Mathematics

Matrices and Determinants-1

Session

• Matrix

• Types of Matrices

• Operations on Matrices

• Transpose of a Matrix

• Symmetric and Skew-symmetric Matrix

• Class Exercise

Session Objectives

Matrix

A matrix is a rectangular array of numbers, real or complex.

11 12 1j 1n

21 22 2 j 2n

i1 i2 ij in

m1 m2 mj mn

a a .. .. a .. .. a

a a .. .. a .. .. a

.. .. .. .. .. .. .. ..

.. .. .. .. .. .. .. ..A

a a .. .. a .. .. a

.. .. .. .. .. .. .. ..

a a .. .. a .. .. a

ColumnRow

Element of mth row and jth column

Order of a Matrix

A matrix with m rows and n columns has an order m x n.

Examples:

1 2 2 3 14 5 6

A = , B = 3 - 4 , C = 5 4 2 etc.7 - 8 2

5 6 1 6 3

Order of A is 2 ×3

Order of B is 3×2

Order of C is 3×3

Example - 1

A matrix has 16 elements, what is the possible number of columns it can have.

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.

Solution :

Here i can take the values 1 and 2 and j can take thevalues 1, 2 and 3. Hence, the order of the matrix is (2 x 3).

Example-2

Solution :

Now,

Write the matrix given by the rule ij1 1

a .3 i 4 j

Hence, the matrix is

5 31

6 24 3

23 2

11 12 13

21 22 23

1 1 5 1 1 1 1 3a ; a 1; a

3 1 4 1 6 3 1 4 2 3 1 4 3 21 1 4 1 1 3 1 1

a ; a ; a 23 2 4 1 3 3 2 4 2 2 3 2 4 3

Row matrix:3

2 5 -42

Column matrix:

5

2

3

71

Types of Matrices

Zero matrix :

Square matrix:

0 0 0

0 0 0

1 2 3

-4 -7 0

1 6 0

Diagonal matrix: ij ijA = a where a =0for i j,

3 0 0

0 -4 0

0 0 4

Types of Matrices

3

1 0 0

I = 0 1 0

0 0 1

Scalar matrix: ij ij

, ifi = jA = a where a =

0, ifi j,

6 0 0

0 6 0

0 0 6

Identity matrix: ij ij

1, ifi = jA = a where a =

0, ifi j,

Types 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.

Equality of Matrices

Example:

a b 1 2If = , then

c d -3 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 i j i ja b .Example:

1 3 4 2 3 1If A = and B = , then2 3 4 5 4 2

-3 4 -5 1 -4 3

1 3 4 2 3 1 3 6 5A + 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

Example:

1 3 4

If A = , then1 2 3

-1 1 1

1 3 4 2 6 8

2A=2 =1 2 3 2 4 6

-1 1 1 -2 2 2

is a matrix and k is a scalar, thenij m×nI f A = a

ij m×nkA = ka .

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

Find X, if Y= and 2X+Y = 3 2

1 4

.-1 0

-3 2

Y= and 2X+Y = 3 2

1 4

-1 0

-3 2

Example - 3

Solution :

-1 0 3 22X= -

-3 2 1 4

-1- 3 0- 2 -4 -212X= X=

-3-1 2- 4 -4 -22

-2 -1X=

-2 -1

3 2

1 4

-1 0

-3 2

2X + =

Find a matrix C such that A+B+C is a zero matrix,

where A= and B = 2 0 1

3 -1 0

.2 1 -1

0 2 1

2 0 1 2 1 -1 0 0 0+ +C=

3 -1 0 0 2 1 0 0 0

Example - 4

Solution : A + B + C = 0

4 1 0 0 0 0+C=

3 1 1 0 0 0

0 0 0 4 1 0C= -

0 0 0 3 1 1

-4 -1 0C=

-3 -1 -1

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

Multiplication of Matrices

n

thij jk

j=1

th

AB = a b = Sum of the product of elements ofi row of A with

the corresponding elements of k column of B

Example

If A= and B = then AB is given as1 2 2

2 -1 4

3

1

4

,

31 2 2

12 -1 4

4

AB

=3 2 8

6 1 16

=13

21

4 4

1 3 + 2 1 + 2 4

2 3 + (-1 1)

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)

Properties of Multiplication of Matrices

0

0

1+4+1 2+0+0 0+2+2 2

x

0

0

(6 2 4) 2 =0 0+4+4x

x

4x=- 4

x=-1

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

1 2 1 2 0 1 2 =0

1 0 2 x

Solution :

2 cos sinA =

-sin cos

2 2

2 2

cos - sin cos sin +sin cos

-sin cos - cos sin cos - sin

cos2 sin2=

-sin2 cos2

Example - 6

cos sin

-sin cos

.2 cos2 sin2A =

-sin2 cos2

If A= , then show that

cos sin

-sin cos

Solution :

2 3 -2 3 -2A =

4 -2 4 -2

9- 8 -6+4=

12- 8 -8+4

1 -2=

4 -4

3 -2 3k -2kkA=k =

4 -2 4k -2k

Example - 7

3 -2

4 -2

1 0

0 1

2A =k A-2I.If A = and I= , then find k if

1 -2 3k - 2 -2k=

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

Solution Contd.2A =k A-2I

1 2

4 4

3k -2k 1 0= - 2

4k -2k 0 1

Show that A = satisfies the equation

A2 – 12A + I = O.

6 5

7 6

2 6 5 6 5 36+35 30+30 71 60A = = =

7 6 7 6 42+42 35+36 84 71

2 71 60 6 5 1 0A - 12A + I = -12 +

84 71 7 6 0 1

6 5A = , then

7 6

71 60 72 60 1 0= - +

84 71 84 72 0 1

Example - 8

Solution :

-1 0 1 0

0 -1 0 1

0 0=

0 0

Hence , A2 – 12A + I=O

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’ .

Transpose of a Matrix

For Example: Consider the matrix a b

c d

The transpose of the above matrix is a c

b d

Example - 9

5 -1 2 1If A = and B = ,

6 7 3 4

then verify that (A+B)T=AT+BT

Solution:

5 -1 2 1 7 0A+B = + =

6 7 3 4 9 11

T 7 9(A B)

0 11

T T5 6 2 3A = and B =

-1 7 1 4

T T 7 9A +B =

0 11

Hence, (A+B)T=AT+BT

3 42 4 -1

AB= -1 2-1 0 2

2 1

6- 4- 2 8+8-1

-3+0+4 -4+0+2

0 15=

1 -2

T 0 1(AB) =

15 -2

2 4 -1

-1 0 2

3 4

-1 2 ,

2 1

.T(AB)If A = and B = find

Example - 10

Solution :

Solution :

0 2y z 0 x x

AA'= x y -z 2y y -y

x -y z z -z z

2 2 2 2 2 2

2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2

4y +z 2y - z -2y +z

= 2y - z x +y +z x - y - z

-2y +z x - y - z x +y +z

0 2y z

x y -z

x -y z

Find the values of x , y, z if the matrix

obeys the law AA’= I.

Example - 11

'

0

2

x x

A y y y

z z z

'AA = I

1 0 0

= 0 1 0

0 0 1

2 2 2 2 2 2

2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2

4y +z 2y - z -2y +z

2y - z x +y +z x - y - z

-2y +z x - y - z x +y +z

Equating the elements of column 2 , we get

2 2 2x +y +z =1 ...(ii)

...2 2 2x - y - z =0 (iii)

Solution (Cont.)

2y2 – z2 = 0 …(i)

Adding (ii) and (iii), we get

2 2 1 12x 1 x x

2 2

Form (i), z2 = 2y2

Solution (Cont.)

Putting the value of x2 and z3 in (ii), we get

2 2

2

2

1+y +2y =1

21

3y =2

1 1y = Þ y =±

6 6

Putting the value of y2 in (i), we get

2 21 1 1z =2× z = z =

6 3 3

T

2 1 -1

Find x, if x 4 -1 1 0 0 x 4 -1 =0.

2 2 4

2(2x +2x+4x- 8+x+4)=0

22x +7x- 4=0

2x-1 x+4 =0

1x = or x =-4

2

2 1 -1

x 4 -1 1 0 0 = 2x+4 - 2 x - 2 -x - 4

2 2 4

T2x+4- 2 x - 2 -x - 4 x 4 -1 =0

x

2x+2 x- 2 -x - 4 4 =0

-1

Solution :

Example - 12

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.

Symmetric and Skew – Symmetric Matrix

T TA + A A - AA = + ,

2 2

T Twhere A + A is symmetric matrix and A - A is skew - symmetric matrix.

Show that A= is a skew-symmetric matrix.

0 5 3

-5 0 -8

-3 8 0

Solution :

T

0 -5 -3 0 5 3

A = 5 0 8 =- -5 0 -8 =-A

3 -8 0 -3 8 0

Example - 13

As AT = - A, A is a skew – symmetric matrix

Express the matrix as the sum of a

symmetric and a skew- symmetric matrix.

6 1 -5

A= -2 -5 4

-3 3 -1

6 1 -5

A = -2 -5 4

-3 3 -1

Solution :

T

6 -2 -3

A = 1 -5 3

-5 4 -1

Example - 14

T

6 1 -5 6 -2 -31 1

Let P = (A+A )= -2 -5 4 + 1 -5 32 2

-3 3 -1 -5 4 -1

12 -1 -81

P = -1 -10 72

-8 7 -2

Solution Cont.

6 -1/2 -4

P = -1/2 5 7/2

-4 7/2 -1

T

6 1 -5 6 -2 -31 1

Let Q = A- A = -2 -5 4 - 1 -5 32 2

-3 3 -1 -5 4 -1

T6 -1/2 -4

P = -1/2 5 7/2 =P

-4 7/2 -1

0 3 -21

Q = -3 0 -12

2 -1 0

Therefore, P is symmetric and Q is skew- symmetric . Further, P+Q = A

Hence, A can be expressed as the sum of a symmetric and a skew -symmetric matrix.

Solution Cont.

0 3/2 -1

Q = -3/2 0 1/2

1 -1/2 0

T0 3/2 -1

Q =- -3/2 0 1/2 =-Q

1 -1/2 0

T TP = P and Q = -Qa

THANK YOU