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Mathematics
Session
Matrices and Determinants-1
Session Objectives
•
Matrix
•
Types of Matrices
•
Operations on Matrices
•
Transpose of a Matrix
•
Symmetric and Skew-symmetric Matrix
•
Class Exercise
Matrix
A matrix is a rectangular array of
numbers, real or complex.
Row
 a11

 a 21
 ..

..

A
 a i1

 ..
a
 m1

a12
a 22
..
..
..
..
..
..
..
a1 j
a2j
..
..
..
..
..
..
..
..
ai2
..
..
..
..
..
a ij
..
..
..
..
..
..
..
..
..
..
a mj
..
..
..
..
a m2





.. 
a in 

.. 
a mn 


a1n
a 2n
..
Element of mth row and jth column
Column
Order of a Matrix
A matrix with m rows and n columns
has an order m x n.
Examples:
4
A=
7
5
-8
1
6

 , B = 3
2
5

Order of A is 2 × 3
Order of B is 3 × 2
Order of C is 3 × 3
2
-4
6

2


,
C
=

5

1


3
4
6
1

2  etc.
3 
Example - 1
A matrix has 16 elements, what is the
possible number of columns it can have.
Solution :
The possible orders for the matrix are
(1 x 16), (2 x 8), (4 x 4), (8 x 2),(16 x 1)
So, the number of possible columns are 16, 8, 4, 2 and 1.
Example-2
Write the matrix given by the rule aij 
1
1

.
3i 4 j
Solution :
Here i can take the values 1 and 2 and j can take the
values 1, 2 and 3. Hence, the order of the matrix is (2 x 3).
Now,
1
1
5

 ;
3 1 4 1 6
1
1
4
a21 

 ;
3  2 4 1 3
a11 
5
6
Hence, the matrix is 
4
 3
1
1

 1;
3 1 4  2
1
1
3


 ;
32 42 2
1
1
3


3 1 4  3 2
1
1


2
32 43
a12 
a13 
a22
a23
1
3
2
3
2

2

Types of Matrices
3


2
5
-4
Row matrix: 

2


5
 
2

Column matrix:  
3
 
 7 
 1
Types of Matrices
0 0 0
Zero matrix : 

0 0 0
 1 2 3
Square matrix:  -4 -7 0 


 1 6 0


Diagonal matrix: A = aij  , where aij = 0for i  j
3 0 0


0
-4
0


0 0 4


Types of Matrices
, if i= j
Scalar matrix: A = aij  , where aij = 
0, if i  j
6
0

0
0
6
0
0
0
6 
1, if i= j


Identity matrix: A = aij  , where aij = 
0, if i  j
1

I3 =  0
0

0
1
0
0

0
1

Equality of Matrices
Two matrices A = [aij] and B = [bij] are
equal, if they have the same order and
aij = bij for all i and j.
Example:
a
If 
c
b  1
=
d   -3
 2
 , then
4 
a = 1, b = -2, c = -3 and d = 4
Addition of Matrices
If A= [aij] and B= [bij] are two matrices of the
same order, then their sum A + B is a matrix
whose (i, j)th element is ai j  bi j .
Example:
1 3 4
2 3 1 




If A =  2 3 4  and B =  5 4 2  , then
 -3 4 -5 
 1 -4 3 




 1 3 4  2 3 1   3 6 5 

 
 

A + B = 2 3 4  + 5 4 2  7 7 6 
 -3 4 -5  1 -4 3   -2 0 -2 

 
 

Multiplication of a Matrix by a Scalar
If A = aij 
m×n
kA = kaij 
m×n
is a matrix and k is a scalar, then
.
Example:
 1 3 4
If A = 1 2 3  , then
 -1 1 1 


 1 3 4  2 6 8
2A = 2 1 2 3  =  2 4 6 
 -1 1 1   -2 2 2 

 

Properties of Addition
If the order of the matrices A, B and C is same, then
(i) A + B = B + A
(Commutativity)
(ii) (A + B) + C = A + (B + C)
(Associativity)
(iii) If m and n are scalars, then
(a) m(A + B) = mA + mB
(b) (m + n)A = mA + nA
Example - 3
 -1 0 
3 2 
Find X, if Y= 
and
2X+Y
=

.

 -3 2 
1 4 
Solution :
3 2 
Y= 
 and
1
4


3 2 
2X +  1 4  =


 -1 0 
2X+Y = 

 -3 2 
 -1 0 


 -3 2 
 -1 0   3 2 
 2X = 
-

-3
2

 1 4
 -1- 3 0 - 2 
1  -4 -2   X =  -2 -1


 2X = 
  X= 

-2 -1

-3
-1
2
4
-4
-2
2




Example - 4
Find a matrix C such that A+B+C is a zero matrix,
 2 1 -1
2 0 1 
where A= 
 and B =  0 2 1  .


 3 -1 0 
Solution :
A+B+C=0
 2 0 1   2 1 -1
 0 0 0

+
+C
=
 



 3 -1 0   0 2 1 
 0 0 0
 4 1 0
 0 0 0

 +C= 

3
1
1


0 0 0
 0 0 0  4 1 0
 -4 -1 0 
 C= 
-
  C=

0 0 0  3 1 1
 -3 -1 -1
Multiplication of Matrices
Let A= [aij]m x n be a m x n matrix and B = [bij]n x p
be a n x p matrix , i.e. , the number of columns of A
is equal to the number of rows of B. Then their
product AB is of order m x p and is given as
n
 AB =  aijb jk = Sum of the product
of elements of ith row of A with
j=1
the corresponding elements of k th column of B
Example
3
1 2 2 
 
If A= 
 and B =  1  , then AB is given as
 2 -1 4 
 4
 
3
1
2
2

  
AB  
  1
2
-1
4

  4
 
1  3 + 2  1 + 2  4 


 2  3 + (-1  1)  4  4 
3  2  8 
=

6

1

16


 13 
= 
 21
Properties of Multiplication of Matrices
If both sides are defined, then
(i)
A(BC) = (AB)C
(Associativity)
(ii) A ( B + C ) = AB + AC and (A + B) C = AC + BC
( Multiplication is distributive over addition)
Example - 5
 1 2 0  0 

 
Find x , if 1 2 1  2 0 1  2  = 0.
 1 0 2  x 

 
Solution :
 1 2 0  0 

2
1 2 1  2 0 1 
  = 0
 1 0 2  x 

 
 0
 
 1+4+1 2+0+0 0+2+2  2   0
 x
 
 0
 
 (6 2 4)  2  = 0  0+4+4x   0
 x
 
 4x =- 4
 x =-1
Example - 6
 cos sin 
 cos2 sin2 
2
If A= 
,
then
show
that
A
=


.
-sin

cos



 -sin2 cos2 
Solution :
 cos sin   cos sin 
A =

 
 -sin cos   -sin cos 
2

cos2 - sin2

 -sin cos -cos sin
 cos2 sin2 
=

 -sin2 cos2 
cos sin +sin cos 

cos2 -sin2

Example - 7
1 0
 3 -2 
2
If A = 
 and I=  0 1  , then find k if A =k A-2I.


 4 -2 
Solution :
 3 -2   3 -2 
A2 = 


 4 -2   4 -2 
 9-8
=
 12 - 8
-6+ 4   1 -2 
=
-8+ 4   4 -4 
 3 -2   3k -2k 
kA =k 
=

4
-2
4k
-2k

 

Solution Contd.
A2 =k A-2I
 1 2   3k -2k   1 0 

=
 -2 

4

4
4k
-2k

 
  0 1
-2k 
 1 -2   3k -2

=

4
-4
4k
-2k
-2

 

Comparing the corresponding elements of the two matrices ,
we get 3k-2 = 1, -2k = -2 , 4 = 4k , -4 = -2k –2
Taking any of the four equations, we get k=1
Example - 8
6
Show that A = 
7
A2 – 12A + I = O.
5
 satisfies the equation
6
Solution :
 6 5  6 5   36+35 30+30   71 60 
6 5
2
,
then
A
=



=
=

7
6
7
6
7
6
42+
42
35+36




 
  84 71 
A=
 71 60 
 6 5 1 0
 A2 - 12A + I = 
-12


+

84
71
7
6



 0 1
 71 60   72 60   1 0 
=
-
+

84
71
84
72

 
 0 1
 -1 0   1 0  =  0 0 

 
 0 0

 0 -1  0 1  
Hence , A2 – 12A + I=O
Transpose of a Matrix
A matrix obtained by changing rows into columns
or columns into rows is called transpose of the
matrix ( say A ). If the matrix is A, then its
transpose is denoted as AT or A’ .
a b

c
d


For Example: Consider the matrix 
a c
The transpose of the above matrix is 

b
d


Example - 9
 5 -1 
2 1 
If A = 
and
B
=


,
6 7 
3 4
then verify that (A+B)T=AT+BT
Solution:
7 9 
 5 -1   2 1   7 0 
T
A +B = 
+ 
=
  (A  B)  

6
7
3
4
9
11
0
11

 
 



 5 6
T 2 3 
AT = 
and
B
=


-1
7
1
4




Hence, (A+B)T=AT+BT
7 9 
 A T + BT = 

0
11


Example - 10
3
4
 2 4 -1 
T


If A = 
 and B =  -1 2  , find (AB) .
 -1 0 2 
 2 1


Solution :
 3 4
2
4
-1



AB= 
-1
2


 -1 0 2   2 1 


8+8-1 
 6- 4-2


 -3+0+4 -4+0+2 
 0 15 
=

 1 -2 
0 1
(AB)T = 

15 -2 
Example - 11
 0 2y z 
Find the values of x , y, z if the matrix  x y -z 




’
obeys the law AA = I.
 x -y z 
Solution :
 0 x x


A '   2y y y 
 z z z 


 0 2y z  0 x x 



AA' =  x y -z  2y y -y 
 x -y z  z -z z 



 4y2 +z2

=  2y2 - z2
 -2y2 +z2

AA' = I
2y2 - z2
x2 + y2 +z2
x2 - y2 - z2
-2y2 +z2 

x2 - y2 - z2 
x2 + y2 +z2 

Solution (Cont.)
 4y2 + z2

  2y2 - z2
 -2y2 + z2

2y2 - z2
x2 + y2 + z2
x2 - y2 - z2
-2y2 + z2   1 0 0 
 

x2 - y2 - z2  =  0 1 0 
x2 + y2 + z2   0 0 1 
Equating the elements of column 2 , we get
2y2 – z2 = 0
…(i)
x2 +y2 +z2 =1
...(ii)
x2 - y2 - z2 = 0
...(iii)
Adding (ii) and (iii), we get
2x
2
 1 x
2
1
1
 x  
2
2
Form (i), z2 = 2y2
Solution (Cont.)
Putting the value of x2 and z3 in (ii), we get
1
+ y2 +2y2 =1
2
1
 3y2 =
2
1
1
 y2 = Þ y = ±
6
6
Putting the value of y2 in (i), we get
1
1
1
z2 = 2×  z2 =  z =
6
3
3
Example - 12
 2 1 -1
T


Find x, if  x 4 -1  1 0 0   x 4 -1 = 0.
2 2 4 


Solution :
 2 1 -1
 x 4 -1  1 0 0  = 2x+4 -2 x -2 -x - 4 
2 2 4 


2x + 4 - 2
x - 2 -x - 4  x
4 -1 = 0
x
 
 2x +2 x - 2 -x - 4   4  = 0
 -1
 
 (2x2 +2x+4x-8+x+4)=0
T
2x2 +7x - 4 = 0
 2x -1 x + 4  = 0
 x=
1
or x = -4
2
Symmetric and Skew – Symmetric Matrix
A square matrix A is called a symmetric matrix, if
AT = A.
A square matrix A is called a skew- symmetric
matrix, if AT = - A.
Any square matrix can be expressed as the sum of
a symmetric and a skew- symmetric matrix.
A + AT
A - AT
A=
+
,
2
2



where A + A T is symmetric matrix and A - A T
 is skew - symmetric matrix.
Example - 13
0 5 3


Show that A=  -5 0 -8  is a skew-symmetric matrix.
 -3 8 0 


Solution :
 0 -5 -3 
0 5 3




A T =  5 0 8  = -  -5 0 -8  = -A
 3 -8 0 
 -3 8 0 




As AT = - A, A is a skew – symmetric matrix
Example - 14
1 -5 
6
Express the matrix A =  -2 -5 4  as the sum of a
 -3 3 -1 


symmetric and a skew- symmetric matrix.
Solution :
6

A =  -2
 -3

1
-5
3
-5 

4
-1 
 6 -2 -3 


 A T =  1 -5 3 
 -5 4 -1 


 6
1
1 
Let P = (A + A T )=  -2
2
2 
 -3
12 -1 -8 
1

 P =  -1 -10 7 
2

 -8 7 -2 
1
-5
3
-5   6
 
4 + 1
-1   -5
-2
-5
4
-3  

3 
-1  
Solution Cont.
-1/2 -4 
 6


 P =  -1/2
5
7/2 
 -4
7/2 -1 

-1/2 -4 
 6


 P T =  -1/2
5
7/2  = P
 -4
7/2 -1 

 6 1 -5   6 -2 -3  
1
1 
 

Let Q =  A - A T  =  -2 -5 4  -  1 -5 3  
2
2 
 -3 3 -1   -5 4 -1  
 0 3 -2 
1

 Q =  -3 0 -1 
2

 2 -1 0 
Solution Cont.
3/2 -1 
 0


 Q =  -3/2
0
1/2 
 1
-1/2 0 

3/2 -1 
 0


 QT = -  -3/2
0
1/2  = -Q
 1
-1/2 0 

PT = P and QT = -Qa
Therefore, P is symmetric and Q is skewsymmetric . Further, P+Q = A
Hence, A can be expressed as the sum of a
symmetric and a skew -symmetric matrix.
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