Determinants - PowerPoint Presentation

Slide1

Determinants Slide2

Determinants

If is a square matrix of order 1,

then |A| = | a

11 | = a11

If is a square matrix of order 2, then

|A| =

= a

11

a

22

– a

21a12Slide3

ExampleSlide4

Solution

If A =

is a square matrix of order 3, then

[Expanding along first row]Slide5

Example

[Expanding along first row]

Solution :Slide6

MinorsSlide7

Minors

M

11

= Minor of a

11

= determinant of the order 2 × 2 square sub-matrix is obtained by leaving first row and first column of A

Similarly, M

23

= Minor of a

23

M

32

= Minor of a

32

etc.Slide8

CofactorsSlide9

Cofactors (Con.)

C

11

= Cofactor of a11 = (–1)1 + 1 M

11 = (–1)1 +1

C

23

= Cofactor of a

23 = (–1)

2 + 3 M23 =

C

32

= Cofactor of a

32

= (–1)

3 + 2

M

32

= etc.Slide10

Value of Determinant in Terms of Minors and CofactorsSlide11

Properties

1.

If

each element of a row (or column) of a determinant is zero, then its value is zero.

Proof

:

Expanding

the determinant along the

row containing only zeros:Slide12

Properties

2

.

If all the elements of a row (or column) is multiplied by a non-zero number k, then the value of the new determinant is k times the value of the original determinant.

Proof

:

expand

elong the row multiplied by Slide13

Properties (Con.)

4. If each element of any row (or column) consists of two or more terms, then the determinant can be expressed as the sum of two

(

or more) determinants.

Pro

of:

expand

along the row containing

the sum:Slide14

Properties of Determinants

2. If any two rows (or columns) of a determinant are interchanged, then the value of the determinant is changed by minus sign.

Pro

of:

first

we proove it for neighbourh rows:Slide15

Cont

. Interchanging 2

rows (columns

)Suppose, that we change the i. and j. rows, and j-i = k.i i ji+1 j ii+2 i+1 i+1…. i+2 i+2I+k=j …. …Then we need k neighbourh change to position the j. row right below the i. (see 2nd

column) One more change and j stands at

the earlier position of i, and i is the row right after j. Again, i must be interchanged by its neighbours

k times,

so the sum of neighbourgh row changes is: k+k+1=2k+1 – an odd number, so because by each change the deterimant changes it sing, after odd number of changes, its changes

its sign. Slide16

Properties

If

any two rows (or columns) of a determinant are identical,

then its value is zero.

Proof: if we

interchanged the identical rows, the determinant remains

the

same, but formally it does changes its sign, so :D = ( - D ), which is possible only if D=0Slide17

5. The value of a determinant is unchanged, if any row (or column) is multiplied by a number and then added to any other row (or column).

Properties (Con.)

Proof

:Slide18

Properties (Con.)

If

A is an

upper

(lower) triangular

matrix, then the determinant is equal

to

the product of the elements of the main diagonal: Proof: Slide19

1. The value of a determinant remains unchanged, if its rows and columns are interchanged

.

(A

matrix and its

transpose have

the same detereminant)Slide20

Row(Column) Operations

Following are the notations to evaluate a determinant:

Similar notations can be used to denote column

operations by replacing R with C.

(i) R

i

to denote ith row

(ii) R

i

Rj to denote the interchange of ith and jth rows.

(iii) R

i Ri + lRj to denote the addition of l times the elements of jth row to the corresponding elements of ith row.(iv) lRi to denote the multiplication of all elements of ith row by l.Slide21

Evaluation of Determinants

If a determinant becomes zero on putting is the factor of the determinant.

, because C

1

and C

2

are identical at x = 2

Hence, (x – 2) is a factor of determinant .Slide22

Sign System for Expansion of Determinant

Sign System for order 2 and order 3 are given bySlide23

Example-1

Find the value of the following determinants

(i) (ii)

Solution :Slide24

Example –1 (ii)

(ii)Slide25

Evaluate the determinant

Solution :

Example - 2Slide26

Example - 3

Evaluate the determinant:

Solution:Slide27

Now expanding along C

1

, we get

(a-b) (b-c) (c-a) [- (c

2

– ab – ac – bc – c

2

)]

= (a-b) (b-c) (c-a) (ab + bc + ac)Solution Cont.Slide28

Without expanding the determinant,

prove that

Example-4

Solution :Slide29

Solution Cont.Slide30

Prove that : = 0 , where

w

is cube root of unity.

Example -5

Solution :Slide31

Example-6

Prove that :

Solution :Slide32

Solution cont.

Expanding along C

1

, we get

(x + a + b + c) [1(x

2)] = x2 (x + a + b + c)

= R.H.S