Class: B.Sc CS.

Subject: Discrete Mathematics

Unit-3

RAI UNIVERSITY, AHMEDABAD

UNIT-III: Basics Of Matrix

Definition: Matrix

A ractangular array of m rows and n columns, enclosed by brackets [ ] is

called a matrix of order m×n. A matrix of order 3×3 is expressed as,

A=[a11 a12 a13

a21 a22 a23

a31 a32 a33]

An element a ij denotes, ith row and jth column

a23 denotes, 2nd row and 3rd column.

Matrices are denoted by capital letters A,B,C,……..etc.

Types of Matrices:

1. Row Matrix:

A matrix having only single row is called row matrix. Its order is 1×n .

For example,

A=[1 4 ]1× 2

A=[ 2 4 −3 ]1× 3

2. Column Matrix:

A matrix having single column is called column matrix. Its order is n×1.

For example,

A=[21]2×1

A=[ 4−26 ]

3× 1

3. Square Matrix:

A matrx in which the number of rows is equal to number columns is called a

square matrix.

∴m=n.

∴ No. of rows = No. of columns

For example,

A=[ 1 3−4 2]

2×2

A=[−3 2 12 3 13 1 −5]

3×3

4. Null Matrix:

A matrix whose all elements are zero, is called null matrix.

For example,

A=[0 00 0]

A=[0 0 00 0 00 0 0]

5. Unit Matrix or Identity Matrix:

A matrix in which all the elements of its principal diagonal are unity(one)

and remaining elements are zero is called unit matrix.

It is denoted by I.

I=[1 00 1]

I=[1 0 00 1 00 0 1]

6. Diagonal Matrix:

A square matrix in which the elements on the principal diagonal are non zero

and all the other elements are zero, is called a diagonal matrix.

For example,

A=[1 00 −5]

A=[3 0 00 −1 00 0 2 ]

7. Scalar Matrix:

A diagonal matrix in which all the elements of its principal diagonal are

equal is called scalar matrix.

For example,

A=[5 0 00 5 00 0 5]

A=[−2 00 −2]

8. Transpose Matrix:

For a given matrix A, if rows and column are interchanged ,the new matrix

obtained A’ is called transpose of a matrix.

Transpose of matrix A is denoted by A ' or AT .

For example,

A=[ 2 1 47 6 −34 1 0 ] ∴ A '=AT=[ 2 7 4

1 6 14 −3 0]

( A' )'=A

9. Symmetric Matrix:

A square matrix A=[aij ] is said to be symmetric, if a ij=a ji of each pair (i, j) .

For Symmetric matrix A'=A

For example,

A=[4 2 02 −3 50 5 6 ] ∴ A '=[4 2 0

2 −3 50 5 6 ]

10. Skew Symmetric Matrix:

A square matrix A=[aij ] is said to be skew symmetric if a ij=−a ji for each pair

(i, j).

For Skew symmetric matrix A'=−A.

For example,

A=[ 0 3 5−3 0 4−5 −4 0] A'=[0 −3 −5

3 0 −45 4 0 ]

11. Singular Matrix:

For a square matrix, if value of its determinant is zero, it is called singular

matrix.

For example,

A=|6 29 3| ∴|A|= (6×3 )− (9×2 )=18−18=0

A=|1 2 34 1 53 6 9|∴|A|=1 ( 9−30 )−2 (36−15 )+3 (24−3 ).

|A|=−21−42+63

|A|=0

If |A|=0 is called singular matrix.

If |A|≠0 is called Non singular matrix.

12. Equal Matrix:

For two matrices, if their corresponding elements are equal, they are called

equal matrices.

For example,

A=[ 2 3−4 5 ] B=[ 2 3

−4 5]∴ A=B

13. Negative matrix:

For two matrices, if their corresponding elements are equal but opposite,

they are called Negative matrix.

For example,

A=[1 −23 −5]∴ (−A )=[−1 2

−3 5]14. Orthogonal matrix:

For square matrix A if Product of matirx A & its transpose matrix A’ (i.e.

AA’) is Identity matrix (I).

A=[cosθ −sinθsinθ cosθ ]

A'=[ cosθ sinθ−sinθ cosθ ]

A A'=[1 00 1 ]=I

15. Upper Triangular Matrix:

For square matrix A if all the elements below the main diagonal are zero

then it is called a Upper Triangular Matrix.

For example,

A=[1 4 50 −2 30 0 3]

16. Lower Triangular Matrix:

For square matrix A if all the elements above the main diagonal are zero

then it is called a Lower Triangular Matrix.

For example,

A=[1 0 04 2 03 −1 6]

17. Trace of Matrix:

For square matrix A sum of all diagonal elements are called trace of matrix

A.

For example,

A=[−1 2 73 5 −81 2 7 ]

tr ( A )=−1+5+7=11

18. Idempotent Matrix:

Matrix A is said to be Idempotent matrix if matrix A satisfy the equation

A2=A .

For example

A=[ 2 −2 −4−1 3 41 −2 −3]

19. Involuntary Matrix:

Matrix A is said to be Involuntary matrix if matrix A satisfy the equation

A2=I . Since I 2=I always. Therefore Unit matrix is involuntary.

20. Conjugate Matrix:

Let A=[ 1+i 2+3 i 47+2 i −i 3−2i ]

Conjugate of matrix A is A

A=[ 1−i 2−3 i 47−2i i 3+2 i ]

Note: Transpose of the conjugate of a matrix A is denoted by Aθ .

21.Unitary Matrix:

A square matrix A is said to be unitary if Aθ A=I

For example,

A=[ 1+i2

−1+i2

−1−i2

1−i2

] , Aθ=[ 1−i2

1−i2

−1−i2

1+i2

], A . Aθ=I

22. Hermitian Matrix:

A square matrix A=(aij) is called Hermitian matrix, if every i-jth element of

A is equal to conjugate complex j-ith element of A.

In other words, a ij=a ji

For example,

[ 1 2+3 i 3+i2−3 i 2 1−2i3−i 1+2 i 5 ]

23. Skew Hermitian Matrix:

A square matrix A=(aij) will be calledd a Skew Hermitian matrix if every i-

jth element of A is equal to negative conjugate complex of j-ith element of

A.

In other words , a ij=−a ji

For example, [ i 2−3i 4+5i−(2+3 i) 0 2i−(4−5i) 2 i −3 i ]

24. Minor:

The minor of an element in a third order determinant is a second order

determinant obtained by deletinng the row and column which contain that

element.

For example,

A=| 1 2 13 4 5

−1 1 2|Minor of element 2 = | 3 5

−1 2|

Minor of element 3 = |2 11 2|

Operations On Matrices and it’s Properties:

Addition of Matrices :

If A and B be two matrices of the same order, then their sum, A+B is defined

as the matrix ,each element of which is the sum of the corresponding

elements of A and B.

Thus if A=[4 2 51 3 −6 ] , B=[1 0 2

3 1 4 ]Then A+B=[4+1 2+0 5+2

1+3 3+1 −6+4 ] A+B=[5 2 7

4 4 −2]. If A=[aij ] ,B= [bij ] then A+B= [aij+bij ]

Properties Of Matrix Addition:

Only matrices of the same order can be added or subtracted.

i. Commutative law: A + B = B + A

ii. Associative law: A + (B + C) = (A + B) + C

Subtraction of Matrices:

The difference of two matrices is a matrix, each element of which is obtained

by subtracting the elements of the second matrix from the

Corresponding element of the first.

A−B=[a ij−a ji]

Thus [8 6 41 2 0]−[3 5 1

7 6 2]¿ [8−3 6−5 4−1

1−7 2−6 0−2]¿ [ 5 1 3

−6 −4 −2] Scalar Multiple of a Matrix:

If a matrix is multiplied by a scalar quantity K, then each element is

multiplied by k,

i.e. A=[ 2 3 44 5 66 7 9]

3 A=3[3×2 3×3 3×43×4 3×5 3×63×6 3×7 3×9]=[ 6 9 12

12 15 1818 21 27 ]

Multiplication:

The product of two matrices A and B is only possible if the number of

columns in A is equal to the number of rows in B.

Let A=[aij ] be an m×n matrix and B=[b ij ] be an n× p matrix. Then the product

AB of these matrices is an m× p matrix C=[c ij ] where,

c ij=a i1b1 j+ai2b2 j+ai3b3 j+…+a¿bnj

Properties of Matrix Multiplication:

a) Multiplication of matrix is not commutative.

AB≠ BA

b) Matrix multiplication is associative , if conformability is assured.

A (BC )=( AB )C

c) Matrix multiplication is distributive with respect to addition.

A (B+C )=AB+AC

d) Multiplcation of matrix A by unit matrix.

AI=IA=A

e) Multiplicative inverse of a matrix exists if |A|≠0.

A . A−1=A−1 . A=I

f) If A is a square then A× A=A2 , A× A × A=A3

g) A0=I

h) I n=I , where n is positive integer.

i) ( AB )'=B ' A'

j) ( AB )−1=B−1 A−1

Example-1. If A=[2 2 22 1 −31 0 4 ], B=[3 3 3

3 0 59 9 −1], C=[ 4 4 4

5 −1 5−7 8 −1]

Find 2 A−3 B+C .

Solution:

2 A−3 B+C=2[2 2 22 1 −31 0 4 ]−3 [3 3 3

3 0 59 9 −1]+ [ 4 4 4

5 −1 5−7 8 −1]

2 A−3 B+C=[4 4 44 2 −62 0 8 ]+[ −9 −9 −9

−9 0 −15−27 −27 3 ]+[ 4 4 4

5 −1 5−7 8 −1].

2 A−3 B+C=[ 4−9+4 4−9+4 4−9+44−9+5 2+0−1 −6−15+5

2−27−7 0−27+8 8+3−1 ].2 A−3 B+C=[ −1 −1 −1

0 1 −16−32 −19 10 ].

Example-2. If A=[1 2 13 4 2] ,B=[3 −2 4

1 5 0] find matrixX from X+A+B=0

Solution: X+A+B=0

∴ X+[1 2 13 4 2]+[3 −2 4

1 5 0 ]=0.

∴ X+[4 0 54 9 2]=[0 0 0

0 0 0] ∴ X=[−4 0 −5

−4 −9 −2] Example-3: show that any square matrix can be expressed as the sum of

two matrices, one symmetric and the other anti-symmetric.

Solution: Let A be a given square matrix.

Then A=12

(A+A' )+ 12(A−A ')

Now, ( A+A ' )'=A '+ (A ' )'=A'+A=A+A'

∴ A+A ' is a symmetric matrix.

Also, ( A−A ' )'=A'−( A ' )'=A '−A=−(A−A' )

∴ ( A−A' )∨12(A−A ') is an anti symmetric matrix.

∴ A=12

( A+A ' )+ 12(A−A ')

Example-4.Express A=[1 −2 −33 0 55 6 1 ] as the sum of a lower triangular

matrix and upper triangular matrix.

Solution: Let A=L+U

[1 −2 −33 0 55 6 1 ]=[ a 0 0

b c 0d e f ]+[1 p q

0 1 r0 0 1]

[1 −2 −33 0 55 6 1 ]=[ a+1 0 0

b+0 c+1 0+rd+0 e+0 f +1]

Equating the corresponding elements on both the sides, we get

a+1=1 p=−2 q=−3

b=3 c+1=0 r=5

d=5 e=6 f +1=1

On Solving these equations, we get

a=0 p=−2 q=−3

b=3 c=−1 r=5

d=5 e=6 f=0

Hence L=[0 0 03 −1 05 6 0]∧U=[1 −2 −3

0 1 50 0 1 ]

Example-5 If A=[1 2 34 5 6] and B=[1 2

2 11 2] find AB and BA .

Solution: AB=¿ [1 2 34 5 6] [1 2

2 11 2]

AB=[ (1×1 )+(2×2 )+(3×1) (1×2 )+(2×1 )+(3×2)( 4×1 )+(5×2 )+(6×1) (4×2 )+ (5×1 )+(6×2)]

AB=[ (1+4+3) (2+2+6)(4+10+6) (8+5+12)]

AB=[ 8 1020 25]

Now BA=¿ [1 22 11 2] [1 2 3

4 5 6 ]

A=[ (1×1 )+(2×4) (1×2 )+(2×5) (1×3 )+(2×6)(2×1 )+(1×4) (2×2 )+(1×5) (2×3 )+(1×6)(1×1 )+(2×4) (1×2 )+(2×5) (1×3 )+(2×6)]

A=[(1+8) (2+10) (3+12)(2+4) (4+5) (6+6)(1+8) (2+10) (3+12)]

A=[9 12 156 9 129 12 15]

Example-6. If A=[ 1 2−2 3] ,B=[2 1

2 3 ] and C=[−3 12 0].

Verify that ( AB )C=A (BC ) and A (B+C )=AB+AC .

Solution: AB=[ 1 2−2 3]×[2 1

2 3] AB=[ (1 ) (2 )+(2 )(2) (1 ) (1 )+(2 )(3)

(−2 ) (2 )+(3 )(2) (−2 ) (1 )+(3 )(3)]

AB=[6 72 7 ]

BC=[2 12 3]×[−3 1

2 0] BC=[−6+2 2+0

−6+6 2+0] BC=[−4 2

0 2] AC=[ 1 2

−2 3]×[−3 12 0]

AC=[−3+4 1+06+6 −2+0]

AC=[ 1 112 −2]

B+C=[2 12 3]+[−3 1

2 0] B+C=[2+(−3) 1+1

2+2 3+0 ] B+C=[−1 2

4 3]i. ( AB )C=[6 7

2 7 ]×[−3 12 0]

( AB )C=[−18+14 6+0−6+14 2+0 ]

( AB )C=[−4 68 2 ] .……………………………………………..(1)

& A (BC )=[ 1 2−2 3]×[−4 2

0 2] A (BC )=[−4+0 2+4

8+0 −4+6] A (BC )=[−4 6

8 2 ] ……………………………………………….(2)

Thus from (1) and (2), we get

( AB )C=A (BC )

ii. A (B+C )=[ 1 2−2 3][−1 2

4 3 ] A (B+C )=[−1+8 2+6

2+12 −4+9] A (B+C )=[ 7 8

14 5] ………………………………………….(3)

AB+AC=[ 6+1 7+12+12 7−2]

AB+AC=[ 7 824 5 ] …………………………………………..(4)

Thus from (3) and (4), we get

A (B+C )=AB+AC

Example-7. If A=[1 2 22 1 22 2 1] show that A2−4 A−5 I=0 where I, 0 are the unit

matrix and the null matrix of order 3 respectively. Use this result to find

A−1 .

Solution: Here, we have A=[1 2 22 1 22 2 1]

A2=[1 2 22 1 22 2 1][1 2 2

2 1 22 2 1]

A2=[9 8 88 9 88 8 9]

A2−4 A−5 I=[9 8 88 9 88 8 9]−4 [1 2 2

2 1 22 2 1]−5[1 0 0

0 1 00 0 1]

A2−4 A−5 I=[9−4−5 8−8+0 8−8+08−8+0 9−4−5 8−8+08−8+0 8−8+0 9−4−5]

A2−4 A−5 I=[0 0 00 0 00 0 0]

A2−4 A−5 I=0⟹5 I=A2−4 A

Both the side multiplying by A−1 , we get

5 A−1=A−4 I

5 A−1=[1 2 22 1 22 2 1]−4 [1 0 0

0 1 00 0 1]

5 A−1=[−3 2 22 −3 22 2 −3 ]

A−1=15 [−3 2 2

2 −3 22 2 −3 ]

Adjoint of a square Matrix:

Let the determinant of the square matrix Abe |A|=[a1 a2 a3

b1 b2 b3

c1 c2 c3]

The matrix formed by the co-factors of the elements in

|A|is [A1 A2 A3

B1 B2 B3

C1 C2 C3]

Where A1=|b2 b3

c2 c3|=b2c3−b3 c2 ,

A2=−|b1 b3

c1 c3|=−b1c3−b3 c1 ,

A3=|b1 b2

c1 c2|=b1c2−b2c1 ,

B1=−|a2 a3

c2 c3|= −a2c3+a3c2 ,

B2=|a1 a3

c1 c3|=a1c3−a3 c1 ,

B3=−|a1 a2

c1 c2|=−a1c2+a2 c1 ,

C1=|a2 a3

b2 b3|=a2b3−a3b2 ,

C2=−|a1 a3

b1 b3|=−a1b3+a3b1 ,

C3=|a1 a2

b1 b2|=a1b2−a2b1 ,

Then the transpose of the matrix of co-factors

[A1 B1 C1

A2 B2 C2

A3 B3 C3]

Is called the adjoint of the matrix A and is written as adj A .

Note: For 2×2 order matrix Adj A is defined as:

If A=[a1 a2

b1 b2] then Adj A = [b2 −a2

b1 a1]

i.e. Change location of elelment of principal diagonal

and change sign of elements of subsidary diagonal.

Property of Adjoint Matrix:

The product of a matrix A and its adjoint is equal to unit matrix multiplied by

the determinant A .

Example-1: If A=[5 27 3] then find A−1 .

Solution: Since here, given matrix A is of the Order 2×2.

∴adj A=[ 3 −2−7 5 ]

Example-2: For matrix A=[2 1 50 3 −12 5 0 ] find co-factor matrix adj A .

Solution: Let A=[aij ]

First we will find co-factors of each element.

A11=|3 −15 0 |=0−(−5 )=5

A12=−|0 −12 0 |=−[0−(−2 ) ]=−2

A13=|0 32 5|=(0−6)=−6

A21=−|1 55 0|=−(0−25 )=25

A22=|2 52 0|=(0−10)=−10

A23=−|2 12 5|=−(10−2 )=−8

A31=|1 53 −1|=(−1−15)=−16

A32=−|2 50 −1|=−(−2−0 )=2

A33=|2 10 3|=(6−0 )=6

∴adj A=[ A11 A21 A31

A12 A22 A32

A13 A23 A33] = [ 5 25 −16

−2 −10 2−6 −8 6 ]

Inverse of a Matrix:

If A and B are two square matrices of the same order, such that

AB=BA=I

Then B is called the inverse of A i.e. B=A−1 and A is the invese of B.

Condition for a square matrix A to possess an inverse is that matrix A is

non-singular.

i.e. |A|≠0

If A is square matrix and B its inverse, then AB=I . Taking determinant of

both sides, we get

|AB|=|I|∨|A||B|=I

From this relation it is clear that |A|≠0

i.e. the matrix A is non singular.

To find the inverse matrix with the help of adjoint matrix:

We know that A . (AdjA )=|A|I

A .1

|A|( Adj A )=I (Provided |A|≠0¿ ………………………..(1)

Since A . A−1=I …………………………………………….(2)

From (1) and (2), we have

∴ A−1= 1|A|

(Adj A )=I

Example-1: If A=[ 1 2−1 3 ] find A−1 .

Solution: |A|=| 1 2−1 3|=3−(−2 )=5≠0

∴ A−1is possible.

Adj A=[3 −21 1 ]

∴ A−1=adj A|A|

A−1=15 [3 −2

1 1 ]

A−1=[ 35

−25

15

15

]

Example-2: If A=[1 −1 12 −1 01 0 1] then find A−1 .

Solution: Let given matrix is A .

∴|A|=|1 −1 12 −1 01 0 1|

∴|A|=1 (−1−0 )+1 (2−0 )+1[0−(−1 )]

∴|A|=−1+2+1=2≠0

∴ A−1is possible.

First we will calculate, co-factors of each element.

A11=|−1 00 1|=(−1−0)=−1

A12=−|2 01 1|=[−(2−0 )]=−2

A13=|2 −11 0 |=[0−(−1 )]=1

A21=−|−1 10 1|=[−(−1−0 )]=1

A22=|1 11 1|=(1−1)=0

A23=−|1 −11 0 |=−[0−(−1)]=−1

A31=|−1 1−1 0|=[0−(−1 )]=1

A32=−|1 12 0|=[−(0−2 )]=2

A33=|1 −12 −1|=[−1−(−2 )]=1

∴adj A=[ A11 A21 A31

A12 A22 A32

A13 A23 A33] = [−1 1 1

−2 0 21 −1 1]

A−1=AdjA

|A| ¿12×[−1 1 1

−2 0 21 −1 1]

A−1=¿ [−12

12

12

−1 0 112

−12

12]

Example-3. If a matrix A satisfies a relation A2+A−I=0 prove that A−1

exists and that A−1=I+A , I being an identity matrix.

Solution: Here, A2+A−I=0

∴ A2+A=I

∴ A ( A+ I )=I

∴|A||A+ I|=|I|

∴|A|≠0 and so A−1 exists.

Again A2+A−I=0⟹ A2+A=I

Multiplying (1) by A−1, we get

A−1 ( A2+A )=A−1 I⟹ A+ I=A−1

∴ A−1=I+A

Reference Book and Website Name:

1. Polytechnic Mathematics -1 by Nirav Prakashan.2. Introduction to Engineering Mathematics-1 by H.K. Dass and Dr.Rama

Verma. (S.Chand)3. Engineering Mathematics ( Pearson Fourth Edition)

4. A Textbook of Engineering mathematics by N.P.Bali and Dr.Manish

goyal

5. http://aleph0.clarku.edu/~djoyce/ma130/elementary.pdf

EXERCISE-3

Q-1.Evaluate the following Questions:

1. If A=[0 2 01 0 31 1 2] ,B=[1 2 1

2 1 00 0 3] find ( i )2 A+3 B ( ii ) 3 A−4 B

2. Express [1 2 03 7 15 9 3]as a sum of symmetric matrix and skew-symmetric

matrix.

3. If A=[2 31 0], B=[4 1

2 −3] prove that ( A+B )T=AT+BT .

4. If A=[2 −1 03 2 −45 1 9 ] and B=[ 17 −1 3

−24 −1 −16−7 1 1 ] and 4 A+3C=B , then find matrix

C .

Q-2. Evaluate the following Questions:

1. If A=[0 1 21 2 32 3 4] and B=[ 1 −2

−1 02 −1] Obtain the product AB and explain why

BA is not defined.

2. If A=[1 2 −13 0 24 5 0 ] and B=[1 0 0

2 1 00 1 3] verify that ( AB )'=B ' A' .

3. Compute AB if A=[1 2 34 5 6] and B=[2 5 3

3 6 44 7 5 ].

4. Verify that A=13 [ 1 2 2

2 1 −2−2 2 −1] is Orthogonal.

Q-3. Evaluate the following Questions:

1. If A=[2 5 33 1 21 2 1] then find Adj A .

2. If A=[1 1 21 9 31 4 2] then find A−1 .

3. If A=[1 1 11 2 31 4 9] ,B=[2 5 3

3 1 21 2 1] ,then show that ( AB )−1=B−1 A−1 .

4. If A=[−4 −3 −31 0 14 4 3 ] Prove that Adj A=A .