Matrix Algebra
Matrix: A system of any mn numbers arranged in a rectangular array of m rows and n columns is
called a matrix of order m  n . A matrix is usually denoted by a single capital letter, namely A,
B, C, … … or by the symbols  aij  ,  aij  , aij .
The matrix of order
m n
is written as:
 a11
a
 21
 ...

 am1
a12
a22
...
am 2
a1n 
... a2 n 
... ... 

... amn  mn
...
 1 2 3 
1 
1 2 3
 


Example: A   2 3 1 ; B   2  ; C  1 2 313 ; D  
.
4 5 6  23

 3  31
 3 1 2  33
Difference between a matrix and a determinant:
The differences between a matrix and a determinant are as follows:
Matrix
Determinant
1. A matrix cannot be reduced to a single 1. A determinant can be reduced to a single
number.
number.
2. In a matrix, the number of rows may not be
equal to the number of columns.
3. An interchange of rows or columns gives a
different matrix.
1 2  1 2 3
4. Examples: 
 ; 
.
3 4   4 5 6 
2. In a determinant, the number of rows must
be equal to the number of columns.
3. An interchange of rows or columns gives the
same determinant with +ve or –ve sign.
1 2 3
1 2
4. Examples:
; 2 1 0 .
3 4
2 3 4
Complex Matrix: Any matrix having complex elements is called a complex matrix.
3
 2  i 2i

3i
1 .
Example: A   2
 3 1  2i 2i 
Conjugate of a Complex Matrix: The matrix obtained from any given matrix A of order
m n
with complex elements aij , by replacing its elements by the corresponding conjugate complex
numbers is called the complex conjugate or conjugate of A and is denoted by A .
3
2i
3 
 2  i 2i
2  i



3i
1 then A   2
3i
1  .
Example: If A   2
 3 1  2i 2i 
 3 1  2i 2i 
Real Matrix: A matrix A is called real if each element is a real number or it satisfies the relation
A  A.
 1 2 3 


Example: A   2 3 1 .
 3 1 2 
Imaginary Matrix: A matrix A is called imaginary if each element is imaginary or it satisfies the
relation A   A .
 i 2i 3i 

3i i  .
Example: A   2i
 3i i
2i 
.
Rectangular Matrix: A matrix A of order m  n is called a rectangular matrix if the number of
rows and the number of columns are not equal i.e, m  n .
1 2 
1 2 2 


Example: A  
; B  3 4  .

2 4 1
5 6 
Square Matrix: A matrix A of order m  n is called a square matrix if the number of rows and the
number of columns are equal i.e, m  n .
 1 2 3 
1 2 


Example: A  
; B   2 3 1

3 4 
 3 1 2 
Horizontal Matrix: A matrix A of order m  n is called a horizontal matrix if the number of rows
is less than the number of columns, i.e, m  n .
1 2 2 
Example: A  
.
2 4 1
Vertical Matrix: A matrix A of order m  n is called a horizontal matrix if the number of rows is
more than the number of columns, i.e, m  n .
1 2 


Example: A  3 4  .
5 6 
Row Matrix: A matrix A is called a row matrix row vector if it contains only one row.
Example: A  1 2 3 .
Column Matrix: A matrix A is called a column matrix or column vector if it contains only one
column.
1 
 
Example: A   2  .
 3 
Null Matrix: A matrix, rectangular or square, each of whose elements is zero is called a Zero
matrix or Null matrix and is denoted by O.
0 0 0 
0 0 0 


Example: A  0 0 0  ; B  
.
0 0 0 

0 0 0 
Identity Matrix: A square matrix whose elements aij  0 when i  j and aij  1 when i  j is called
the Identity matrix or Unit matrix and is denoted by Ior U.
1 0 0 


Example: I  0 1 0 
0 0 1 
.
Diagonal Matrix: A square matrix whose elements aij  0 when i  j is called a Diagonal
matrix. The elements aij when i  j are known as diagonal elements and the line along which
they lie is known as the principal diagonal or leading diagonal.
1 0 0 


Example: A  0 2 0 
0 0 3
Scalar Matrix: If in a square matrix A all the diagonal elements are equal to a nonzero element
aand all the remaining elements are equal to zero then it is called a Scalar matrix.
2 0 0


Example: A   0 2 0 
 0 0 2
Triangular Matrix: A square matrix whose elements aij  0 when i  j or i  j is called a
Triangular matrix.
1 3 2 4 
1 2 3 




Example: A  0 1 2 3  ; B  0 2 4 
0 0 5 7 
0 0 3 
Upper Triangular Matrix:A square matrixAwhose elements aij  0 when i  j is called an
upper triangular matrix.
2
0
Example: A  
0

0
3 4 5
1 3 4 
0 2 5

0 0 7
Lower Triangular Matrix: A square matrix A whose elements aij  0 i  j is called a lower
triangular matrix.
2
3
Example: A  
4

5
0 0 0
1 0 0 
3 2 0

4 5 7
Sub-matrix: A matrix, which is obtained from a given matrix by deleting any number of rows and
columns or both, is called a sub-matrix of the given matrix.
1 2 3 4 


Example: If A  5 6 7 8  be a matrix then
0 7 5 2 
1 2 3 
5 6 7  ;  3 4 

 7 8  etc. are its sub-matrices.

0 7 5  
Minor of a matrix: Let A be an m  n matrix. The determinant of the square sub-matrix of order
r  r obtained
by deleting  m  r  rows and  n  r  columns from A, is called a minor of order r
of A.
1 2 2 3 1 3
1 2 3
Example: If A  
then
are minors of order 2 of A.
;
;

4 5 5 6 4 6
4 5 6
Equality of matrices: Two matrices A   aij  and B  bij  are said to equal iff


They are of the same dimensions i.e, they are of the same order.
The elements in the corresponding positions of the two matrices are same.
Order or dimension of a matrix: The order or dimension of a matrix is given by stating the
number of rows and the number of columns in the matrix.
 1 2 1 3 


Example: A   4 3 2 0  is a matrix of order 3  4 .
 2 1 0 4 
Addition and subtraction of matrices: If the two matrices A and B are of the same order, then
only their addition and subtraction is possible and these matrices are said to be Conformable for
Addition or Subtraction.
given by A   aij  and B  bij  , then the
matrix A  B is defined as the matrix each element of which is the sum of the corresponding
elements of A and B i.e. A  B  aij  bij  , where i  1, 2, 3,....., m and j  1, 2,3,....., n .
Addition: If A and B be two matrices of order
m n
1 2
 2 3
1  2 2  3  3 5
Example: If A  
then A  B  
and B  



.
3 4 
 4 5
3  4 4  5   7 9 
given by A   aij  and B  bij  , then the
matrix A  B is defined as the matrix each element of which is obtained by subtracting the elements
of B from the corresponding elements of A i.e. A  B  aij  bij  , where i  1, 2, 3,....., m and
Subtraction: If A and B be two matrices of order
m n
j  1, 2,3,....., n .
1 2
 2 3
1  2 2  3  1 1
and B  
Example: If A  
then A  B  



.
3 4 
 4 5
3  4 4  5  1 1
Multiplication of matrices:
If A and B be two matrices such that the number of columns in A is equal to the number rows in
B i.e. if A   aij  and B  b jk  are m  n , n  p matrices then the product of the matrices A and
B denoted by AB is defined as matrix
C  cik 
n
  aij b jk
j 1
In the matrix product AB, the matrix A is called the pre-multipliers and B is called the postmultipliers.
1 0 
Example: If A  1 2 and B  
 then AB  1  4 0  6  5 6 .
 2 3
Transpose of a matrix: The matrix obtained from any given matrix A by interchanging its rows
and columns is called its transpose. The transpose of A, is denoted by
At or A' .
1 2 
1 0 1 


t
Example: If A  0 1  , then A  
.
2 1 2 

1 2
Commutative matrices: If A and B are two square matrices such that AB=BA, then A and B are
called commutative matrices or are said to commute.
Anti-commutative matrices: If A and B are two square matrices such that AB  BA , then A and
B are called anti-commutative matrices or are said to anti-commute.
NOTE: The transpose of the product of two matrices is equal to the product of their
t
transpose in reverse order, i.e.  AB   Bt At .
Determinant of a square matrix:
Answer: The determinant whose elements are exactly the same as those of a square matrix A, is
called the determinant of the square matrix A and denoted by A .
1 1
1 1
Example: If A  
; then A 
.

0 2
0 2 
Singular matrix: A square matrix A is said to be a singular matrix, if the determinant ofAis zero,
i.e. A  0 .
1 2
 1 2 
 0.
Example: Let A  
; then A 

1 2
 1 2 
Non-Singular matrix: A square matrix A is said to be a non-singular matrix, if the determinant
of A is not zero, i.e. A  0 .
1 2
 1 2
Example: Let A  
; then A 
 0.

1 2
 1 2 
Symmetric matrix: A square matrix A   aij  is said to be a symmetric matrix if aij  a ji for all
values of i and j i.e. At  A .
a

Example: Let A   h
 g
g
a
f  ; then At   h
 g
c 
h
b
f
h
b
f
g
f   A .
c 
Here A is A symmetric matrix.
Skew-Symmetric matrix: A square matrix A   aij  is said to be a skew-symmetric matrix if
aij  a ji for all values of i and j i.e. At   A .
h
0

Example: Let A   h 0
  g  f
g
h
 0 h  g 
0




t
f  ; then A   h 0  f     h 0
 g f
  g  f
0 
0 
g
f   A .
0 
Here A is a skew- symmetric matrix.
NOTE: Every square matrix can be expressed uniquely as the sum of a symmetric matrix
and a skew-symmetric matrix.
 Let A be a square matrix of order n and AT be the transpose of A. Then we can write,
A
1
1
A  AT    A  AT 

2
2
where
1
A  AT  = Symmetric matric.

2
1
A  AT  = Skew- symmetric matrix.

2
Hermitian matrix: A square complex matrix A   aij  is said to be a Hermitian matrix if it is
nn
 
t
equal to the transpose of its conjugate complex i.e. if A  A .
 l
Example: A  
  i
  i 
m 
Skew-Hermitian matrix: A square complex matrix A   aij  is said to be a Skew-Hermitian
nn
 
t
matrix or anti-Hermitian matrix if A   A .
1  i
 i
Example: A  
0 
 1  i
Unitary matrix: A square complex matrix A   aij 
is said to be a Unitary matrix if
nn
 
t
AA*  A* A  I where A  A .
*
Example: A 
1 1 i 
1  1 1 i
 i 1 ; B 


2
3 1  i 1 

Normal matrix: A square complex matrix A   aij 
is said to be a Normal matrix if
nn
 
t
AA*  A* A where A  A .
*
1 
 2  3i
Example: A  
1  2i 
 i
Idempotent matrix: A square matrix A   aij  is said to be an Idempotent matrix if A2  A .
nn
 1 3 5 


Example: A   1 3 5
 1 3 5 
Periodic matrix: A square matrix A   aij  is said to be a Periodic matrix if Ak 1  A , where
nn
k is a positive integer.
 1 2 6 


Example: A   3 2 9  is a periodic matrix of period 2.
 2 0 3
Involutory matrix: A square matrix A   aij  is said to be an Involutory matrix if A2  I .
nn
4 3
 2 1
Example: A  
; B


 5 4
 3 2 
Nilpotent matrix: A square matrix A   aij  is said to be a Nilpotent matrix of order m , if
nn
m
m 1
A  O and A
 O , where m is a positive integer and O is the null matrix. If m is the least
m
positive integer such that A  O , then m is called the index of the nilpotent matrix A.
0 1 
Example: A  
 is a Nilpotent matrix of index 2.
0 0
Orthogonal matrix: A square matrix A   aij  is said to be an orthogonal matrix if AAt  I ,
nn
t
where I is an identity matrix and A is the transposed matrix of A.
 cos 
Example: A  
  sin 
2
1 2
3 3  3


2 1
2 
sin  

; B
3 3
3 
cos  


2  2  1
3
3 
 3
Trace of a matrix: Let A be a square matrix of order n. Then the sum of the elements of A lying
along the principal diagonal is called the trace of A. We shall write trace of A as tr.A.
 cos 
Example: Let A  
  sin 
sin  

cos  
So that tr.A= cos   cos  = 2cos  .
 1 2 3 
 3 1 2 




Problem-01: If A   5 0 2  and B   4 2 5  then find A  B and 2A  B .
 1 1 1 
 2 0 3




Solution: The given matrices are,
 1 2 3 
 3 1 2 




A   5 0 2  and B   4 2 5 
 1 1 1 
 2 0 3




 1 2 3   3 1 2 

 

Now, A  B =  5 0 2    4 2 5 
 1 1 1   2 0 3 

 

 1  3 2  1 3  2 


= 5  4 0  2 2 5 
 1  2 1  0 1  3 


 4 1 1


= 9 2 7 
 3 1 4 


 1 2 3   3 1 2 

 

Again, 2A  B = 2  5 0 2    4 2 5 
 1 1 1   2 0 3 

 

 2 4 6   3 1 2 

 

= 10 0 4    4 2 5 
 2 2 2   2 0 3 

 

 2  3 4  1 6  2 


= 10  4 0  2 4  5 
 2  2 2  0 2  3 


 1 5 8 


=  6 2 1  .
 0 2 1 


1 2
2 3 5
Problem-02: Let 𝐴 = (
) and 𝐵 = (2 3). Find AB and BA. Hence show that 𝐴𝐵 ≠
3 −1 4
3 4
𝐵𝐴.
Solution: The given matrices are,
Now,
1
2 3 5
𝐴𝐵 = (
) (2
3 −1 4
3
2
3)
4
=(
2 + 6 + 15 4 + 9 + 20
)
3 − 2 + 12 6 − 3 + 16
=(
23 33
)
13 19
1
𝐵𝐴 = (2
3
2
2 3 5
)
3) (
3 −1 4
4
2+6
=(4+9
6 + 12
8
= (13
18
1
3
5
Hence 𝐴𝐵 ≠ 𝐵𝐴. [Showed]
3−2
6−3
9−4
13
22)
31
5+8
10 + 12)
15 + 16
1 4 0
3 2 1
 3 2 1






Problem-03: If A   2 5 0  , B   1 2 3  and C   1 2 3  then find AB and BC .
 3 6 0
 4 5 6
7 8 9






Solution: The given matrices are,
1 4 0
3 2 1
 3 2 1






A   2 5 0  , B   1 2 3  and C   1 2 3 
 3 6 0
 4 5 6
7 8 9






1 4 0  3 2 1



Now, AB   2 5 0   1 2 3 
 3 6 0  4 5 6



 1.3  4.1  0.4 1.2  4.2  0.5 1.1  4.3  0.6 


  2.3  5.1  0.4 2.2  5.2  0.5 2.1  5.3  0.6 
 3.3  6.1  0.4 3.2  6.2  0.5 3.1  6.3  0.6 


 3  4  0 2  8  0 1  12  0 


  6  5  0 4  10  0 2  15  0 
 9  6  0 6  12  0 3  18  0 


 7 10 13 


  11 14 17  .
15 18 21


 3 2 1  3 2 1



Again, BC   1 2 3   1 2 3 
 4 5 6 7 8 9



 3.3  2.1  1.7 3.2  2.2  1.8 3.1  2.3  1.9 


  1.3  2.1  3.7 1.2  2.2  3.8 1.1  2.3  3.9 
 4.3  5.1  6.7 4.2  5.2  6.8 4.1  5.3  6.9 


6 48
369 
 927


  3  2  21 2  4  24 1  6  27 
12  5  42 8  10  48 4  15  54 


 18 18 18 


  26 30 34  .
 59 66 73 


1 2
) then find 𝐴2 + 3𝐴 + 𝐼.
−3 0
Problem-04: If 𝐴 = (
1 2
Solution: Given, 𝐴 = (
)
−3 0
1 2
1 2
Now, 𝐴2 = (
)(
)
−3 0 −3 0
=(
1 × 1 + 2(−3)
1×2+2×0
)
(−3) × 1 + 0 × (−3) (−3) × 2 + 0 × 0
=(
−5 2
)
−3 −6
−5
−3
So, 𝐴2 + 3𝐴 + 𝐼 = (
1 2
1
2
) + 3(
)+(
−3 0
0
−6
0
)
1
=(
3 6
1 0
−5 2
)+(
)+(
)
−9 0
0 1
−3 −6
=(
−1
8
) (Ans.)
−12 −5
2
Problem-05: If 𝐴 = (1
1
1 3
−2 2) then find the value of 𝐴3 − 2𝐴2 − 9𝐴.
2 1
2 1 3
Solution: Given, 𝐴 = (1 −2 2)
1 2 1
2 1
Now, 𝐴2 = (1 −2
1 2
8
= (2
5
3 2
2) (1
1 1
4+1+3
1 3
−2 2) = (2 − 2 + 2
2+2+1
2 1
6 11
9
1)
−1 8
8 6 11 2
So, 𝐴 = 𝐴 . 𝐴 = (2 9
1 ) (1
5 −1 8
1
3
2
1 3
−2 2)
2 1
2−2+6 6+2+3
1 + 4 + 4 3 − 4 + 2)
1−4+2 3+4+1
16 + 6 + 11 8 − 12 + 22
=( 4+9+1
2 − 18 + 2
10 − 1 + 8
5 + 2 + 16
33
Now, 𝐴 − 2𝐴 − 9𝐴 = (14
17
3
2
33
= (14
17
24 + 12 + 11
33
6 + 18 + 1 ) = (14
15 − 2 + 8
17
18 47
8 6 11
2
−14 25) − 2 (2 9
1 ) − 9 (1
23 21
5 −1 8
1
18
−14
23
47
16
25) − ( 4
21
10
33 − 16 − 18
= ( 14 − 4 − 9
17 − 10 − 9
−1 −3
= ( 1 −14
−2
7
12 22
18
18 2 ) − ( 9
−2 16
9
18 − 12 − 9
−14 − 18 + 18
23 + 2 − 18
−2
5)
−4
18 47
−14 25)
23 21
1 3
−2 2)
2 1
9
27
−18 18)
18
9
47 − 22 − 27
25 − 2 − 18 )
21 − 16 − 9
(Ans.)
1 −2 3
Problem-06: If 𝐴 = ( 2
3 −1) then find the value of 𝐴2 − 3𝐴 + 9𝐼3 .
−3 1
2
1 −2 3
Solution: Given, 𝐴 = ( 2
3 −1)
−3 1
2
1 −2 3
1 −2 3
Now, 𝐴2 = 𝐴. 𝐴 = ( 2
)
(
3 −1
2
3 −1)
−3 1
2
−3 1
2
1−4−9
= ( 2+6+3
−3 + 2 − 6
−2 − 6 + 3 3 + 2 + 6
−12
−4 + 9 − 1 6 − 3 − 2 ) = ( 11
6 + 3 + 2 −9 − 1 + 4
−7
−12
𝐴 − 3𝐴 + 9𝐼3 = ( 11
−7
2
−5 11
4
1)
11 −6
1 −2 3
1 0
−5 11
3 −1) + 9 (0 1
4
1 ) − 3( 2
−3
1
2
0 0
11 −6
−12 − 3 + 9
= ( 11 − 6 + 0
−7 + 9 + 0
−5 + 6 + 0 11 − 9 + 0
−6 1
4−9+9
1+3+0 ) =( 5 4
11 − 3 + 0 −6 + 6 + 9
2 8
1 2
0
) and 𝐵 = (
3 −4
1
Problem-07: Let 𝐴 = (
5
). Prove that (𝐴𝐵)𝑇 = 𝐵 𝑇 𝐴𝑇 .
7
1 2
0
Solution: Given, 𝐴 = (
) and 𝐵 = (
3 −4
1
5
)
7
0
0)
1
2
4 ) (Ans.)
−3
1 3
)
2 −4
So, 𝐴𝑇 = (
0
5
and 𝐵 𝑇 = (
1
)
7
1 2
2
19
0 + 2 5 + 14
0 5
)(
)=(
)=(
)
3 −4 1 7
−4 −13
0 − 4 15 − 28
𝐴𝐵 = (
Now, (𝐴𝐵)𝑇 = (
2
19
−4
) … … … … … … … … … . . (1)
−13
0 1 1
)(
5 7 2
Again, 𝐵 𝑇 𝐴𝑇 = (
0+2
0−4
2
3
)=(
)=(
5 + 14 15 − 28
19
−4
−4
) … … … … . (2)
−13
From (1) & (2) , we get (𝐴𝐵)𝑇 = 𝐵 𝑇 𝐴𝑇 .
 1 2 3 


Problem-08: Express A   0 1 2  as the sum of a Symmetric and Skew-symmetric
 4 5 3 


matrices.
Solution: The given matrix is,
 1 2 3 


A   0 1 2 
 4 5 3 


 1 0 4 


The transpose of A is A   2 1 5 
 3 2 3 


t
The symmetric matrix of A is,
 1 2 3   1 0 4  
1
1 
 

t
A  A    0 1 2    2 1 5  

2
2 
 

 4 5 3   3 2 3  
 2 2 7 
1

  2 2 7 
2

 7 7 6 

 1 1

 1
1


 7  7
2
2
7 
2 

7

2

3 

The skew-symmetric matrix of A is,
 1 2 3   1 0 4  
1
1 
 

t
A  A    0 1 2    2 1 5  

2
2 
 

 4 5 3   3 2 3  
 0 2 1
1

  2 0 3 
2

 1 3 0 

1
0

  1 0


 1  3
2
2
1
 
2

3 
2 

0 

7  
1

0
1  
 1 1


2
2

 

7 
3 

1   1 0
Therefore, A   1
.
2 
2 

 

 7  7 3   1  3 0 
2
2
2
 2

1 0 6 


Problem-09: Express A   4 4 1 as the sum of a Symmetric and Skew-symmetric matrices.
 0 9 6 


Solution: The given matrix is,
1 0 6 


A   4 4 1
 0 9 6 


1 4 0 


t
The transpose of A is A   0 4 9 
 6 1 6 


The symmetric matrix of A is,
 1 0 6   1 4 0  
1
1 
 

t
A  A    4 4 1   0 4 9  

2
2 
 

 0 9 6   6 1 6  
6 
2 4
1

  4 8 10 
2

 6 10 12 
1 2 3 


  2 4 5 
 3 5 6 


The skew-symmetric matrix of A is,
 1 0 6   1 4 0  
1
1 
 

t
A  A    4 4 1   0 4 9  

2
2 
 

 0 9 6   6 1 6  
 0 4 6 
1

  4 0 8
2

 6 8 0 
 0 2 3 


  2 0 4
 3 4 0 


 1 2 3   0 2 3 

 

Therefore, A   2 4 5    2 0 4  .
 3 5 6   3 4 0 

 

Inverse of a Matrix: A square matrix A is said to be invertible if there exists a unique matrix B
such that AB=BA=I, where I is the identity matrix. Then B is called the inverse of A and it is
denoted by B= A1 .
Mathematically,
A1 
adj. A
A
; where, A  0
Note:1.A matrix A has inverse iff it is non-singular. i.e. A  0 .
2. The inverse of a matrix, if it exists, is unique.
Cofactors of a square matrix: If A be any
 a11

a
i.e, A   21
 ...

 an1
a12
a22
...
an 2
n  n square matrix
... a1n 

... a2 n 
... ... 

... ann 
The determinant of A is,
A
a11
a12
... a1n
a21
...
an1
a22
...
an 2
... a2 n
... ...
... ann
Then the cofactor of its any entry aij is defined as,
Aij   1
i j
 det er min ent of submatrix madeby 

.
 delating ith rowand jth column of A 
 a11

a
Adjoint matrix: Let A   21
 ...

 an1
a12
a22
...
an 2
... a1n 

... a2 n 
be any
... ... 

... ann 
of entries aij , then the matrix of cofactors from A is,
n  n square matrix and Aij be the cofactors
 A11

 A21
 ...

 An1
... A1n 

... A2 n 
... ... 

... Ann 
A12
A22
...
An 2
The transpose of this matrix is called the adjoint of A and is denoted by adj(A).
 A11

A
i.e, adj ( A)   21
 ...

 An1
A12
A22
...
An 2
t
... A1n   A11
 
... A2 n   A12

... ...   ...
 
... Ann   A1n
A21
A22
...
A2 n
... An1 

... An 2 
.
... ... 

... Ann 
Block Matrix: A matrix A may be partitioned into a system of similar matrices, called blocks,
by a set of horizontal and vertical lines. The matrix A is then called a block matrix.
1 2 3


Problem-01: Find the adjoint matrix of the matrix A   2 3 1  .
 3 1 2


Solution: The given matrix is,
1 2 3


A  2 3 1
 3 1 2


The cofactors of each elements of A are,
A11 
3 1
1 2
 6 1
5
A23  
1
A12  
;
   4  3 ;
 1
2
3 1
  1  6 
5
2 1
3 2
A13 
2 3
A21  
3 1
 29
 7
;
2
1
3
2
   4  3 ;
 1
A22 
1 3
3 2
 29
 7
;
A31 
2 3
3 1
 29 ;
 7
A32  
1
3
2 1
A33 
  1  6  ;
1 2
2 3
 3 4 .
 1
5
The matrix of cofactors is,
 5 1 7 


 1 7 5 
 7 5 1 


 5 1 7 


 adj. A   1 7 5 
 7 5 1 


t
 5 1 7 


  1 7 5 
 7 5 1 


This is required adjoint matrix.
 4 3 3 


Problem-02: Find the adjoint matrix of the matrix A   1 0 1  .
4 4 3


Solution: The given matrix is,
 4 3 3 


A 1 0 1 
4 4 3


The cofactors of each elements of A are,
A11  4 ; A12  1 ; A13  4
A33  3 .
The matrix of cofactors is,
;
A21  3 ; A22  0 ; A23  4 ; A31  3 ; A32  1 ;
 4 1 4 


 3 0 4 
 3 1 3 


 4 1 4 


 adj. A   3 0 4 
 3 1 3 


t
 4 3 3 


 1 0 1 
 4 4 3


This is required adjoint matrix.
 2 3 4


Problem-03: Find the inverse of the matrix A   4 3 1  .
1 2 4


Solution: The given matrix is,
 2 3 4


A  4 3 1
1 2 4


The determinant of A is,
2 3 4
A4 3 1
1 2 4
 2 12  2   3 16  1  4  8  3
 20  45  20
 5
Since, A  0 . So the given matrix is a non-singular matrix and it has an inverse matrix.
The cofactors of each elements of A are,
A11  10 ; A12  15 ; A13  5
A33  6 .
;
A21  4 ; A22  4 ; A23  1 ; A31  9 ; A32  14 ;
The matrix of cofactors is,
 10 15 5 


 4 4 1 
 9 14 6 


 10 15 5 


 adj. A   4 4
1 
 9 14 6 


t
 10 4 9 


  15 4 14 
 5 1 6 


Therefore,
A1 
adj. A
A
 10 4 9 
1

   15 4 14 
5

 5 1 6 
This is required inverse matrix.
1 2 3 
1 1 1 




Problem-04: If A  1 3 5  and B  1 2 3  then find AB 1 .
1 4 9 
1 5 12 




Solution: The given matrices are,
1 2 3 
1 1 1 




A  1 3 5  and B  1 2 3 
1 4 9 
1 5 12 




The determinant of B is,
1 1 1
B 1 2 3
1 4 9
 118  12   1 9  3  1 4  2 
 662
2
Since, B  0 . So it is a non-singular matrix and it has an inverse matrix.
The cofactors of each elements of B are,
B11  6 ; B12  6 ; B13  2 ; B21  5 ; B22  8 ; B23  3 ; B31  1 ; B32  2 ;
B33  1 .
The matrix of cofactors is,
 6 6 2 


 5 8 3 
 1 2 1 


 6 6 2 


 adj. A   5 8 3 
 1 2 1 


t
 6 5 1 


  6 8 2 
 2 3 1 


 B 1 
adj.B
B
 6 5 1 
1

  6 8 2 
2

 2 3 1 
Therefore,
1 2
1
AB  1 3
2
1 5
 0
1
  2
2
 0
1
3   6 5 1 


5   6 8 2 
12   2 3 1 
2 0

4 0
1 3 
0
1

  1
2

 0 1

2
0 
0 

3 
2
This is required answer.
 1 1

0 1
Problem-05: Find the inverse of the matrix A  
2 1

 3 2
0 2

1 1
by using block matrix.
2 1

1 6
Solution: The given matrix is,
 1 1

0 1
A
2 1

 3 2
0 2

1 1
2 1

1 6
Now we form the block matrix M   A I  and reduce M to echelon form.
 1 1

0 1
M 
2 1

 3 2
 1 1

0 1

0 3

0 1
0 2
1 1
2
1
1
6
0 2
1 1
1
0
0 0
1 0
2 3
1 0
2 0 1
3 0 0
1 0 0 0

0 1 0 0
0 0 1 0

0 0 0 1
0

0  R3'  R3  2 R1
0  R4 '  R4  3R1

1
 1 1 0 2

0 1 1 1

 0 0 1 0

0 0 0 1
 1 1

0 1

0 0

0 0
0 2
1 1
1
0
0
1
0 0

0 0  R3'  R3  3R2
2 3 1 0  R4 '  R4  R2

3 1 0 1 
1
0
0
1
0

0  R3'   R3
2 3 1 0  R4 '  R4  R2

3 1 0 1 
1
0
0
1
0
0
In echelon form, the left half of M is in triangular form; hence A is invertible. Further row reduce
M to row canonical form.
1

0

0

0
1 2 0
1 0 1
1

0

0

0
1 0 0
1 0 0
1

0

0

0
0 1
0 0
0
1
0

0  R1'  R1  2 R2
2 3 1 0  R2 '  R2  R3

3 1 0 1 
1 2
2 2
0
1
3 4
5 3
2
1
0 1 0
0 0 1
0

1  R1'  R1  2 R3
2 3 1 0  R2 '  R2  R4

3 1 0 1 
0 0 0
1 0 0
2 1
5 3
1
1
0 1 0
0 0 1
1

1 '
R1  R1  R2
2 3 1 0 

3 1 0 1 
The final block matrix is in the form  I A1 
 2 1 1 1


5 3 1 1 
 A1  
 2 3 1 0 


 3 1 0 1 
This is required inverse matrix.
NOTE: Any matrix A of order more than 3  3 has an inverse matrix if it does not contain zero
row or zero column in its echelon form.
Also, Any matrix A of order more than 3  3 has an inverse matrix if the echelon form of A
reduces into triangular form.
 1 1 2

3 0 2
Problem-06: Find the inverse of the matrix A  
 2 1 1

1 0 1
1

2
by using block matrix.
1

1
Solution: The given matrix is,
 1 1 2

3 0 2
A
 2 1 1

1 0 1
1

2
1

1
Now we form the block matrix M   A I  and reduce M to echelon form.
 1 1 2

3 0 2
M 
 2 1 1

1 0 1
 1 1 2 1

0 3 4 1

 0 3 5 1

 0 1 1 0
1
2
1
1
1 0
3 1
2 0
1 0
 1 1 2 1

0 3 4 1

 0 0 1 0

0 0 1 1
1
3
 1 1 2 1

0 3 4 1

 0 0 1 0

0 0 0 1
1
3
1
0
1
1
1 0 0 0

0 1 0 0
0 0 1 0

0 0 0 1
0 0 '
 R2  R2  3R1
0 0 '
R3  R3  2 R1
1 0
'
 R4  R4  R1
0 1
0 0

0 0  R3'  R3  R2
1 1 0  R4 '  3R4  R2

1 0 3 
0
1
0 0

0 0 '
R4  R4  R3
1 1 0 

2 1 3 
0
1
In echelon form, the left half of M is in triangular form; hence A is invertible. Further row reduce
M to row canonical form.
 1 1 2

0 3 4

0 0 1

0 0 0
0
0
0
1
0 2 1 3  '
 R1  R1  R4
2 1 1 3  '
R2  R2  R4
1 1 1 0 
'
 R3   R3
1 2 1 3 
 1 1

0 3

0 0

0 0
 1 1

0 1

0 0

0 0
1

0

0

0
1 0
0 1
1 3 

3 3  R1'  R1  2 R3
1 1 1 0  R2 '  R2  4 R3

1 2 1 3 
0 0
0 0
2
2
0 0
0 0
1 0
0 1
0 0 0
1 0 0
0 1 0
0 0 1
2
6
0
3
1 3 

1 1  ' 1
R2  R2
1 1 1 0 
3

1 2 1 3 
0
1
0 2 

1 1  '
R1  R1  R2
1 1 1 0 

1 2 1 3 
0
2
1
1
The final block matrix is in the form  I A1 
 0 1 0 2 


2 1 1 1 
 A1  
 1 1 1 0 


 1 2 1 3 
This is required inverse matrix.
Exercise-01
1.
2.
3.
4.
5.
1 2
3 5 9
Given, 𝐴 = (3 6), 𝐵 = (
). Calculate 𝐴𝐵 and 𝐵𝐴. Is 𝐴𝐵 = 𝐵𝐴? [Ans. No]
6 −2 1
5 8
−20 20
3 −2
Find the matrix B if 𝐴 = (
) and 𝐴2 + 3𝐴 + 𝐵 = 0. [Ans. 𝐵 = (
)]
10 −30
−1 4
1 0 1
0 1 1
If 𝐴 = (1 1 0) and 𝐵 = (1 1 1) then prove that 𝐴3 − 𝐴2 + 𝐴 − 2𝐵 = 0.
0 1 1
1 1 1
1 2
1
If 𝐴 = (0 1 −1) then show that 𝐴3 − 3𝐴2 − 𝐴 + 9𝐼 = 0.
3 −1 1
1 0 1
0 1 1
If 𝐴 = (1 1 0) and 𝐵 = (1 1 1). Verify that
0 1 1
1 1 1
(𝐴 + 𝐵)𝑇 = 𝐴𝑇 + 𝐵 𝑇 .
(i)
(𝐴𝐵)𝑇 = 𝐵 𝑇 𝐴𝑇 .
(ii)
1 2 4


6. Find the Symmetric and Skew-symmetric parts of the matrix A   6 8 1  .
3 5 7


1 1
1
4
3


 1 1
0
1
7. Find the Symmetric and Skew-symmetric parts of the matrix A  
.
2

3
0

6


5


1 
 1 1 1
 1 2 3 


8. Find the adjoint matrix of the matrix A   2 3 1 .
 3 1 2 


 1 0 3


9. Find the adjoint matrix of the matrix A   2 1 0  .
3 2 1


 3 3 4 


10. Find the inverse of the matrix A   2 3 4  .
 0 1 1 


 1 2 3 


11. Find the inverse of the matrix A   2 1 0  .
 4 2 5 


−1 2 −3
2 1 −1
12. If 𝐴 = ( 2
1
0 )and 𝐵 = (0 2 1 ) then find A1B .
4 −2 5
5 2 −3
1 2 3


13. If A   2 3 4  then show that AA1  I .
1 5 7


 1 3 3 1 


1 1 1 0 

14. Find the inverse of the matrix A 
by using block matrix.
 2 5 2 3 


 1 1 1 0 
 0 1 2 3 


1 1
4
4

15. Find the inverse of the matrix A 
by using block matrix.
1 3 7
9


 1 2 4 6 
Rank of a Matrix
Echelon Form of a Matrix: A matrix A is said to in echelon form if
1. All the non-zero rows, if any precede the zero rows.
2. The number of zero preceding the first non-zero element in a row is less than the
number of such zero in the succeeding row.
 0 6 0 2 0 


Example:  0 0 0 4 5  ;
0 0 0 0 2


5 4 0 1


0 0 2 9 .
0 0 0 0


Reduced Echelon Form of a Matrix: A matrix A is said to be in reduced echelon form if
1. All the non-zero rows, if any precede the zero rows.
2. The number of zero preceeding the first non-zero element in a row is less than the
number of such zero in the succeeding row and
3. The first non-zero element in a row is unity.
1

0
Example: 
0

0
3 4 5 6

1 2 3 4
.
0 1 2 3

0 0 0 0
Row Reduced Echelon Form of a Matrix: A matrix A is said to be in row reduced echelon form
if
1. A is an echelon form,
2. Each pivot element is 1 and
3. Each pivot element is the only non-zero entry in its columns.
1

0
Example: 
0

0
0 0 5 6

1 0 3 4
.
0 1 2 3

0 0 0 0
Pivot Element: The first non-zero entry of a non-zero row is called the pivot element of that row.
1 2 3


Example:  0 5 6  ; here, 1, 5 and 7 are the pivot elements of the 1st, 2nd and 3rd rows
0 0 7


respectively.
Normal Form of a Matrix: Every non-zero matrix A of order m  n can be reduced by the
application of elementary row and column operations into equivalent matrix of one of the
following forms:
 I O
1.  r
,
O O
I 
2.  r  , 3.  I r
O
O  and4.  I r 
where I r is r  r identity matrix and O null matrix of any order.
These four forms are called normal or canonical form of A.
Elementary Operations or Transformation on a Matrix: Let A be any matrix. Consider the
following operations on the matrix A.
1. Interchange of any two rows or two columns of A.
2. Multiply each of the element of a row or a column of A with a non-zero scalar.
3. Add scalar multiple of one row of A to another row or one column of A two another column.
Each of the above operations are called elementary transformation or operations of A.
Rank of a Matrix: Let M be a matrix. The number of pivot elements in an echelon form of M is
called the rank of M and it is denoted by   M  .
Alternatively, the number of non-zero rows in a row echelon form of a matrix is called the rank
of the matrix.
2
0
Problem 01: Find rank of the matrix A = .
2
2
-1
3
3
5
3
4
7
11
4
1
. by row reduction .
5
6
Solution
Step 1 Transform the matrix A to row echelon form.
2
0
.
2
2
-1
3
3
5
3
4
7
11
4
1
.
5
6
R3 <== R3 - R1
R4 <== R4 - R1
~
2
0
.
0
0
-1
3
4
6
3
4
4
8
4
1
.
1
2
2
0
.
0
0
-1
3
4
6
3
4
4
8
4
1
.
1
2
2
0
.
0
0
R3 <== 3 x R3 - 4 x R2
R4 <== R4 - 2 x R2
~
-1
3
0
0
3
4
-4
0
4
1
.
-1
0
Step 2 Rank equals number of nonzero rows in the echelon form of the matrix.
Therefore, rank of A = 3.
1
2
Problem 02: Find rank of the matrix A = .
1
2
3
6
3
6
-2
-4
-2
1
-1
-2
. by row reduction .
1
-1
Solution
Step 1 Transform the matrix A to row echelon form.
1
2
.
1
2
3
6
3
6
-2
-4
-2
1
-1
-2
.
1
-1
1
0
.
0
0
3
0
0
0
-2
0
0
5
-1
0
.
2
1
R2 <== R2 - 2 x R1
1
0
.
0
0
R3 <== R3 - R1
R4 <== R4 - 2 x R1
~
R2 <==> R4
~
1
0
.
0
0
3
0
0
0
3
0
0
0
-2
5
0
0
-2
0
0
5
-1
0
.
2
1
-1
1
.
2
0
Step 2 Rank equals number of nonzero rows in the echelon form of the matrix.
Therefore, rank of A = 3.
3
Problem 03: Find rank of the matrix A = . 0
1
-2
2
-2
0
2
-3
-1
1 . .
2
0
1
2
1
Solution
Step 1 Transform the matrix A to row echelon form.
3
0
.
1
0
-2
2
-2
1
0
2
-3
2
-1
1
.
2
1
3
0
.
0
0
-2
2
-4
1
0
2
-9
2
-1
1
.
7
1
3
0
.
0
0
-2
2
0
0
0
2
5
2
-1
1
.
-9
1
R3 <== 3 x R3 - R1
~
3
0
.
0
0
-2
2
-4
1
0
2
-9
2
-1
1
.
7
1
3
0
.
0
0
-2
2
0
0
0
2
5
2
-1
1
.
-9
1
R3 <== R3 + 2 x R2
R4 <== 2 x R4 - R2
~
R4 <== 5 x R4 - 2 x R3
~
3
0
.
0
0
-2
2
0
0
0
2
5
0
-1
1
.
-9
23
Step 2 Rank equals number of nonzero rows in the echelon form of the matrix.
Therefore, rank of A = 4.
1
2
Problem 04: Find rank of the matrix A = . 3
4
5
2
3
5
7
8
3
4
6
10
9
4
5
7
13
10
5
6
4 . .
16
3
Solution
Step 1 Transform the matrix A to row echelon form.
. 1
2
3
4
5 .
R2 <== R2 - 2 x R1
R3 <== R3 - 3 x R1
. 1
2
3
4
5 .
2
3
4
5
3
5
7
8
4
6
10
9
5
7
13
10
1
0
. 0
0
0
2
-1
-1
-1
-2
3
-2
-3
-2
-6
4
-3
-5
-3
-10
1
0
. 0
0
0
2
-1
0
0
0
3
-2
1
0
2
4
-3
2
0
4
6
4
16
3
5
-4
-11 .
-4
-22
5
-4
7 .
0
14
R4 <== R4 - 4 x R1
0
0
0
0
R5 <== R5 - 5 x R1
~
-2
-3
-2
-6
1
0
. 0
0
0
R3 <== R3 - R2
R4 <== R4 - R2
R5 <== R5 - 2 x R2
~
1
0
. 0
0
0
R5 <== R5 - 2 x R3
~
-1
-1
-1
-2
2
-1
0
0
0
-3
-5
-3
-10
2
-1
0
0
0
3
-2
1
0
2
3
-2
1
0
0
4
-3
2
0
4
4
-3
2
0
0
-4
-11
-4
-22
5
-4
7 .
0
14
5
-4
7 .
0
0
Step 2 Rank equals number of nonzero rows in the echelon form of the matrix.
Therefore, rank of A = 3.
1
1
Problem 05: Find rank of the matrix A = .
3
2
2
3
8
1
-3
-2
-7
-9
-2
0
-2
-10
1
0
.
0
0
2
1
2
-3
-3
-4
. .
-11
-3
Solution
Step 1 Transform the matrix A to row echelon form.
1
1
.
3
2
2
3
8
1
-3
-2
-7
-9
-2
0
-2
-10
-3
-4
.
-11
-3
R2 <== R2 - R1
R3 <== R3 - 3 x R1
R4 <== R4 - 2 x R1
~
-3
1
2
-3
-2
2
4
-6
-3
-1
.
-2
3
1
0
.
0
0
2
1
2
-3
-3
1
2
-3
-2
2
4
-6
-3
-1
.
-2
3
1
0
.
0
0
R3 <== R3 - 2 x R2
R4 <== R4 + 3 x R2
~
2
1
0
0
-3
1
0
0
-2
2
0
0
-3
-1
.
0
0
Step 2 Rank equals number of nonzero rows in the echelon form of the matrix.
Therefore, rank of A = 2.
0
1
Problem 06: Find rank of the matrix A = .
0
1
1
0
1
0
0
1
0
1
-1
-1
. by row reduction .
1
0
Solution
Step 1 Transform the matrix A to row echelon form.
0
1
.
0
1
1
0
1
0
0
1
0
1
-1
-1
.
1
0
1
0
.
0
1
0
1
1
0
1
0
0
1
-1
-1
.
1
0
1
0
.
0
0
0
1
1
0
1
0
0
0
-1
-1
.
1
1
R1 <==> R2
~
1
0
.
0
1
0
1
1
0
1
0
0
1
-1
-1
.
1
0
R4 <== R4 - R1
~
1
0
.
0
0
0
1
1
0
1
0
0
0
-1
-1
.
1
1
R3 <== R3 - R2
~
1
0
.
0
0
0
1
0
0
1
0
0
0
-1
-1
.
2
1
Step 2 Rank equals number of nonzero rows in the echelon form of the matrix.
Therefore, rank of A = 4.
Problem-07: Find the Rank, Echelon form(EF), Row reduced echelon form(RREF) and Normal
 1 2 1 2 1 


form(NF) of the following matrix. A   2 4 1 2 3 
 3 6 2 6 5 


Solution: Given that,
 1 2 1 2 1 


A   2 4 1 2 3 
 3 6 2 6 5 


Now we shall reduce the matrix to echelon form by the elementary row operations.

 1 2 1 2 1 


A   2 4 1 2 3 
 3 6 2 6 5 


 1 2 1 2 1 


  0 0 3 6 1 
 0 0 5 12 2 


 1 2 1 2 1


  0 0 3 6 1
 0 0 0 6 1


R2 '  R2  2 R1
R3'  R3  3R1
R3'  3R3  5R2
This matrix is in echelon form.
Since it contains 3 non-zero rows so the rank of the matrix A is,   A  3.
Again,


 1 2 1 2
1 


1 

 0 0 1 2

3 

1
0 0 0 1  
6

1
R2 '  R2
3
1
R3'   R3
6
4 

1 2 0 0
3 


1 

 0 0 1 2

3 


 0 0 0 1  1 
6

R1'  R1  R2
4 

1 2 0 0 3 


 0 0 1 0 0 

1
0 0 0 1  
6

R2 '  R2  2 R3
This matrix is in row reduced echelon form.
Problem-08: Find the Rank, Echelon form(EF), Row reduced echelon form(RREF) and Normal
 3 2 0 1


0 2 2 1

form(NF) of the following matrix. A 
 1 2 3 2 


0 1 2 1 
Solution: Given that,
 3 2 0 1


0 2 2 1

A
 1 2 3 2 


0 1 2 1 
Now we shall reduce the matrix to echelon form by the elementary row operations.
 3 2 0 1


0 2 2 1

 A
 1 2 3 2 


2 1
0 1
 3 2

0 2

0 4

0 1
0 1 

2 1
R3'  R1  3R3
9 7 

2 1
 3 2 0 1 


0 2
2 1  R3'  R3  2 R2


0 0
5 9  R4 '  R2  2 R4


 0 0 2 1 
 3 2

0 2

0 0

0 0
1 

1 
R4 '  5R4  2 R3
5 9 

0 23 
0
2
This matrix is in echelon form.
Since it contains 4 non-zero rows so the rank of the matrix A is,   A  4.
Again,
1  2
0 1 
3
3

0
1
1 1 
2 

0
0
1  9 

5
0
0
0
1 

1

0

0

0

0 2
3
1
1
0
1
0
0
0 

1 
2 
 9 
5
1 
1

0


0
0


5 
1 0 23 
10 

9
0 1 
5
0 0
1 
1

0

0

0
0 0 0

1 0 0
0 1 0

0 0 1
0 0
R1' 
R1
R2 ' 
R2
R3 
R3
'
R4 ' 
3
2
5
 R4
R1'  R1 
23
2 R2
6
R1'  R1 
2 R3
3
R2  R2  R3
'
R1'  R1 
R2 '  R2 
6 R4
5
23R4
R3'  R3 
9 R4
10
5
3
This is row reduced echelon form of the matrix A.
This is also normal form of the matrix A.
i.e,  I 4 
Problem-09: Find the Rank, Echelon form(EF), Row reduced echelon form(RREF) and Normal
form(NF) of the following matrix.
1

2
A
1

2
3 2 1 

6 4 2 
3 2 1 

6 1 1 
Solution: Given that,
1

2
A
1

2
3 2 1 

6 4 2 
3 2 1 

6 1 1 
Now we shall reduce the matrix to echelon form by the elementary row operations.
1

2
 A
1

2
3 2 1 

6 4 2 
3 2 1 

6 1 1 
1

0

0

0
3 2 1

0 0 0
0 0 2

0 5 1
1

0

0

0
3 2 1

0 5 1
0 0 2

0 0 0
R2 '  R2  2 R1
R3'  R3  R1
R4 '  R4  2 R1
R2  R4
This matrix is in echelon form.
Since it contains 3 non-zero rows so the rank of the matrix A is,   A  3.
1

0

0

0
3 2 1

1
0 1
5
0 0 1

0 0 0 
1
R2 '  R2
5
1
R3'  R3
2

1


 0

0
0

3
3 0  
5

1 
0 1
5 

0 0 1 
0 0 0 
R1'  R1  2 R2
1

0

0

0
3 0 0

0 1 0
0 0 1

0 0 0
3
R1'  R1  R3
5
1
R2 '  R2  R3
5
This is row reduced echelon form of the matrix A.
1

0

0

0
0 0 0

0 1 0
0 0 1

0 0 0
C2 '  C2  3C1
1

0

0

0
0 0 0

1 0 0
0 0 1

0 0 0
C2  C3
1

0

0

0
0 0 0

1 0 0
0 1 0

0 0 0
C3  C4
 I O
 3

O O
This is the normal form of the matrix A.
Problem-10: Find the Rank, Echelon form(EF), Row reduced echelon form(RREF) and Normal
form(NF) of the following matrix.
1

1
A
2

3
2 3 

1 4 
3 4 7 3 

8 1 7 8 
3
4
1
3
Solution: Given that,
1

1
A
2

3
2 3 

1 4 
3 4 7 3 

8 1 7 8 
3
4
1
3
Now we shall reduce the matrix to echelon form by the elementary row operations.
1

1
 A
2

3
2 3 

1 4 
3 4 7 3 

8 1 7 8 
3
4
1
3
 1 3 1 2 3 


0 1 2 1 1 

 0 3 6 3 3 


 0 1 2 1 1 
1

0

0

0
3 1 2 3 

1 2 1 1 
0 0 0 0

0 0 0 0
R2 '  R2  R1
R3'  R3  2 R1
R4 '  R4  3R1
R3'  R3  3R2
R4 '  R4  R2
This matrix is in echelon form.
Since it contains 2 non-zero rows so the rank of the matrix A is,   A  2.
1

0

0

0
0 5 5 0 

1 2 1 1
0 0 0 0

0 0 0 0
R1'  R1  3R2
This is row reduced echelon form of the matrix A.
1

0

0

0
1

0

0

0
0 0 0 0

1 2 1 1
0 0 0 0

0 0 0 0
0 0 0 0

1 0 0 0
0 0 0 0

0 0 0 0
C3'  C3  5C1
C4 '  C4  5C1
C3'  C3  2C2
C4 '  C4  C2
C5'  C5  C2
 I O
 2

O O
This is the normal form of the matrix A.
Problem-11: Find the Rank, Echelon form(EF), Row reduced echelon form(RREF) and Normal
form(NF) of the following matrix.
1

2
A  3

3
5

2 1 2

1 2 1
2 3 2

3 3 3
3 5 3 
Solution: Given that,
1

2
A  3

3
5

2 1 2

1 2 1
2 3 2

3 3 3
3 5 3 
Now we shall reduce the matrix to echelon form by the elementary row operations
1

2
 A  3

3
5

1

0
 0

0
0

2 1 2

1 2 1
2 3 2

3 3 3
3 5 3 
2
1
3 0
4 0
3 0
7 0
1 2

 0 3
 0 0

0 0
0 0

1
0
0
0
0
2

3 
4 

3 
7 
2

3 
0

0
0 
R2 '  R2  2 R1
R3'  R3  3R1
R4 '  R4  3R1
R5'  R5  5R1
R3'  3R3  4 R2
R4 '  R4  R2
R5'  3R5  7 R2
This matrix is in echelon form.
Since it contains 2 non-zero rows so the rank of the matrix A is,   A  2.
1

0
 0

0
0

2 1 2

1 0 1
0 0 0

0 0 0
0 0 0 
1
R2 '   R2
3
1

0
 0

0
0

0 1 0

1 0 1
0 0 0

0 0 0
0 0 0 
R1'  R1  2 R2
1

0
 0

0
0

0 1 0

1 0 1
0 0 0

0 0 0
0 0 0 
R1'  R1  2 R2
This is row reduced echelon form of the matrix A.
1

0
 0

0
0

0 0 0

1 0 0
0 0 0

0 0 0
0 0 0 
C3'  C3  C1
C4 '  C4  C2
 I O
 2

O O
This is the normal form of the matrix A.
Exercise: 02
1. Find the Rank, Echelon form(EF), Row reduced echelon form(RREF) and Normal
form(NF) of the following matrices:
1 2 3 2 1


i)  3 1 5 2 1  ,
 7 8 1 2 5 


 3 2 1 3 0


ii)  1 2 4 6 7 
 0 2 2 4 4


 1 3 1

0 11 5
iv) 
 2 5 3

4 1 1
 1 1 2

3 0 2
v) 
 2 1 1

1 0 1
 2

5
vii) 
 3

 1
2

3
1

5
1 3 4

0 1 1
1 2 3

0 1 3
1

1
viii) 
3

2
2 3
3 2
1

2
1

1
2
0
3 

4 
8 7 2 11

1 9 10 3 
 4 2 8 0 4 


iii)  1 1 9 0 5 
 0 7 5 0 2 


 3 2 0 1


0 2
2 1

vi)
 1 2 3 2 


2 1
0 1
 1 3 1 


7 5 6
ix)  2 2 0 


4 2 3
 3 9 3 


System of Linear Equations
Linear Equation: An equation in which the power of each unknown is one is called a linear
equation. The general form of a linear equation is defined as,
a1 x1  a2 x2  a3 x3  ... ...  an xn  b ... … … (1)
where,
a1 , a2 , a3 , … … …, an and b real numbers and x1 , x2 , x3 , … … … xn are unknowns(or
variables) which is to be determined.
If b  0 then the equation (1) is called a homogeneous linear equation and if b  0 then it is called
a non-homogeneous linear equation.
Example: 1. ax  by  c (non-homogeneous) represents a straight line.
2. ax  by  cz  d  0 (homogeneous) represents a plane.
Solutions of Linear Equation: A solution of the linear equation
a1 x1  a2 x2  a3 x3  ... ...  an xn  b
inn variables is a sequence of n numbers, 1 ,  2 ,  3 , … … … ,  n such that the equation is satisfied
when we substitute
x1  1 , x2   2 , … … … , xn   n .
The set of all such solutions of this linear equation is called a solution set.
System of Linear Equations: A finite set of linear equations is known as a system of linear
equations. So,
a11 x1  a12 x2  ... ...  a1n xn  b1 
a21 x1  a22 x2  ... ...  a2 n xn  b2 

 where, aij , bi  R, m  2
... ... ... ... ... ... ... ... ... ... ... ... 
am1 x1  am 2 x2  ... ...  amn xn  bm 

is a system of m linear equations in n variables
If
x1 , x2 , x3 , … … … xn .
bi  0 , then the above system is called a homogeneous system of linear equations and if bi  0 ,
then it is called a non-homogeneous system of linear equations.
Classification of System of Linear Equations: Regarding the nature of solutions, systems of
linear equations are classified as follows:
1. Inconsistent: A system of linear equations is called an inconsistent if it has no solution.
2. Consistent: A system of linear equations is called consistent if it has one or more solution.
It is also classified as,
a). Unique: A system of linear equations is called unique if it has only one solution.
b).Redundant: A system of linear equations is called redundant if it has more than one
solution.
Solution of System of Linear Equations: A system of m linear equations in n unknowns is,
a11 x1  a12 x2  ... ...  a1n xn  b1 
a21 x1  a22 x2  ... ...  a2 n xn  b2 

 ... ... ... (1) where, aij , bi  R, m  2
... ... ... ... ... ... ... ... ... ... ... ... 
am1 x1  am 2 x2  ... ...  amn xn  bm 

A sequence of numbers 1 ,  2 ,  3 , … … …  n is called solution of the system of linear equations
(1) if
x1  1 , x2   2 , … … … , xn   n is a solution of every equations in the system.
Free Variables: If a system of m linear equations in n unknowns is,
a11 x1  a12 x2  ... ...  a1n xn  b1 
a21 x1  a22 x2  ... ...  a2 n xn  b2 

 ... ... ... (1) where, aij , bi  R, m  2
... ... ... ... ... ... ... ... ... ... ... ... 
am1 x1  am 2 x2  ... ...  amn xn  bm 

and its echelon form is,
a11 x1  a12 x2  a13 x3  ... ...  a1n xn  b1 
a22 x2  a13 x3  ... ...  a2 n xn  b2 

 ... ... ... (2) where, aij , bi  R, m  2
... ... ... ... ... ... ... ... ... ... ... 
am3 x3  ... ...  amn xn  bm 

The variables which do not appear at the beginning of any equation of (2) are called free variables.
Trivial and non-trivial Solution: A homogeneous system of m linear equations in n unknowns
is,
a11 x1  a12 x2  ... ...  a1n xn  0 
a21 x1  a22 x2  ... ...  a2 n xn  0 

 … … … (1)
... ... ... ... ... ... ... ... ... ... ... ... 
am1 x1  am 2 x2  ... ...  amn xn  0 

The above system of linear equations (1) has a solution, namely zero n-tuple O   0,0,0,... ...,0, 
called zero or trivial solution and any other solution, if it exists, is called non-zero or non-trivial
solution.
Augmented matrix: Consider a non-homogeneous system of m linear equations in n unknowns
is,
a11 x1  a12 x2  ... ...  a1n xn  b1 
a21 x1  a22 x2  ... ...  a2 n xn  b2 

 ... ... ... (1) where, aij , bi  R, m  2
... ... ... ... ... ... ... ... ... ... ... ... 
am1 x1  am 2 x2  ... ...  amn xn  bm 

we can write it as AX  B .
 a11

a
Where, A   21
 ...

 am1
a12
a22
...
am 2
... a1n 
 b1 
 x1 

 
 
... a2 n 
x2 
b

, X 
and B   2 
 ... 
 ... 
... ... 

 
 
... amn 
 bm 
 xn 
Here, A is called the coefficient matrix, X is called the column matrix of the variables and B is
called the column matrix of the constants.
The matrix  A
 a11

a
B    21
 ...

 am1
a12
...
a1n
a22
...
am 2
... a2 n
... ...
... amn
b1 

b2 
is called augmented matrix of the system of
... 

bm 
linear equations (1). The augmented matrix is also denoted by A* or  A, B  or  A B  .
NOTE:
1. If the coefficient matrix and augmented matrix have the same rank, then the system of
linear equations is said to be consistent.
2. If the coefficient matrix and augmented matrix have the different rank, then the system of
linear equations is said to be inconsistent and it has no solution.
3. If the coefficient matrix and augmented matrix have the same rank and the rank is equal to
the number of variables then the system of linear equations has a unique solution.
4. If the coefficient matrix and augmented matrix have the same rank and the rank is less than
the number of variables then the system of linear equations has infinite number of solutions.
Gauss Elimination Method: Consider a non-homogeneous system of m linear equations in n
unknowns is,
a11 x1  a12 x2  ... ...  a1n xn  b1 
a21 x1  a22 x2  ... ...  a2 n xn  b2 

 ... ... ... (1) where, aij , bi  R, m  2
... ... ... ... ... ... ... ... ... ... ... ... 
am1 x1  am 2 x2  ... ...  amn xn  bm 

we can write it as AX  B .
 a11

a
Where, A   21
 ...

 am1
... a1n 
 b1 
 x1 

 
 
... a2 n 
x2 
b

, X 
and B   2 
 ... 
 ... 
... ... 

 
 
... amn 
 bm 
 xn 
a12
a22
...
am 2
The augmented matrix is,
 a11

 a21
 ...

 am1
a12
... a1n
a22 ... a2 n
... ... ...
am 2 ... amn
b1 

b2 
... 

bm 
reduce this matrix into the following form,
 a11'

 0
 ...

 0
a '12 ... a '1n
a22 ' ... a '2 n
... ... ...
0
... amn '
l1 

l2 
... 

lm 
then the reduced system is,
1.x1  a '1n .x2  ... ...  a '1n .xn  l1
0.x1  1.x2  ... ...  a '2 n .xn  l2
... ... ... ... ... ... ... ... ... ... ... ...
0.x1  0.x2  ... ...  1.xn  lm
Now, by back substitution, we solve for,
xn , xn 1 , … …, x1 .
This process which eliminates unknowns from succeeding equations is known as Gauss
elimination.
NOTE:
1. If an equation
0 x1  0 x2  ... ...  0 xn  b, b  0 occurs, then the system is inconsistent and
has no solution.
2. If an equation
0 x1  0 x2  ... ...  0 xn  0 occurs, then the equation can be deleted without
affecting the solution.
3. If the number of equations is equal to the number of variables, then the system has a unique
solution.
4. If the number of equations is less than the number of variables, then the system has a
infinitely many solutions.
Gauss Jordan Elimination Method: Consider a non-homogeneous system of m linear equations
in n unknowns is,
a11 x1  a12 x2  ... ...  a1n xn  b1 
a21 x1  a22 x2  ... ...  a2 n xn  b2 

 ... ... ... (1) where, aij , bi  R, m  2
... ... ... ... ... ... ... ... ... ... ... ... 
am1 x1  am 2 x2  ... ...  amn xn  bm 

we can write it as AX  B .
 a11

a
Where, A   21
 ...

 am1
a12
a22
...
am 2
... a1n 
 b1 
 x1 

 
 
... a2 n 
x2 
b

, X 
and B   2 
 ... 
 ... 
... ... 

 
 
... amn 
 bm 
 xn 
The augmented matrix is,
 a11

 a21
 ...

 am1
a12
b1 

b2 
... 

bm 
... a1n
a22 ... a2 n
... ... ...
am 2 ... amn
reduce this matrix into the following form,
1

0
 ...

0
0 ... 0
1 ... 0
... ... ...
0 ... 1
l1 

l2 
... 

lm 
then the reduced system is,
1.x1  0.x2  ... ...  0.xn  l1
0.x1  1.x2  ... ...  0.xn  l2
... ... ... ... ... ... ... ... ... ... ...
0.x1  0.x2  ... ...  1.xn  lm
Thus, we get
x1  l1 , x2  l2 , … … xn  lm .
Problem-01: Show that the following system of linear equations is not consistent
x  y  z  3
3x  y  2 z  2
2x  4 y  7z  7
Solution: The given system of linear equations is,
x  y  z  3 

3x  y  2 z  2  ... … … (1)
2 x  4 y  7 z  7 
the system (1) can be written as,
AX  B … … … (2)
1 1 1 
 x
 3 


 
 
where, A   3 1 2  , X   y  and B   2  .
2 4 7 
z
 7


 
 
The augmented matrix is,
1 1 1
 A , B    3 1 2
2 4 7

3 

2 
7 
1 1 1

  0 2 5
0 2 5

3  '
 R  R2  3R1
7  2'
R  R3  2 R1
13  3
1 1 1

  0 2 5
0 0 0

3 

7  R3'  R3  R2
20 
which is the echelon form of the augmented matrix.
The reduced system is,
x  y  z  3

 2 y  5z  7 
0  20 
Since, 0  20 occurs, which is not possible so the given system of linear equations is inconsistent.
(Showed)
Problem-02: Show that the following system of linear equations is consistent
2 x  y  3z  8
x  2 y  z  4
3x  y  4 z  0
and find the solution.
Solution: The given system of linear equations is,
2 x  y  3z  8 

 x  2 y  z  4  ... … … (1)
3x  y  4 z  0 
the system (1) can be written as,
AX  B … … … (2)
 2 1 3 
 x
8


 
 
where, A   1 2 1  , X   y  and B   4  .
 3 1 4 
z
0


 
 
The augmented matrix is,
 2 1 3
 A , B    1 2 1
 3 1 4

8

4
0 
 2 1 3

 0 3
5
 0 5 17

8  '
 R  2 R2  R1
16  2'
R  2 R3  3R1
24  3
 2 1 3

 0 3
5
 0 0 76

8 

16  R3'  3R3  5R2
152 
which is the echelon form of the augmented matrix.
The reduced system is,
2 x  y  3z  8 

3 y  5 z  16 
 76 z  152 
By back substitution we get, z  2 , y  2 , x  2 .
Hence the given system is consistent and the solution is,
x  2, y  2, z  2.
Problem-03: Show that the following system of linear equations is not consistent
x  y  2z  w  5
2 x  3 y  z  2w  2
4 x  5 y  3z
7
Solution: The given system of linear equations is,
x  y  2 z  w  5

2 x  3 y  z  2w  2  … … … (1)
4 x  5 y  3z
 7 
the system (1) can be written as,
AX  B … … … (2)
x
1 1 2 1 
5
 
y


 
where, A   2 3 1 2  , X    and B   2  .
z
4 5 3 0 
7
 


 
w
 
The augmented matrix is,
1 1 2 1
 A , B    2 3 1 2
4 5 3 0

5

2
7 
1 1 2 1

  0 1 5 4
 0 1 5 4

 '
 R2  R2  2 R1
 R '  R  4R
3
1
13  3
1 1 2 1

  0 1 5 4
0 0 0 0

5

8  R3'  R3  R2
5 
5
8
which is the echelon form of the augmented matrix.
The reduced system is,
x  y  2z  w  5 

y  5 z  4w  8
0  5 
Since, 0  5 occurs, which is not possible so the given system of linear equations is inconsistent.
(Showed)
Problem-04: Solve the following system of linear equations
5 x  6 y  4 z  15
7 x  4 y  3z  19
2 x  y  6 z  46
Solution: The given system of linear equations is,
5 x  6 y  4 z  15 

7 x  4 y  3z  19  … … … (1)
2 x  y  6 z  46 
the system (1) can be written as,
AX  B … … … (2)
 5 6 4 
 x
 15 


 
 
where, A   7 4 3  , X   y  and B   19  .
2 1 6 
z
 46 


 
 
The augmented matrix is,
 5 6 4
 A , B    7 4 3
2 1 6

 5 6 4

  0 62 43
 0 17 22

4
 5 6

  0 62 43
 0 0 2095

15 

19 
46 
15  '
 R  5R2  7 R1
10  2'
R  5R3  2 R1
200  3
15 

10  R3'  62 R3  17 R2
12570 
which is the echelon form of the augmented matrix.
The reduced system is,
5 x  6 y  4 z  15 

62 y  43z  10 
2095 z  12570
By back substitution, we get z  6 , y  4 , x  3 .
Hence the required result is, x  3 , y  4 , z  6 .
Problem-05: Solve the following system of linear equations
x  y  2 z  s  3t  1
2 x  y  2 z  2s  6t  2
3x  2 y  4 z  3s  9t  3
Solution: The given system of linear equations is,
x  y  2 z  s  3t  1 

2 x  y  2 z  2s  6t  2  … … … (1)
3x  2 y  4 z  3s  9t  3 
the system (1) can be written as,
AX  B … … … (2)
 x
 
 1 1 2 1 3 
1
 y

 X z
 
where, A   2 1 2 2 6  ,
and B   2  .
 
 3 2 4 3 9 
 3


 
s
t
 
The augmented matrix is,
 1 1 2 1 3
 A , B   2 1 2 2 6
 3 2 4 3 9

3
 1 1 2 1

  0 3 6 0
0
 0 1 2 6 18

1

2
3 
1 '
 R  R2  2 R1
0  2'
R  R3  3R1
0  3
3
 1 1 2 1

  0 3 6
0
0
 0 0 0 18 54

 1 1 2 1 3

  0 1 2 0 0
0 0 0 1 3

1

0  R3'  3R3  R2
0 
1  R2'   1 R2
3

0
1
'
0  R3   R3
18
which is the echelon form of the augmented matrix.
The reduced system is,
x  y  2 z  s  3t  1 

y  2z
 0
s  3t  0 
There are 3 equations in 5 unknowns, so there are (5-3)=2free variables which are z and t. Thus
the system is consistent with an infinite number of solutions.
We put
za
and t  b . So by back substitution we have, s  3b , y  2a , x  1 .
Hence the required result is, x  1 , y  2a , z  a , s  3b , t  b .
Problem-06: Solve the following system of linear equations
2 x  3 y  5z  t  3
3x  4 y  2 z  3t  2
x  2 y  8z  t  8
7 x  9 y  z  8t  0
Solution: The given system of linear equations is,
2 x  3 y  5z  t  3 
3 x  4 y  2 z  3t  2 

 … … … (1)
x  2 y  8z  t  8 
7 x  9 y  z  8t  0 

the system (1) can be written as,
AX  B … … … (2)
2

3
where, A  
1

7
1
 x
 3

 
 
3
y
2

, X
and B    .
z
 8 
2 8 1

 
 
9 1 8
t
 0 
3 5
4 2
The augmented matrix is,
2

3
 A , B   
1

7
1

3

2

7
3 5
4 2
1
3
2 8 1
9 1 8
2 8 1
4 2 3
3 5
9 1
1
8
3

2 
8

0
8

2 
R1  R3
3

0
8 1
1 2

0 2 22 6

 0 1 11 3

 0 5 55 15
8  '
 R2  R2  3R1
26  '
R3  R3  2 R1
13  '
 R4  R4  7 R1
56 
8 1
1 2

0 2 22 6

0 0
0
0

0
0
0 0
8 

26  R3'  2 R3  R2
0  R4 '  2 R4  5R2

18 
1

0

0

0
2 8 1
1 11 3
0
0
0
0
0
0
8

13  '
1
R2   R2
0
2

18 
which is the echelon form of the augmented matrix.
The reduced system is,
x  2 y  8z  t  8 
y  11z  3t  13


0  0
0  18

or,
x  2 y  8z  t  8 

y  11z  3t  13
0  18
Since, 0  18 occurs, which is not possible so the given system of linear equations is inconsistent
and it has no solution.
Problem-07:Solve the following system of linear equations
x  2 y  2z  t  0
2 x  5 y  3z  t  1
3x  8 y  4 z  t  2
x  5 y  z  2t  3
Solution: The given system of linear equations is,
x  2 y  2z  t  0 
2 x  5 y  3z  t  1 
 … … … (1)
3x  8 y  4 z  t  2 
x  5 y  z  2t  3
the system (1) can be written as,
AX  B … … … (2)
1

2
where, A  
3

1
2 2 1
 x
0

 
 
5 3 1
y
1

, X
and B    .
z
 2
8 4 1

 
 
5 1 2
t
 3
The augmented matrix is,
1

2
 A , B   
3

1
1

0

0

0
2 2 1
5 3 1
8 4 1
5 1 2
2 2 1
1 1 1
2
3
2
3
2
3
0

1
2

3
0 '
 R2  R2  2 R1
1 '
R3  R3  3R1
2 '
 R4  R4  R1
3
1

0

0

0
2 2 1
1 1 1
0
0
0
0
0
0
0

1  R3'  R3  2 R1
0  R4 '  R4  3R2

0
which is the echelon form of the augmented matrix.
The reduced system is,
x  2 y  2z  t  0 
y  z  t  1 

0 0
0  0 
or,
x  2 y  2 z  t  0

y  z  t  1
There are 2 equations in 4 unknowns, so there are (4-2) = 2 free variables which are z and t. Thus
the system is consistent with an infinite number of solutions.
We put
z a
and t  b . So by back substitution we have, y  1  a  b , x  4a  3b  2 .
Hence the required result is, x  4a  3b  2 , y  1  a  b ,
z  a , t  b.
Exercise: 03
1. Solve the following system of linear equations
3x  5 y  7 z  13
4 x  y  12 z  6
2 x  9 y  3z  20
2. Solve the following system of linear equations
x y z  3
x  2 y  3z  4
x  4 y  9z  6
3. Solve the following system of linear equations
2x  3y  4z  1
3x  4 y  5 z  1
5x  7 y  2 z  3
4. Determine for what value of a the following system of linear equations has
x y  z 1
2 x  3 y  az  3
x  ay  3z  2
i). no solution, ii). more than one solution, iii). a unique solution.
5. Determine for what values of  and  the following system of linear equations has
2x  3y  z  5
3x  y   z  2
x  7 y  6z  
i). no solution, ii). more than one solution, iii). a unique solution.
6. Solve the following system of linear equations
x y z t  4
x  2 y  2z  t  4
x  4 y  9 z  2t  16
Eigenvalues and Eigenvectors
Eigenvalues and Eigenvectors: Let A be an n-square matrix. A scalar  is called an eigenvalue
or characteristic root of A if there exists a non-zero column vector v such that,
Av  v
or,  A   I  v  0
where, I is the n-square identity matrix. Every vector satisfying this relation is called an
eigenvector or characteristic vector of A associated with the eigenvalue  .
Characteristic Matrix: If A be an n-square matrix and  be an eigenvalue of A , then the matrix
A   I n or  I n  A is called a characteristic matrix of A.
Characteristic Polynomial: If A be an n-square matrix and  be an eigenvalue of A, then the
determinant of the characteristic matrix
A   I n or  I n  A is called a characteristic polynomial of
A.
i.e,   A   I n
or,    I n  A .
Characteristic Equation: If A be an n-square matrix and  be an eigenvalue of A, then the
equation A   I n  0 or  I n  A  0 is called a characteristic equation of A.
NOTE: If A is an n  n triangular matrix (upper triangular, lower triangular, or diagonal), then the
eigenvalues of A are the entries on the main diagonal of A.
 a11

0
Example: If A  
 ...

 0
a12
a22
...
0
... a1n 

... a2 n 
, then its eigenvalues are,
... ... 

0 ann 
a11 , a22 , … … , ann .
 3 2 2 


Problem-01: Find the eigenvalues of the matrix A   6 5 2  .
 7 4 4 


Solution: The given matrix is,
 3 2 2 


A   6 5 2 
 7 4 4 


The characteristic matrix of A is,
 3 2 2 
1 0 0




A   I   6 5 2     0 1 0 
 7 4 4 
0 0 1




 3 2 2    0 0 

 

  6 5 2    0  0 
 7 4 4   0 0  

 

2
2 
 3  


  6
5
2 
 7
4
4   

The characteristic polynomial of A is,
  A  I
3  
2
 6
5
7
4
2
2
4
  3     5    4     8  26  4     14  224  7 5   
  3     20  9   2  8  2  24  6  14   2  24  35  7 
  3      2  9  12   2  6  10   2  7  11
 3 2  27  36   3  9 2  12  12  20  14  22
  3  6 2  11  6
The characteristic equation of A is,
A  I  0
  3  6 2  11  6  0
  3  6 2  11  6  0
  3   2  5 2  5  6  6  0
  2    1  5    1  6    1  0
    1   2  5  6   0
    1   2   3  0
   1, 2, 3
The eigenvalues of A are 1, 2, 3.
 1 3 3 


Problem-02: Find the eigenvalues of the matrix A   3 5 3  .
 6 6 4 


Solution: The given matrix is,
 1 3 3 


A   3 5 3 
 6 6 4 


The characteristic matrix of A is,
 1 3 3 
1 0 0




A   I   3 5 3     0 1 0 
 6 6 4 
0 0 1




 1 3 3    0 0 

 

  3 5 3    0  0 
 6 6 4   0 0  

 

3
3 
1  


 3
5  
3 
 6
6
4   

The characteristic polynomial of A is,
  A  I

1 
3
3
3
5  
3
6
6
4
 1     5    4     18  33  4     18  318  6  5   
 1      2    20  18  3 12  3  18  3  18  30  6 
 1      2    2   9  18  36  18
  2    2   3   2  2  9  18  36  18
  3  12  16
The characteristic equation of A is,
A  I  0
  3  12  16  0
  3  12  16  0
  3  2 2  2 2  4  8  16  0
  2    2  2    2  8    2   0
    2    2  2  8  0
    2   2   4   0
   2,  2, 4
The eigenvalues of A are -2, -2, 4.
1 2 2


Problem-03: If A   1 2 1 , then find its eigenvalues.
 1 1 4 


Solution: The given matrix is,
1 2 2


A   1 2 1
 1 1 4 


The characteristic matrix of A is,
1 2 2
1 0 0




A   I   1 2 1    0 1 0 
 1 1 4 
0 0 1




 1 2 2   0 0 

 

  1 2 1   0  0 
 1 1 4   0 0  

 

2
2 
1  


 3
2   1 
 1
1
4   

The characteristic polynomial of A is,
  A  I
1 
2
 3
2
1
1
2
1
4
 1     2    4     1  2 4     1  21   2   
 1    8  6   2  1  2  3     2  3   
 1     9  6   2   6  2  6  2
 1      2  6  9 
 1      3   3
The characteristic equation of A is,
A  I  0
 1      2  6  9   0
 1      3   3  0
   1, 3, 3
The eigenvalues of A are 1, 3, 3.
 3 1 1 


Problem-04: Find the eigenvalues of the matrix A   7 5 1  .
 6 6 2 


Solution: The given matrix is,
 3 1 1 


A   7 5 1 
 6 6 2 


The characteristic matrix of A is,
 3 1 1 
1 0 0




A   I   7 5 1     0 1 0 
 6 6 2 
0 0 1




 3 1 1    0 0 

 

  7 5 1    0  0 
 6 6 2   0 0  

 

1
1 
 3  


  7
5
1 
 6
6
2   

The characteristic polynomial of A is,
  A  I
3  
1
 7
5
6
6
1
1
2  
  3     5    2     6  7  2     6  42  6 5   

  3     

  3      2  3  10   6  14  7  6    42  30  6 
2
 3  4    7  8   12  6 
 3 2  9  12   3  3 2  4  7  8  12  6
  3  12  16
The characteristic equation of A is,
A  I  0
  3  12  16  0
  3  12  16  0
  3  4 2  4 2  16  4  16  0
  2    4  4    4   4    4   0
    4    2  4  4   0
    4   2   2  0
   2,  2, 4
The eigenvalues of A are -2, -2, 4.
 1 0 2 


Problem-05: Find the eigenvalues of the matrix A   0 0 0  .
 2 0 4 


Solution: The given matrix is,
 1 0 2 


A 0 0 0 
 2 0 4 


The characteristic matrix of A is,
 1 0 2 
1 0 0




A  I   0 0 0     0 1 0
 2 0 4 
0 0 1




 1 0 2    0 0 

 

  0 0 0  0  0
 2 0 4   0 0  

 

2 
1   0


 0

0 
 2
0 4   

The characteristic polynomial of A is,
  A  I
1  0
 0

2
0
2
0
4
 1      4     0  0  2  0  2 
 1     4   2   4
 4   2  4 2   3  4
  3  5 2
The characteristic equation of A is,
A  I  0
  3  5 2  0
  3  5 2  0
  2    5  0
   0, 0, 5
The eigenvalues of A are 0, 0, 5.
2 1 0


Problem-06: Find the eigenvalues of the matrix A   3 2 0  .
 0 0 4


Solution: The given matrix is,
2 1 0


A  3 2 0
 0 0 4


The characteristic matrix of A is,
2 1 0
1 0 0




A  I   3 2 0    0 1 0
 0 0 4
0 0 1




 2 1 0  0 0 

 

  3 2 0   0  0 
 0 0 4  0 0  

 

2

 3
 0

0 

0 
4   
1
2
0
The characteristic polynomial of A is,
  A  I

2
1
0
3
2
0
0
0
4
  2     2    4     0  3  4     0  0
  2     4     3 4   
2


 4    2     3
2
  4     4  4   2  3
  4      2  4  1
The characteristic equation of A is,
A  I  0
  4      2  4  1  0
 4    0 or,  2  4  1  0
   4 or ,  
4  16  4
2
or ,  
4  12
2
or ,  
42 3
2
or,   2  3
   4, 2  3
The eigenvalues of A are 4, 2  3 .
1 2 3


Problem-07: Find the eigenvalues of the matrix A   0 1 2  .
 0 2 1 


Solution: The given matrix is,
1 2 3


A  0 1 2
 0 2 1 


The characteristic matrix of A is,
1 2 3
1 0 0




A  I   0 1 2    0 1 0
 0 2 1 
0 0 1




1 2 3  0 0 

 

  0 1 2   0  0 
 0 2 1   0 0  

 

2
3 
1  


  0 1 
2 
 0
2 1   

The characteristic polynomial of A is,
  A  I
1 
2
3
 0
1 
2
0
2
1 


 1    1     4
2
 1      2  2  1  4 
 1      2  2  5
The characteristic equation of A is,
A  I  0
 1      2  2  5  0
 1    0 or,  2  2  5  0
  1 or ,  
2  4  20
2
or ,  
2  16
2
or ,  
or ,  
2  16i 2
2
2  4i
2
or,   1  2i
   1, 1  2i
The eigenvalues of A are 1, 1  2i .
1 0 0 


Problem-08: Find the eigenvalues of the matrix A   0 1 1 .
0 1 1 


Solution: The given matrix is,
1 0 0 


A   0 1 1
0 1 1 


The characteristic matrix of A is,
1 0 0 
1 0 0




A   I   0 1 1    0 1 0 
0 1 1 
0 0 1




1 0 0   0 0 

 

  0 1 1   0  0 
0 1 1   0 0  

 

0
0 
1  


  0 1   1 
 0
1 1   

The characteristic polynomial of A is,
  A  I
1 
 0
0
0
1 
1
1
1 
0


 1    1     1
2
 1      2  2  2 
The characteristic equation of A is,
A  I  0
 1      2  2  2   0
 1    0 or,  2  2  2  0
  1 or ,  
2  4 8
2
or ,  
2  4
2
or ,  
2  4i 2
2
or ,  
2  2i
2
or ,   1  i
   1, 1  i
The eigenvalues of A are 1, 1  i .
Exercise:
1.
2.
3.
4.
2

Find the eigenvalues of the matrix A   2
1

1

Find the eigenvalues of the matrix A   0
0

 2

Find the eigenvalues of the matrix A   2
 1

0

Find the eigenvalues of the matrix A   0
4

2 1 

8 2 
2 2 
2 3

2 3
0 2 
2 3 

1 6 
2 0 
1 0

0 1
17 8 
Ans : 0, 3,  7.
Ans : 1, 2, 2.
Ans : 5,  3,  3.
Ans : 4, 2  3.