the entries of A are
3.
( zero except (possibly) for those on the main diagonal a
11
, a
22
 a nn
), the superdiagonal ( a
12
, a
23
 a n

1, n
), and the subdiagonal
( a
21
, a
32
 a
,

1
). upper-triangular if a ij

0 for i
 j , i.e., all entries below the main diagonal are zero.
4.
lower-triangular if a ij

0 for i
 j , i.e., all entries above the main diagonal are zero.
5.
orthogonal if all its entries are real and A
T 
A

1
, i.e., AA
T  T 
I .
6.
unitary if it has complex entries and A *

A

1
, i.e., AA *

*

.
Exercises
1.
Classify the following matrices as symmetric, skew-symmetric, Hermitian, or none of these: a.

 
1

3
3 4



b.

 e x e
 ix e ix e
 x

 c.




0 1 1
1 0 1
1 1 0




1
 i d.




1
2

3 i 0
 i e.




0 1 1

1 0 1

1

1 0




3 i

1 



 f.





1 1 1

1 1 1


1

1 1

 g.




1
2

3 i
1

 i i i i
3

1 




2.
Find the value of x such that A x
2 
9
 4

2 x x
2 
2 x

1
0


is a.
symmetric b.
skew-symmetric
3.
Show that every skew-symmetric matrix A

[ a ij
] of order n has all main diagonal entries zero, i.e., a
11
 a
22
 a nn

0 .
4.
Show that every Hermitian matrix has only real numbers along the main diagonal.
5.
Find the determinant of the following: a.


1

1
3 3

 b.

0 1 1

1 0 1



 1 1 0



c.

 e x e
 ix e ix e
 x

 d.

  cos
 sin
 sin
 cos


 e.

 e x e e x
2 e
2 x
2 x

 f.





1 2

2
1 3 0


0

2 1


6.
Determine which of the following are invertible and find the inverse: a.
b.

 
1

1
3 3




1

3 3
1

 c.




0 1 1
1 0 1
1 1 0



 d.




0


1
1

1
0
1
1
1 0



 e.

  cos
 sin
 sin
 cos


 f.




1 x
0 1
0 0 x x
1
2



 g.

1



0 a b
1
0 0 c
1




, where a , b , c are any three complex numbers. h.



 a
0
0 b
0 0
0
0 c





7.
Show that the inverse of every 3 3 upper-triangular matrices with non-zero diagonal entries is also upper triangular.
Hint: If the i th row of A is multiplied by a non-zero constant p , then the i th column of A

1
gets divided by p . Use this property to reduce A to a matrix of the same form as that in Exercise 6g.
8.
Find the inverse of a general n n diagonal matrix with non-zero diagonal entries.
9.
Classify the following matrices as orthogonal, unitary, or neither: a.

1 0

 0 1  b.

0 0 1

0 1 0



 1 0 0

 c.

 
1

1
1 0

 d.


1
1
1

1 

 e.




1
1
2
2

1
1
2
2



 f.
g.




1
1
2
2
 i i

  cos
 sin
 sin
 cos

2
2






Elementary Row and Column Operations
A matrix is said to be in row-reduced echelon form (or reduced row echelon form) if it satisfies the following conditions:
1.
In each row, the first non-zero entry (called the leading entry) should be1.
2.
The leading entry of every row is to the left of the leading entry, if any, of all rows below it.

Note that zero rows (i.e., rows with all entries 0) are allowed, but the second condition implies that all zero rows must occur below all non-zero rows (i.e., rows with at least one non-zero entry).
Example
Among the matrices given below, A and C are in row-reduced echelon form, while B and D are not (why?).
A
 



1 0 2

1
0 0 1 4
0 0 0 0




B
 



1 0 2

1
0 0 0 0
0 0 1 4




C





1
0
2
1
0
0
0
4

3
1


5
3


0 0 0 1 1 2


D





1
0
0
2
1
1
0
0
0
0
4
5

3
4
1



5
3
1




The following three operations performed on a matrix are called elementary row operations:
1.
Interchange : Interchange any two rows. R i

R j
denotes interchanging of the i th and j th rows
2.
Scaling : Multiply every entry of a row by the same non-zero scalar. R i
 kR i denotes multiplying the i th row by the scalar k .
3.
Row Addition : Add a multiple of one row of the matrix to another row.
R i

R i
 kR j denotes adding k times the j th row, to the i th row.
Note : These definitions are motivated by analogous operations that can be performed on a given system of linear equations to obtain equivalent systems of linear equations.
We can use elementary row operations to transform a matrix into its rowreduced echelon form.
Example
A 



4
3

8 0
5 0
0
4

6
4

20
12
2 6 0 9 5

14





~





1
2
3

2
6
0
5 0
0
0
4
9

6
5
1


5
12
14




R
1

R
1
/ 4

1 2 0 0

1

5
~ 0
 0
1
2
0
0
4
9
3
7


3
4




R
2

R
2

3 R
1
R
3

R
3

2 R
1

1 2 0 0

1

5
~ 0

1 0 4 3

0 0 0 1 1 2
3 



R
3

R
3

2 R
2
Elementary column operations can also be defined, similar to elementary row operations. Applying these operations, we can find the analogous columnreduced echelon form of a matrix.
Example
A 



4
3

8 0
5 0
0
4

6
4

20
12
2 6 0 9 5

14




~



4 0 0 0 0 0
3 1 0 4 3

3



 2 2 0 9 7

4


C
2

C
2

2 C
1
C
5

C
5

C
1
C
6

C
6

5 C
1
~





4 0 0 0 0 0

3 1 0
2 2 0
0 0 0
1 1 2



C
4

C
4

4 C
2
C
C
5
6


C
C
5
6


3
3
C
C
2
2
~





4 0 0 0 0 0
3 1 0
2 2 0
0
1
0
0
0
0




C
5

C
5

C
4
C
6

C
6

2 C
4
~





4 0 0 0 0 0
2
3 1
2
0
1
0
0
0 0
0 0




C
3

C
4

All the elementary row and column operations can be defined in terms of multiplication by special matrices called elementary matrices.
Definition
An elementary matrix E is a square matrix that generates an elementary row operation on a matrix A under the multiplication EA .
Example:
Let
Then,
E
 



0 1 0
1 0 0
0 0 1




, A

 

1
4
2
5
3
6
7 8 9




EA





0 1 0
 
1 2 3
1 0 0
 
4 5 6
0 0 1
 
 
7 8 9
 
 
4 5
1 2
6
3


7 8 9


.
.
Thus, left-multiplication by E has the effect of interchanging the first and second rows of A .
Now, AE





1
4 5 6
 
1 0 0
7
2
8
3
9
 
 
 
0
0
1
0
0
1
 
 
2
5
8
1
4
3
6
7 9




.
Right-multiplication by E , therefore, interchanging the first and second columns of A .
Construction of Elementary Matrices
An elementary operation E of order n n that performs a specified elementary row operation is obtained by performing the same operation on the identity matrix of equal order.
Example

Let A

1 2 3 4


5 6 7 8
9 10 11 12




. To construct an elementary matrix, say E
1
, that interchanges the first and third rows of A , consider interchanging the first and third rows of I , we obtain E
1
I

1 0 0


0 1 0
0 0 1







0 0 1
0 1 0


1 0 0


.
. By
Then,





0 0 1 1 2 3 4 9
0 1
1 0
0
0
 
 
 

5 6 7 8
9 10 11 12









10 11 12
5 6 7 8
1 2 3 4


 .
To construct an elementary matrix E
2
that subtracts 5 times the first row of A from the second row of A , perform the same operation on I :
E
2




1 0 0

5 1 0

0 0 1


Then





1

5
0
0
1 0
 
5 6 7 8
0
0
 
1 2 3 4
1
 
  9 10 11 1 2






1 2 3 4

 0

4

8

1

9 10 11 12
2 



.
However, E
1
cannot be used for interchanging the first and third columns of A , as it is not possible to right-multiply A having order 3 4 by E
1
having order
. Therefore, we need to obtain another elementary matrix, say E
3
, from the identity matrix of order 4 by interchanging the first and third columns (or equivalently,
Thus, E
3
 




0 0 1 0
0 1 0 0
1 0 0 0
0 0 0 1




. rows).

Then, AE
3





1 2 3 4
5 6 7 8
9 10 11 12



 



0 0 1 0
0 1 0 0
1 0 0 0




0 0 0 1



3


7
11
2
6
10
1
5
9
4
8
12




Rank and Inverse Using Row Operations
The rank of a matrix is the order of its largest non-singular square submatrix. It is also possible to define and compute the rank in a different manner.
Definition
The row rank of a matrix is the number of non -zero rows in the echelon form of that matrix.
The column rank of a matrix is the number of non -zero columns in the echelon form of that matrix.
Example
A
 



2

1 3
1 0

1
0 2 1




1 1 4
~




1
2
0

0
2
1

1
3
1




R
1

R
2
1 1 4
~




1
0
0

0
2
1

1
1
1




R
2

R
2

2 R
1
R
4

R
4

R
1
 0 1 3

~

1 0 1



0

1 1
0 0 1




R
3

R
3

2 R
2
R
4

R
4

R
2
~
 0 0 4
 1 0 1



0 1

0 0 1
1


 R
2
 
R
2

R
4

R
4

4 R
3
 0 0 0
As there are three non-zero rows in the row-reduced echelon form of A , the row rank of A is 3. Now, we find the column rank.
A
 



2

1 3
1 0

1
0 2 1




1 1 4
~





1 2 3
0 1

1
2 0 1




C
1

C
2
1 1 4
~





1 0 0
0 1 1
2 4 5




C
2

C
2

2 C
1
C
3

C
3

3 C
1
1 3 7
~




 
1 0 0
0 1 0
2
1
4
3
1
4




C
3

C
3

C
2
As there are three non-zero columns in the column-reduced echelon form of A , the column rank of A is 3. We see that the row rank and column rank of A are equal. Indeed, this is true in general, as stated in the following theorem.

Theorem
The row rank of a matrix is the same as the column rank, and the common value is the rank of the matrix.
It is also possible to find the inverse of a non-singular matrix using elementary row (or column) operations. Consider a square matrix A of order n , and the identity matrix I of the same order. We define an augmented matrix [ | ] by appending the columns of I to the columns of A , on the right. Then we reduce A in [ | ] to the identity matrix of order n by performing a series of elementary row operations on the augmented matrix. When A has been reduced to the identity matrix, I would have been reduced to A

1
. If it is not possible to reduce A to the identity matrix, then its row reduced form contains at least one zero row, which means that A is singular and hence has no inverse.
Example
1.
A

1 2

 3 4 


1 2 1 0
3 4 0 1


~
~
~

1 2 1 0
 0

2

3 1


R
2

R
2

3 R
1

1 2 1
 
0

0 1 3 / 2 1 / 2


R

1 0
 

2

1

0 1 3 / 2 1 / 2


2

R
2
/ 2
R
1

R
1

2 R
2

 
2 1
3
2 2




2.
A


0 1 1
1 1 1
1 1 3




[ | ]
 



0 1 1 1 0
1 1 1 0 1
0
0
1 1 3 0 0 1







1 1 1 0 1 0

1 1 3 0
0 0
0 1

 R
1

R
2

1 1 1 0 1 0

 0
1
0
1
2
1
0

0 0
1 1





1 1 1 0 1 0
~ 0 1 1 1
 0 0 1 0

0 0
1 / 2 1 / 2





1 1 1 0 1 0
~ 0 1
 0
0 1
0 1 0
1 / 2

1 / 2

1 / 2
1 / 2





1 0 0

1 1 0
~ 0 1
 0
0
0 1
1
0
1 / 2

1 / 2

1 / 2
1 / 2





A

1  






1 1
1
0

1
2

1 1
0
1
2 2
2






Exercises
Determine the rank of the following matrices by using elementary row operations:
1.




1

2 0
1 1 1
2

1 1




Ans. Rank = 2

2.
3.
4.




1 1 0
2

2 1
1

1 0




Ans. Rank = 3

1 1 1 1

1

1 1 2



 3 1 3 4


Ans. Rank = 2





1 3 1 1
1 1 2 2
2 1

2 1




Ans. Rank = 4
1 2

2 2
5.

1 1 1

1 0 1 
 1 1 0




Ans. Rank = 3
Determine the inverse of the following matrices by using elementary row operations:
1.
2.
3.
4.
A





1 1 0
2

2 1
1

1 0




Ans.
A

1  





1
2
0
1
2
1
2
0

1
0 1

2
2






A


1 1 0

1

1 1 
 1

1 0




Ans.
A

1 






1
2
0
1
2
1
2
0

1
0 1

1
2






A
 




1
2
3
1 1 2
1

1
2




Ans.
A

1
A

1 1 1


1 0 1
1 1 0




Ans.
A

1
 



4

7

5

3
2

4
5
3
4




 




1
1
1

1
0
1

1
0
1





5.
A
 



5 8 1
0
4
2
3

1
1




Ans A

1




5

11

4 9 5
6
8

17

10




Rank of a matrix and solving system of equations
The rank of a matrix A is the maximum number of linearly independent row vectors (or column vectors) in A. A simple application of rank comes from the fact that a linear system of equations AX=B has a solution if and only if the rank of the coefficient matrix A is the same as the rank of the matrix [A|B] where [A|B] denotes the coefficient matrix augmented by column vector B.
Problem : Find the rank of the following matrix using elementary row transformations.
A =






1
2
1
1
2
3
1
1

1
0
3
1



0
4



2
1


.
Solution
: Clearly, rank (A) ≤ min {4, 4} = 4.
Apply row transformations,
Perform, R
2

R
2
+ R
1
, R
3

R
3
-2R
1
, R
4

R
4
-R
1
A


1

 0


0
0
2
5

3

1

1

1
5
2



0
4



2
1


Perform, R
2

R
4,
R
2

- R
2
A


1

 0


0
0
2
1

3
5

1

2
5

1


0
1
2


4 




Perform, R
3

R
3
+ 3R
2
, R
4

R
4
- 5R
2
, R
3

-R
3
A


1

 0


0
0
2
1
0
0

1

2
1
9


0
1
1


9 



Perform, R
4

R
4
-9R
3
A


1

 0


0
0
2
1
0
0

1

2
1
0

0
1
0
1





The above matrix is an echelon form of a matrix and the number of non zero rows in it is 3. Therefore the rank (A) = 3.
Example : Find the rank of the matrix A =

9




1
2
7
9
2
4
7
9

3


6


7 
Solution : Perform, R
1

R
1
9
, R
3

R
3
2
, R
4

R
4
7
A


1

 1


1
1
1
2
2
1
1

3


3


1 
Perform, R
3

R
3
-R
2
, R
4

R
4
-R
1
, R
2

R
2
-R
1
.

A


1

 0


0
0
1
1
0
0
1
2



0


0 
The above matrix is in echelon form and the number of nonzero rows in it is 2 and so the rank (A) = 2.
Example : Find the rank of the matrix by using elementary row transformations.
Solution : Perform, R
1

R
4

1

 4


3
2

3

1

2

2
7
1
3
5
6
1
4
3





Perform, R
2

R
2
-4R
1
, R
3

R
3
-3R
1
, R
4

R
4
-2R
1
.

1

 0


0
0

3
11
7
4
7

27

18

9



6
23



14
9


Perform, R
4

(R
4
+R
3
) – R
2

2

 4


3
1

2

1

2

3
5
1
3
7
3
1
4
6






1

 0


0
0

3
11
7
0
7

27

18
0


6
14
0
23






Perform, R
2
 R
2
11
, R
3
 R
3
11


1


0


0
0

3
1
1
0
7


27
18
11
7
0

6
23
11

2
0






Perform, R
3

R
3
-R
2

1

 0


0
 0

3
1
0
0
7

9
27
11
77
0

6
23
11
1
11
0






Perform, R
3

77
9
R
3


1


0


0
0

3
1
0
0
7

27
11
1
0

6
23
11
7
9
0






The above form is an echelon form and hence the rank is = 3.
Linear Equations-Consistency:
Definitions : A system of m linear equations in n unknowns x
1
, x
2
, …, x n
is a set of equations of the form a
11 x
1
+ a
12 x
2
+ … + a
1n x n
= b
1 a
21 x
1
+ a
22 x
2
+ … + a
2n x n
= b
2
… ... ... _______ (*) say.

...
… ... a m1 x
1
+ a m2 x
2
+ ... + a mn x n
= b m
.
The a ij are given numbers, which are called the coefficients of the system. The b i are also given numbers.
If the b i
are all zero, then (*) is called a homogeneous system. If at least one b i
is not zero, then (*) is called a non-homogeneous system .
A solution of (*) is a set of numbers x
1
, x
2
, .., x n which satisfy all the m equations. If the system (*) is homogeneous, it has at least one trivial solution x
1
= 0, x
2
= 0… x n
=
0.
From the definition of matrix multiplication, we can write the above system (*) as
AX = B where A =

 a a
11 a
12
....
a
1n


21 a
22
....
a
....
....
....
....
2n



....
a m1
....
a m2
....
....
....
a mn






 is the coefficient matrix
X =
 x
1


 x
2


 

 x n
 n 1 and B =

 b
1


 b
2

 

 b m
 
Note that X has n elements where as B has m elements.
The matrix A B or [A:B] =





 a a
11
21
....
a a
12
22
....
.....
....
a
..... a
1 n
2 n
....
b b
1
2
....
 a
....
....
....
....
....
m1 a m2
....
a mn b m







is called the augmented matrix. The matrix A B determines the system (*) completely, since it contains all the given numbers appearing in (*).
Note: The matrix equation AX = B need not always have a solution. It may have no solution or a unique solution or an infinite number of solutions.
Definition : A system of equations having no solution is called an inconsistent system .
A system of equations having one or more solution is called a consistent system .
Observations: Consider a system of non-homogeneous linear equations AX = B.
i) if rank A

rank A B , then system is inconsistent. ii) if rank A = rank A B = number of unknowns, then system has a unique solution. iii) if rank A = rank A B < number of unknowns, then system has an infinite number of solutions.
Example : Test for consistency and solve x + y + z = 6 x – y + 2z = 5
3x + y + z = 8
Solution : The augmented matrix A B =

1 1 1 : 6

1

1 2 : 5



 3 1 1 : 8


Perform R
2

- R + R and R
3

-3R + R
3


A : B
 

1 1 1 : 6



0
0


2
2

1
2 :
:


1
10




Perform R
3

-R + R
3

A : B
 

1 1 1 : 6



0

2 1 :

1
0 0

3 :

9




Now both A and A B have all the three rows non zero. Therefore the rank (A) = 3
= rank A B = the number of independent variables. Therefore the given system of equations is consistent and will have unique solution.
Now convert the prevailing form of [A:B] to a set of equations we get
x + y + z = 6 ……….(i)
– 2y + z = -1………..(ii)
-3z = -9 ………..(iii)
Now from (iii), z = 3, substituting in (ii) we get y = 2. From (i) we get x = 1.
Exercises and Answers
1. Find the rank of the matrix A =

1

 3


4
1
Answer: Rank of A is 3.
7

5
2
5
2. Find the rank of the following matrix A =

1

 2


3
4
Answer: Rank of A is 2.

4
3
2
1





2
1
3
5
1
0
1
2





Gauss Elimination Method

This is the elementary elimination method and it reduces the system of equations to an equivalent upper triangular system which can be solved by back substitution. The method is quite general and is well-adapted for computer operations.
Here we shall explain it by considering a system of three equations for sake of simplicity.
Consider the equations a
1 x + b
1 y + c
1 z = d
1 a
2 x + b
2 y + c
2 z = d
2
…………………… (i) a
3 x + b
3 y + c
3 z = d
3
Step I : To eliminate x from second and third equations.
We eliminate x from the second equation by subtracting

 a
2 a
1


times the first equation from the second equation. Similarly we eliminate x from the third equation by eliminating

 a
3 a
1


times the first equation from the third equation. We thus, get the new system
a
1
x + b
1 y + c
1 z = d
1 b ' y
 c ' z
 d '
2
……………………(ii) b ' y
 c ' z
 d '
3
Here the first equation is called the pivotal equation and a
1
is called the first pivot element.
Step II : To eliminate y from third equation in (ii).
We eliminate y from the third equation of (ii) by subtracting

 b ' b '
3
2

 times the second equation from the third equation. We thus, get the new system

a
1
x + b
1 y + c
1 z = d
1 b ' y
 c ' z
 d '
2
………………..(iii) c '' z
 d ''
3
Here the second equation is the pivotal equation and b '
2
is the new pivot element.
Step III: To evaluate the unknowns
The values of x, y, z are found from the reduced system (iii) by back substitution.
This also can be derived from the augmented matrix of the system (i),
[A|B] =



 a a
1
2 b b
1
2 c c
1
2 a
3 b
3 c
3 d
1 d
2 d
1




…………………(iv)
Using elementary row transformation, reduce the coefficient matrix A to an upper triangular matrix.
That is., [A|B]
~ 


 a
1 b
1 c
0 b '
2 c '
1
2
0 0 c ''
3 d d' d ''
1
2
3




………………..(v)
From (v) the system of linear equations is equivalent to a
1
x + b
1 y + c
1 z = d
1 b ' y
 c ' z
 d'
2

By back substitution we get z, y and x constituting the exact solution of the system (i).
Note: The method will fail if one of the pivot elements a
1
, b '
2
or vanishes. In such cases the method can be modified by rearranging the rows so that the pivots are non-zero. If this is impossible, then the matrix is singular and the equations have no solution.

Example: Solve by Gauss elimination method.
2x + y + 4z = 12
4x + 11y – z = 33
8x – 3y + 2z = 20
The augmented matrix of the system is
[A|B] = 



2 1 4 12
4 11

1 33
8

3 2 20




R
1
R
2
R
3
Perform R
2

R
2
- 2R
1
and R
3

R
3
– 4R
1
[A|B] =




2 1
0 9

4
9
12
9
0

7

14

28




Perform R
2

1
9
3

R
3

7
9
R
2
[A|B] =

2 1

0 1

 0 0

4 12
1 1


3 3


Now we have
2x + 1y + 4z = 12
1y – 1z = 1
3z = 3
Therefore z = 1
Now by the back substitution we get y = 2 and x = 3.
Example: Solve the following system of equations by Gauss elimination method.

5x + y + z + w = 4 x + 7y + z + w = 12 x + y + 6z + w = – 5 x + y + z + 4w = – 6
It is convenient to perform row transformation if the leading entry is 1. We shall write the augmented matrix by interchanging the first equation with the fourth equation.
A B =




1 2
1 7
1 4
1 1

1 1 6 1

5 1 1 1


6

12
5
4




R
1
R
2
R
3
R
4
Perform: R
2

R
2
- R
1
, R
3

R
3
– R
1
and R
4

R
4
-5R
1
A B =




1 1 1 4

6
0 6 0

3 18


0 0 5

3 1


0

4

4

19 34
Perform: R
4

R
4
+
2
3
R
2
A B =




1 1
0 6
0 0
0 0
1
0
5

4

6
3 18



3 1

4

21 46


Perform: R
4

R
4
+
4
5
R
3

A B =


1 1 1
0 6 0




0 0 5

0 0 0



3
117
5
4
3

6
18
1






234
5


Perform R
4

5R
4
A B =




1 1 1
0 6 0
0 0 5
0 0 0


4
3
3

6
18
1





117 234
Now we have
1x + 1y + 1z + 4w = –6
6y + 0z
– 3w
= 18
5z
– 3w
= 1
–117w = 234
Therefore w = –2.
By back substitution we get, z = –1, y = 2, x = 1.
Therefore (x, y, z, w) = (1, 2, –1, –2) is the required solution.
Exercises and Answers
1. Solve the system x - z - 2 = 0 y - z + 2 = 0 x + y - 2z = 0

Answer : Rank (A) = Rank (A:B) = 2 < the number of unknowns = 3. Therefore the system is consistent, and have infinite number of solutions, which involve n – r = 3 –
2 = 1 constant (s).
LU Decomposition Method to solve a System
LU Decomposition of a matrix
Let
A =





 a
11 a
21 a
22
... a
...
a
12
...
...
...
a a n1 a n2
... a
1n
2n
...
nn





 be a nonsingular square matrix. Then A can be factorized into the form LU, where
L =





 l
1 l
21
1 0 ...
0
...
...
... ... ...

 n1 l
0 n2
0 ...
... ...
0
1




and U =





 u
11 u
12
...
u
1n
0
...
0 u
0
22
...
...
...
...
u u
2n
...
nn






, if a

0, a
11 a
12 a
21 a
22 a
11 a
12 a
13

0, a
21 a
22 a
23

0, and so on a
31 a
32 a
33
Such a factorization, when it exists, is unique. Similarly, the factorization LU where
L =





 l l
11
0 ...
...
0 l
21 l
22
...
n1 l
...
n2 l
0
...
n3
...
...
... l
0
...
nn






and U =




1 u
12
...
...
u
1n

 0 1 u
22
...
u
2n
...
...
...
...
...
0 0 0 ...
1





 is also a unique factorization.
Example 1: Factorize the matrix

2 3 1
A = 1


3
2
1
3
2





into the LU form.
Solution: Let





2 3 1
1 2 3
3 1 2



 
 




1 l
21
0
1
0
0
 
 
  u
11
0 u u
12
22 u u
13
23 l
31 l
32
1   0 0 u
33





From the above, we get u
11
= 2, u
12
= 3, u
13
= 1,

l
21
= ½, u
22
= ½, l
32
= - 7 l
31
= 3/2 u
23
= 5/2 u
33
= 18
It follows that



1
L = 1/ 2
3/ 2

0
1
7
0
0 and U = 0 1/2
1








2 3 1
5/2
0 0 18





LU Decomposition Method
Consider the system given by a x a x a x


 a x a x a x


 a x a x a x a x
 b
1 a x
 b
2 a x
 b
1
.
.
n2 2 n3 3 n









………………(1) which can be written in the matrix form
AX =B or
LUX = B ………………(2)
If we set
UX = Y ………………. (3) then equation (2) becomes
LY = B ……………..…(4) which is equivalent to the system l y
 l y
2
 
  
 y y
1
3
 b l y y b
21 1 2 b
1
2
3 
 l y n 1 1
 l y n 2 2
  y n
 b n
…….(5) where

  y
1
Y y
2
  y n
, B =
  b
1 b
2
.
b n
The system (5) can be solved for y
1
, y
2,…, y n
by forward substitution. When Y is known, the system (3) becomes u x u x
12 2 u x
2

11 1
 
  u x
1 n n
 y
1
 u x n y


……………(6) u x nn n
 y n which can be solved for x
1
, x
2
,…, x n
by back substitution process. This method has the advantage of being applicable to solve systems with the same coefficient matrix but different right – hand vectors.
Example 2:
Solve the system of equations
2x + 3y + z = 9
X + 2y + 3z = 6
3x + y + 2z = 8 by the method of LU decomposition.
Solution:
We have
A
 



2
1
3 1
2


3 1 2


 
 
In example 1, we obtained the LU decomposition of A. This is
L



1
1.5

0
7
0
1








2
0.5
1 0 and U = 0 0.5
2.5
0
3
0
1
18




.
If Y = (y
1
, y
2
, y
3
)
T
, then the equation LY = B gives the solution: y
1
= 9, y
2
= 3/2 and y
3
= 5.
Finally, the matrix equation UX = Y, where X = (x, y, z)
T
, gives the required solution

x = 35/18, y = 29/18, z = 5/18.
Tridiagonal Systems:
Consider the system of equations defined by a u
2 1
 b u
1 1 b u
2 2

 c u
1 2 c u
2 3
 d d
1

 a u
3 2
 b u a u
3 3 n n

1

 c u
3 4 b u n n d d


 n

(1)
The matrix of the coefficients is
A



 b c a
0
1
2 b a
1
2
3 c b
0
0 0 0
0 0 0
2
3 c
0
0
3
0
0
0
0 a n
0 0

1 b a n

1 n c b n

1 n






(2)
Matrices of the type, given in equation (2), called the tridiagonal matrices, occur frequently in the solution of ordinary and partial differential equations by finite difference methods.
Thomas method to solve the above system is as follows:
Step I:
Set

1
= b
1
Compute
 i a c i i
 i

1

1 , i

2,3,..., n
Step II:
Set

1
 d
1

1
Compute
 i
 d i
 a
 i i
 i

1 , i

2,3,..., n
Step III:

The solution u i
is obtained from u i
  i
 c u i i
 i

1 , i
 
1, n

2,...,1 where u n
  n
.
Example 1:
Solve the system
2x – y = 0
 x + 2y – z = 0
 y + 2z –u = 0
 z +2u = 1
Solution: b
1
 b
2 b
3 b
4

2 a
2
 a
3
 a
4
 
1 c
1
 c
2
  
3
1 d
1
 d
2
 d
3

0, d
4

1
Applying step I, we get

1

2,

2

3
2
,

3

4
3
,

4

5
.
4
Using step II, we obtain

1

0,

2

0,

3

0,

4

4
.
5
Finally form step III, the solution is u

4
5
, z

3
5
, y

2
5
, x

1
.
5
Exercise
1. Solve the system
2x + y = 0

3x + y + z = 1
2y + z + t = 0
3z – t = 2
2. Solve the system
2x + y =1 x + 2y + z = 3 y + 2z + u = 3
z + 2u =1
ITERATIVE METHODS:
We shall now describe the iterative or indirect methods, which start from an approximation to the true solution and if convergent, derive a sequence of closer approximations – the cycle of computations being repeated till the required accuracy is obtained. This means that in a direct method the amount of computation is fixed, while in an iterative method the amount of computation depends on the accuracy required. In the case of matrices with a large number of zero elements, it will be advantageous to use iterative methods which preserve these elements.
Jacobi’s Method:
Let the system be given by a x a x a x


 a x a x a x


 a x a x a x a x
 b
1 a x
 b
2 a x
 b
1
.
.
n2 2 n3 3 n









In which the diagonal elements a ii do not vanish. If this is not the case, then the equations should be rearranged so that this condition is satisfied. Now, we rewrite the system above as

.
x
1
 b
1
 a
12 a
11 a
11 x
2
 a
13 a
11 x
3
.
x
2
 b
2
 a
21 a
22 a
22 x
1
 a
23 a
22 x
2 x n
 b n
 a n 1 a nn a nn x
1
 a n 2 a nn x
2 a
1 n a
11 x n a
2 n a
22 x n a
,

1 x n

1 a nn
Let x
1
(0)
, x
(0)
2
,....., x
(0) n
be the initial approximation to the solution.
If x
1
, x
2
,....., x n
denotes the approximation to the solution at k th
iteration, then the solution at (k +1) th
iteration is given by
.
.
x
1
( k

1)  b
1
 a
12 a
11 a
11 x
2 x
2
( k

1)  b
2
 a
21 a
22 a
22 x
1 x n
( k

1)  b n
 a a nn a n 1 nn x
1
 a
13 a
11 x
3
 a
23 a
22 x
2
 a a n 2 nn x
2 a
1 n a
11 x n a
2 n x n a
22 a
,

1 x n

1 a nn
This method is due to Jacobi and is called the method of simultaneous displacements.
Gauss Seidel Method:
In Jacobi’s method, to obtain the values at the (k+1) th
iteration, only k th iteration values are used. A simple modification to the above method sometimes yields faster convergence and isdescribed below.
Here, along with the k th
iteration values, all available (k + 1) th iteration values are used. It is clear, therefore, that this method uses an improved component as soon as it is available and it is called the method of successive displacements, or the Gauss – Seidel method.
The Jacobi and Gauss – Seidel methods converge, for any choice of the first approximation x j
(0)
(j= 1, 2, …, n), if every equation of the system satisfies the condition that sum of the absolute values of the coefficients a ij
/a ii is almost equal to, or in at least one equation less than unity, i.e. provided that n  a / a ii
   where the “<” sign should be valid in the case of ‘at least’ one equation. A matrix A satisfying the above condition is known as a diagonally dominant matrix. It can be shown

that the Gauss – Seidel method converges twice as fast as the Jacobi method. The working of the methods is illustrated in the following examples.
Example 3:
10x
1
– 2x
2
– x
3
– x
4
= 3
- 2x
1
+ 10x
2
– x
3
– x
4
= 15
- x
1
– x
2
+ 10x
3
– 2x
4
= 27
- x
1
– x
2
– 2x
3
+ 10x
4
= - 9
To solve these equations by the iterative methods, we rewrite them as follows: x
1
= 0.3 + 0.2x
2
+ 0.1x
3
+ 0.1x
4 x
2
= 1.5 + 0.2x
1
+ 0.1x
3
+ 0.1x
4 x
3
= 2.7 + 0.1x
1
+ 0.1x
2
+ 0.2x
4 x
4
= - 0.9 + 0.1x
1
+ 0.1x
2
+ 0.2x
3
It can be verified that the coefficient matrix is diagonally dominant. The results are given in the following table:
Gauss – Seidel Method n x
1
1 0.3 x
2 x
3 x
4
1.56 2.886 - 0.1368
2 0.8869 1.9523 2.9566 - 0.0248
3 0.9836 1.9899 2.9924 - 0.0042
4 0.9968 1.9982 2.9987 - 0.0008
5 0.99994 1.9997 2.9998 - 0.0001
6 0.9999 1.9999 3.0
7 1.0 2.0 3.0
0.0
0.0
Jacobi Method n
1 x
1
0.3 x
2
1.5 x
3
2.7 x
4
- 0.9
2 0.78 1.74 2.7
3 0.9 1.908 2.916
- 0.18
- 0.108
4 0.9624 1.9608 2.9592 - 0.036
5 0.9845 1.9848 2.9851 - 0.0158
6 0.9939 1.9935 2.9938 - 0.006
7 0.9975 1.9975 2.9976 - 0.0025
8 0.9990 1.9990 2.9990 - 0.0010
9 0.9996 1.9996 2.9996 - 0.0004
10 0.9998 1.9998 2.9998 - 0.0002
11 0.9999 1.9999 2.9999 - 0.0001

12 1.0 2.0 3.0 0.0
From above two tables, it is clear that twelve iterations are required by Jacobi method to achieve the same accuracy as seven Gauss – Seidel iterations.
Example 4:
Solve the system
6x + y + z = 20 x + 4y – z = 6 x – y + 5z = 7 using Jacobi and Gauss – Seidel Method.
(a) Jacobi method:
We rewrite the given system as
20 1 1 x = - y - z = 3.3333 - 0.167y - 0.1667z
6 6 6 y = 1.5 – 0.25x + 0.25z z = 1.4 – 0.2x + 0.2y
In matrix form, the above system may be written as where
X = C + BX
C =





3.3333

1.5
1.4





, B = - 0.25




0 - 0.1667
- 0.1667

0.2
0
0.2
0.25
0




and X= y
 
Assuming
X
(0)
=





3.3333
1.5
1.4





, We obtain
X =





3.3333
 
1.5
1.4



0
+ - 0.25
 
 
0.2
- 0.1667
0
0.2
- 0.1667
0.25
0
 
 
 
 
3.3333
1.5
1.4
 
 
 
2.8499
1.0167
1.0333





X =





3.3333
 
1.5
1.4


+ - 0.25

 
 
0 - 0.1667
- 0.1667
 
2.8499
2.9647
0.2
0
0.2
0.25
0
1.0167
 
 
 
1.0458
1.0333
1.0656





Proceeding in this way, we obtain

X = 1.0012 and X




2.9991

1.0010


(9) 





2.9995
1.0005
1.0004





We, therefore, conclude that
3
  i.e. x = 3, y =1 and z = 1
(b) Gauss Seidel Method:
As before, we obtain the first approximation as



2.8499



1.0333


The subsequent iteration values are given by x
(2)
= 2.9916, y
(2)
= 1.0104, z
(2)
= 1.0038, x
(3)
= 2.9975, y
(3)
= 1.0016, z
(2)
= 1.0008, x
(4)
= 2.9995, y
(4)
= 1.0003, z
(4)
= 1.0002 from which we conclude that x = 3, y = 1 and z = 1.
EIGENVALUES AND EIGENVECTORS
Introduction
We know that many physical systems arising in engineering applications can be represented as discrete models involving matrices. Some key parameters describing physical systems (e.g., the resonance frequency) are closely related to eigenvalues of the matrix representing the system. For example, the natural frequency of the bridge is the eigenvalue of smallest magnitude of a system that models the bridge. The engineers exploit this knowledge to ensure the stability of their constructions. Eigenvalue analysis is also used in the design of car stereo systems, where it helps to reduce the vibration of the car due to the music. In electrical engineering, the application of eigenvalues and eigenvectors is useful for decoupling three-phase systems through symmetrical component transformation.

An eigenvector of an 𝑛 × 𝑛 matrix A is nonzero vector X such that 𝐴𝑋 = 𝜆𝑋 for some scalar 𝝀 . A scalar 𝝀 is called an eigenvalue of A if there is a nontrivial solution X of 𝐴𝑋 = 𝜆𝑋 ; such an X is called an eigenvector corresponding to 𝝀 .

AX
 
X

( A
 
)

O where I is the n th order identity matrix. To find all 𝝀, the determinant of this matrix ( A
 
I ) equated to zero. i.e, A
 
I
 a
11 a
21
  a a
12
22
 
....
....
a
1 n a
2 n
....
....
.... ....
a n 1 a n 2
....
a nn
 

0 (1) is called as the characteristic equation of the matrix A . On expanding the determinant, the characteristic equation takes the form n n
1 n-1
2 n-2 n
(2) where the k i
’s are expressible in terms of the elements a ij
of the matrix A .
The roots of the characteristic equation of A are the eigenvalues, latent roots or the characteristic roots of the matrix A .
The matrix equation (1) represents n homogenous linear equations
( a
11
 
) x
1
 a x
2 a x
21 1

( a
22


.....


) x
2

.....
 a x n a x
2 n n


0
0
.... .... .... ....
a x n 1 1
 a x n 2 2
  a nn
 
) x n

0
(3) which will have a non-trivial solution if and only if the coefficient matrix is singular, i.e, if A
 
I

0 .
The characteristic equation of A has n roots (need not be distinct) and corresponding to each root, the equation (3) will have a non-zero solution, which is the eigenvector or latent vector.
Examples:
1. Find the eigenvalues and eigenvectors of the matrix A =

5 4
 1 2


.

Solution: The characteristic equation is A
 
I

0 i.e,
5
 
4
1 2
 

0 or

2 
7
  
0 or (
 
6)(
  
0
 
6,1
If X

[ , ]
T
( A
 
)
is an eigenvector corresponding to the eigenvalue


5
 
1 2
4
 

 x
1
  

0 .

,then
Corresponding to
 
6 , we have



1 4
1

4

 x
1
  

0 which gives only one equation  
4 x
2

0 .
 x
1
 x
4 1
2 giving the eigenvector (4,1)
T .
Corresponding to
 
1 , we have


4 4
1 1

 x
1
  

0 which gives only one equation x
1
 x
2

0 .
 x
1
1
 x

2
1
giving the eigenvector (1, 1)
T .
2. Find the eigenvalues and eigenvectors of the matrix A =



1 1 3

1 5 1


3 1 1


.
Solution: The characteristic equation is given by A
 
I

1
 
1
1 5
 
3
3
1

0
1 1
  i.e

3 
7

2 
36

0 or (
 
2)(
 
3)(
 
6)

0 .

2,3, 6
If X
  x
1
, x
2
, x
3

T
is an eigenvector corresponding to an eigenvalue

, then we have


A
   

1



1

5
3
1
 
1 1
3
1
  x
1   
  
  

O
For
 
2 , we have the equations 3 x
1 x
2
3 x
3

0 x
1

7 x
2
 x
3

0
3 x
1 x
2
3 x
3

0
Solving, we obtain x

1
 x
2
 x
1 0 1
3 , an eigenvector is ( 1, 0,1)
T .
Also, every non-zero multiple of this vector is an eigenvector corresponding to
  
2 .
Similarly, the eigenvectors corresponding to
 
3 and
 
6 are the arbitrary non-zero multiples of the vectors (1, 1,1)
T and (1, 2,1)
T .
Exercises:
1.Find the eigenvalues and eigenvectors of the matrices: a) 



8

6 2

6 7

4
2

4 3




b) 




2 1

1
1 1

2


1

2 1


c)




2 0 1
0 2 0
1 0 2




Properties of Eigenvalues:
1. Any square matrix A and its transpose A
T have the same eigenvalues.
2. The eigenvalues of a triangular matrix are the diagonal elements of the matrix.
3. The eigenvalues of an idempotent matrix are either zero or unity.
4. The sum of the eigenvalues of a matrix is the sum of the elements of the principal diagonal(trace of a matrix).
5. The product of the eigenvalues of a matrix is equal to its determinant.
6. If

is an eigenvalue of a matrix A , then
1

is the eigenvalue of A

1 .

7. If

is an eigenvalue of an orthogonal matrix, then
1

is also its eigenvalue.
8. If
 
1 2
,....
 n
are the eigenvalues of a square matrix A of order n then A m has the eigenvalues  
1 m m
,
2
,.....
 n m where m is a positive integer.
Cayley-Hamilton Theorem
Statement: Every square matrix satisfies its own characteristic equation. i.e if the characteristic equation for the n th order square matrix A is
A
 
I ( 1) n
 n  k
1
 n

1  k
2
 n

2 
.....
k n
0 , then
 n n
A
 k A n

1  k A n

2 
.....
 
0
…………………. (4)
Multiplying (4) by A

1 , we get n n

1
A
 k A n

2 
.....
 k n

1
I
 k A

1 
0 .

A

1  
1 k n

 n
A n

1  k A
1 n

2 
....
 k n

1
I .
This result gives the inverse of A in terms of n

1 powers of A and is considered as a practical method for the computation of the inverse of the large matrices.
Examples:
1. Verify Cayley-Hamilton theorem for the matrix A =

1 4
 2 3


and hence find its inverse.
Solution: The characteristic equation of A is given by
1
 
4
2 3
 

0 or

2 
4

5 0 .
By Cayley-Hamilton theorem, A satisfies its characteristic equation, i.e.
A
2 
4 A

5 I

O .

Now, A
2 
4 A

5 I



1 4
 
1 4
2 3   2 3



4


1 4
2 3



5


1 0
0 1





9 16

4 16 5 0
8 17 8 12 0 5





0 0
0 0



O
This verifies the theorem.
Multiplying by A

1 , we get A

4 I

5 A

1 
O or
A

1 
1
5
( A

4 )

1
5




1 4
 2 3
 


4 
1 0
0 1






1
5



2
3

4
1


.
2. Using Cayley-Hamilton theorem, find A
8 , if A


1 2
2

1


.
Solution: The characteristic equation of A is given by
1
 
2

2  
0 .
A
8 
( A
2 4
)

( A
2 
5 I

5 )
4

( A
2 
5 )
4 
4( A
2 
I
3
5 ) 5 I

6( A
2 
I
2
5 ) (5 )
2 
4( A
2 
I I
3 
I
4
By Cayley-Hamilton theorem, A
2 
5 I

O .

A
8 
I
4 
625 I .
2


0 or
3. Using Cayley-Hamilton theorem, find A

2 , where A


1 2 0



2
0

0
1

0
1




.
Solution: The characteristic equation of A is given by
1
 
2
0
 
2 
5

5 0 (5)
2

0
0
0


0 or
Multiplying by A

1 , we get 
A 2  
5 I

5 A

1 
O or A

1 
1
5
A
2  
5 I .

Multiplying by
5 A

2   
5 A

1
A I 5
A

1
A 2  
5 I
, we get

6 I

A
2
  
5 A

1 
5 A

2 
O

A

2 
1
5
6 I

A
2 
1
5






6
0
0
6
0
0
 
 
5
0
0
5
0
0 
 0 0 6 0 0 1









1
5




1 0 0
0 1 0
0 0 5




or
4. If A
 



2
1
2
3
1
1
1 2 2




, then find the eigenvalues of A

1 and hence, find its determinant.
Solution: The eigenvalues of A are given by the roots of its characteristic equation, i.e A
 
I
 
3 
7

2 
11

5 0 .
5,1,1 .
If

is an eigenvalue of A , then
1

is an eigenvalue of A

1 .
Therefore, the eigenvalues of A

1 are
1
5
,1,1 .

A

1 
1
5
.
Exercises:
1. Verify Cayley-Hamilton theorem for the following matrices and find its inverse: a)




2
1

2
1

1
1


1

1 2


b)





7 2

2
6

1 2


6 2

1


c)

3 2 4

4 3 2 
 2 4 3




2. If A 



2
1

2
1

2
1


1

1 2


, find A
4 .

3. Find the eigenvalues of A
2 
2
 where A
 

2 3 4


0 4 2
0 0 3




.
Iterative method to obtain largest eigenvalue of a matrix (Rayleigh Power method)
Consider the matrix equation AX
 
X which gives the following homogenous system of equations
( a
11
 
) x
1
 a x
12 2 a x
21 1

( a
22


.....


) x
2

.....
 a x
1 n n a x
2 n n


0
0
.... .... .... ....
a x n 1 1
 a x n 2 2
  a nn
 
) x n

0
In many applications, it is required to compute numerically the largest eigenvalue and the corresponding eigenvector.
We start with a column vector X which is as near the solution as possible and evaluate AX , which is written as 
(1) (1)
X after normalisation, that is, we obtain
 (1) from AX which is largest in magnitude. Hence, the resulting vector X
(1) will have largest component unity. This gives the first approximation  (1) to the largest eigenvalue and
AX
(1) 
X
(1) to the eigenvector. Similarly, we evaluate

(2) (2)
X which gives the second approximation. We repeat this process till
X
(r) 
X
( r

1) becomes negligible. Then,  will be the largest eigenvalue of A and X , the corresponding eigenvector.
This iterative method to find the dominant eigenvalue of a matrix is known as
Rayleigh’s power method.
For the initial approximation, we can take the vector X

[1, 0, 0]
T , if it is not explicitly provided.
Note: The largest eigenvalue of A

1 is the smallest eigenvalue of A .
Examples:
1. Determine the largest eigenvalue and the corresponding eigenvector of the matrix A 



2
1

2
1

0
1


0

1 2


up to 4 iterations.

Solution: Let the initial approximation to the required eigenvector be
X
 
1, 0, 0

T
.
AX


2
0
1


2
1
1

0
2
1
1
0
0
2
  
0
1 2





1
0.5
0




 
(1)
X
(1)
AX
(1)


2
1

2
1

0
1
 
 
 

1
0.5
 
2.5
 
2




2.5

1




0.8



0

1 2   0   0.5
  0.2


(2)
X
(2)
AX
(2)


2

1 0
 
1
 
1 2

1
 
 

0.8
 
0

1 2   0.2
 
2.8
2.8
1.204





2.8





1
1
0.43




 
(3)
X
(3)
AX
(3)


2

1 0
 
1
 
2.9841

1 2

1
 
 

1
 
3.43
0

1 2   0.43
  1.8522




3.43

0.87





1
0.54



  (4)
X
(4)
Hence, the largest eigenvalue is 3.43
and the corresponding eigenvector is

0.87,

1, 0.54

T
.
2. Find the largest eigenvalue and the corresponding eigenvector of the matrix
A
 




1 3

1
3 2 4


1 4 10


with the initial approximation

1, 1, 0

T
. Carry out 4 iterations.
Solution: Given X
 
1, 1, 0

T
we have
AX





1 3

1 1 4

0.8

3 2 4 1

5

5 1


 
1 4 10       0.6

 
(1)
X
(1)
AX
(1) 






1
3
3
2

4
1
1 4 10
 
 
 
 
0.8
1
0.6
 
 
 
3.2
6.8
9.2






0.3478
9.2 0.7391
 1




 
(2)
X
(2)
AX
(2) 





1 3

1
 
0.3478
3 2 4
 
0.7391
 
 
1 4 10
 
  1  
1.5652
6.5217





12.6087




0.1241



1
 
(3)
X
(3)

AX
(2) 





1 3

1
 
0.1241
3 2 4
 
0.5172
 
 
1 4 10
 
  1  
0.6759
5.4069
11.9448






0.0566


1




 
(4)
X
(4)
Hence, the largest eigenvalue is 11.9448
and the corresponding eigenvector is

0.0566, 0.4527, 1

T
Exercises:
1. Find the largest eigenvalue and the corresponding eigenvector of the matrix




1
4

4
3

2
1


6 3 5


using the initial approximation

1, 0, 0

T
. Carry out 4 iterations.
2. Find the dominant eigenvalue and the corresponding eigenvector of the matrix





6
2

3
2

2
1


2

1 3


. Tak