Matrix presentation By DHEERAJ KATARIA

PREPARED BY:-

DHEERAJ KATARIA(007)

Matrix - a rectangular array of variables

or constants in horizontal rows(m) and

vertical columns (n) enclosed in

brackets.

Element - each value in a matrix; either

a number or a constant.

Dimension - number of rows by

number of columns of a matrix.

**A matrix is named by its dimensions.

11 12 13 14

21 22 23 24

31 32 33 34

mn mn mn mn

a a a a

a a a a

a a a a

a a a a

Row

1Row

2Row

3

Row m

Column

1

Column

2

Column

3

Column

4

Examples: Find the dimensions of each

matrix.

1. A =

2 1

0 5

4 8

2. B =

1

2

3

4

0 5 3 13. C =

2 0 9 6

Dimensions: 3x2 Dimensions: 4x1

Dimensions: 2x4

Different types of Matrices

• Column Matrix - a matrix with

only one column.

• Row Matrix - a matrix with

only one row.

• Square Matrix - a matrix that

has the same number of rows

and columns.

row

nmrows

mnmmm

n

n

n

aaaa

a

a

a

aaa

aaa

aaa

A

321

3

2

1

333231

232221

131211

A matrix is a rectangular array of numbers. We

subscript entries to tell their location in the array

Matrices are

identified by

their size.

Matrices and Rows

""

""

columnthj

rowthiaij

A matrix of m rows and n columns is

called a matrix with dimensions m x n.

2 3 4

1.) 11

2

3 8 9

2.) 2 5

6 7 8

103.)

7

4.) 3 4

2 X

3 3 X 3

2 X 11 X 2

To add matrices, we add the

corresponding elements. They must

have the same dimensions.

5 0 6 3

4 1 2 3A B

A + B5 6 0 3

4 2 1 3

1 3

6 4

To subtract matrices, we subtract the

corresponding elements. The matrices

must have the same dimensions.

1 2 1 1

3.) 2 0 1 3

3 1 2 3

1 1 2 ( 1)

2 1 0 3

3 2 1 3

0 3

3 3

5 4

ADDITIVE INVERSE OF A MATRIX:

1 0 2

3 1 5A

1 0 2

3 1 5A

Equal Matrices - two matrices that

have the same dimensions and

each element of one matrix is equal

to the corresponding element of the

other matrix.

*The definition of equal matrices

can be used to find values when

elements of the matrices are

algebraic expressions.

1. 2x

2x 3y

y

12

* Since the matrices are

equal, the corresponding

elements are equal!

* Form two linear

equations.2x y

2x 3y 12 * Solve the

system using

substitution.

Examples: Find the values for x and y

2x y

y 3y 12

4y 12

y 3

2x 3

x 3

2

* Write as linear equations.2. 3x y

x 2y

x 3

y 2

7y 7

y 1

2x y 3

2x 6y 4

2x 1 3

2x 2

x 1

Now check your answer

* Combine like terms.

* Solve using elimination.

3x y x 3

x 2y y 2

x 2y y 2

1 2 1 1 2

1 2 1

1 1

2x y 3

x 3y 2

Scalar

Multiplication:

1 2 3

1 2 3

4 5 6

k

We multiply each # inside our matrix

by k.

1 2 3

1 2 3

4 5 6

k k k

k k k

k k k

812

026

2

14

58

132

y

x

812

026

528

11432

y

x

812

026

56

0432

y

x

812

026

21012

086

y

x

Scalar Multiplication:

6x+8=26

6x=18

x=3

10-2y=8

-2y=-2

y=1

Associative Property of Addition

(A+B)+C = A+(B+C)

Commutative Property of Addition

A+B = B+A

Distributive Property of Addition and Subtraction

S(A+B) = SA+SB

S(A-B) = SA-SB

NOTE: Multiplication is not included!!!

THE IDENTITY MATRIX IS A

SQUARE MATRIX WITH ONE

DOWN THE DIAGONALS

In a 2 X 2 In a 3 X 3

10

01

100

010

001

FINDING THE INVERSE OF A

MATRIX

THE MULTIPLICATIVE IDENTITY

The multiplicative identity for real numbers is the number 1. The property is:

In terms of matrices we need a matrix that can be multiplied by a matrix

(A) and give a product which is the same matrix (A).

If a is a real number, then a x 1 = 1 x a = a.

THE INVERSE OF A MATRIX (A-1)

For an n n matrix A, there may be a B such

that AB = I = BA.

The inverse is analogous to a reciprocal

A matrix which has an inverse is nonsingular.

A matrix which does not have an inverse is

singular.

An inverse exists only if 0A

PROPERTIES OF INVERSE MATRICES

111 -- ABAB

'11 -AA'

AA 11-

THE IDENTITY MATRIX FOR

MULTIPLICATION

Let A be a square matrix with n rows and n columns. Let I be a matrix

with the same dimensions and with 1’s on the main diagonal and 0’s

elsewhere.

Then AI = IA = A

THE MULTIPLICATIVE IDENTITY

1406

7410

2973

9470

B

1000

0100

0010

0001

I

Give the multiplicative identity for matrix B.

This identity matrix is I4.

THE MULTIPLICATIVE INVERSE

10

01

)3(1)1(2)2(1)1(2

)3(1)1(3)2(1)1(3

32

11

12

13

For every nonzero real number a, there is a real number 1/a such that a(1/a)

= 1.

In terms of matrices, the product of a square matrix and its inverse is I.

THE INVERSE OF A MATRIX

Let A be a square matrix with n rows and n columns. If there is an n x

n matrix B such that AB = I and BA = I, then A and B are inverses of

one another. The inverse of matrix A is denoted by A-1.


23

35

53

32BandA

To show that matrices are inverses of one another, show that the

multiplication of the matrices is commutative and results in the identity

matrix.

Show that A and B are inverses.


10

01

)2(5)3(3)3(5)5(3

)2(3)3(2)3(3)5(2

23

35

53

32AB

and


10

01

)5(2)3(3)3(2)2(3

)5)(3()3(5)3)(3()2(5

53

32

23

35BA

FINDING THE INVERSE OF A MATRIX -

METHOD 1

dc

baBandALet

53

21

10

01

53

21

dc

ba

Use the equation AB = I.

Write and solve the equation:

INVERSES – METHOD 1, CONT.

10

01

53

21

dc

ba

10

01

5353

22

dbca

dbca

1235

153

02

053

12

dandbcanda

db

db

ca

ca

INVERSES – METHOD 1, CONT.

13

25

10

01

)1(5)2(3)3(5)5(3

)1(2)2(1)3(2)5(1

13

25

53

21

So the inverse of A =

We can check this by multiplying A x A-1

Find the multiplicative inverse of:

43

21A

21

23

12

13

24

2

11A

2)2(3)4(143

21

)(||

11 AadjA

A

We can check to see if we are correct by multiplying. Remember that AA-1 = I

10

01

)2/1(4)1(3)2/3(4)2(3

)2/1(2)1(1)2/3(2)2(1

21

23

12

43

21

ELEMENTARY ROW OPERATIONS

1. Interchange the order in which the equations are listed.

2. Multiply any equation by a nonzero number.

3. Replace any equation with itself added to a multiple of another equation.

RANK

r is said to be rank of a matrix if it possess these two conditions:-

1). It contains a non-zero minor of order r.

2).All minor of order (r+1) vanishes.

Rank Theorem

Let A be the coefficient matrix of a system of linear equations with

n variables. If the system is consistent, then

Number of free variable = n – rank(A)

RANK OF MATRIX

Equal to the dimension of the largest square sub-matrix of A

that has a non-zero determinant.

Example:

has rank 3

RANK OF MATRIX (CONT’D)

Alternative definition: the maximum number of linearly

independent columns (or rows) of A.

Therefore,

rank is not 4 !Example:

RANK AND SINGULAR MATRICES

ECHELON FORM

A rectangular matrix is in echelon form if it has the following

properties:

1. All nonzero rows are above any rows of all

zeroes.

2. Each leading entry of a row is in a column to

the right of the leading entry of the row above it.

ECHELON FORM

EXAMPLE

ECHELON FORM

A rectangular matrix is in row reduced echelon form if it has

the following properties:

1. It is in echelon form.

2. All entries in a column above and below a

leading entry are zero.

3. Each leading entry is a 1, the only nonzero entry

in its column.

1 0 0 0 0 −20 1 0 0 0 30 0 1 0 0 10 0 0 1 0 40 0 0 0 1 2

REDUCED ROW ECHELON FORM

REDUCED ROW ECHELON FORM

HOMOGENEOUS SYSTEM

AUGMENTED MATRIX

Solve the system of equations using Gauss-Jordan Method

2 0

2 3 1

2 2 3

x y z

x y z

x y z

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

1 2 2

1 1 2

0 1 1

0

1

1

2 3 1

R R R

0 1 1

1 1 2 0

2 3

1

2 1

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

1 1 2 0

0 1 1 1

2 2 1 3

1 3 3

2 2 4 0

2 2 1 3

3

2

0 0 3

R R R

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

1 3 3

2 2 4 0

2 2 1 3

3

2

0 0 3

R R R

0 0

1 1 2 0

0 1 1

3

1

3

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

1 1 2 0

0 1 1 1

0 0 3 3

2 2

0

1

1 1 1

R R

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

2 2

0

1

1 1 1

R R

0 1 1 1

1 1 2 0

0 0 3 3

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

1 1 2 0

0 1 1 1

0 0 3 3

2 1 1

0 1 1 1

1

1 0 1 1

1 2 0

R R R

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

2 1 1

0 1 1 1

1

1 0 1 1

1 2 0

R R R

0 1 1 1

0 0 3 3

1 0 1 1

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

1 0 1 1

0 1 1 1

0 0 3 3

3 3

1

3

0 0 1 1

R R

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

3 3

1

3

0 0 1 1

R R 1 0 1 1

0 1 11

0 0 1 1

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

1 0 1 1

0 1 11

0 0 1 1

3 2 2

0 0 1 1

0 1 1

1

1

0 0 2

R R R

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

3 2 2

0 0 1 1

0 1 1

1

1

0 0 2

R R R

1 0 1 1

0 0 1 1

0 1 0 2

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

1 0 1 1

0 1 0 2

0 0 1 1

3 1 1

0

1 0

0 1

0 0

1

1 0 1 1

R R R

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

3 1 1

0

1 0

0 1

0 0

1

1 0 1 1

R R R

0 1 0 2

0 0 1 1

1 0 0 0

Gauss-Jordan Method


2 0

2 3 1

2 2 3

x y z

x y z

x y z

1 0 0 0

0 1 0 2

0 0 1 1

(0, 2, 1)

Gauss-Jordan Method

The homogeneous equation Ax = 0 has a nontrivial solution

if and only if the equation has at least one free variable.

Basic variables: The variables corresponding to pivot columns

00000

31100

02010

1x 2x 3x4x

Free variables: the others

free is

3

2

free is

4

43

42

1

x

xx

xx

x

A SYSTEM OF LINEAR EQUATIONS

TWO EQUATIONS, TWO UNKNOWNS:

LINES IN A PLANE

THREE POSSIBLE TYPES OF SOLUTIONS

1. No solution


1. A unique solution


1. Infinitely many solutions

THREE EQUATIONS,THREE

UNKNOWNS:

PLANES IN SPACE

Definition of Homogeneous

A system of linear equations is said to be homogeneous

if it can be written in the form Ax = 0, where A is an

matrix and 0 is the zero vector in Rm.

nm

Example:

023

034

0452

321

321

321

xxx

xxx

xxx

Note: Every homogeneous linear system is consistent.

i.e. The homogeneous system Ax = 0 has at least one solution, namely the trivial

solution, x = 0.

The homogeneous equation Ax = 0 has a nontrivial solution

if and only if the equation has at least one free variable.

Basic variables: The variables corresponding to pivot columns

00000

31100

02010

1x 2x 3x4x

Free variables: he others

free is

3

2

free is

4

43

42

1

x

xx

xx

x

Example 3: Describe all solutions for

486

1423

7453

321

321

321

xxx

xxx

xxx

Solutions of Nonhomogeneous Systems

i.e. Describe all solutions of where

Ax b

816

423

453

Aand

b

7

1

4

Geometrically, what does the solution set represent?

Homogeneous

086

0423

0453

321

321

321

xxx

xxx

xxx

Nonhomogeneous

486

1423

7453

321

321

321

xxx

xxx

xxx

1 0-4

30

0 1 0 0

0 0 0 0

1 0-4

3-1

0 1 0 2

0 0 0 0

freex

x

xx

3

2

31

0

3

4

freex

x

xx

3

2

31

2

13

4

1

0

3/4

3

3

2

1

x

x

x

x

1

0

3/4

0

2

1

3

3

2

1

x

x

x

x

THEOREM

CRAMER’S RULE

Example:

Solve the system: 3x - 2y = 10

4x + y = 6

10 2

6 1 222

3 2 11

4 1

x

3 10

4 6 222

3 2 11

4 1

y

The solution is

(2, -2)

CRAMER’S RULE

●Example:Solve the system 3x - 2y + z = 9

x + 2y - 2z = -5x + y - 4z = -2

x

9 2 1

5 2 2

2 1 4

3 2 1

1 2 2

1 1 4

23

23 1 y

3 9 1

1 5 2

1 2 4

3 2 1

1 2 2

1 1 4

69

23 3

CRAMER’S RULE

Example, continued: 3x - 2y + z = 9 x + 2y - 2z = -5

x + y - 4z = -2

z

3 2 9

1 2 5

1 1 2

3 2 1

1 2 2

1 1 4

0

23 0

The solution is

(1, -3, 0)

GAUSSIAN ELIMINATION

To solve a system of equations using Gaussian elimination with matrices, we use the same rules as before.

1. Interchange any two rows.

2. Multiply each entry in a row by the same nonzero constant.

3. Add a nonzero multiple of one row to another row.

A method to solve simultaneous linear equations of the form [A][X]=[C]

Two steps1. Forward Elimination

2. Back Substitution

FORWARD ELIMINATION

735.0

21.96

8.106

7.000

56.18.40

1525

3

2

1

x

x

x

2.279

2.177

8.106

112144

1864

1525

3

2

1

x

x

x

The goal of forward elimination is to transform the coefficient matrix into an upper triangular matrix

FORWARD ELIMINATION

A set of n equations and n unknowns

11313212111 ... bxaxaxaxa nn

22323222121 ... bxaxaxaxa nn

nnnnnnn bxaxaxaxa ...332211

. .

. .

. .

(n-1) steps of forward elimination

Step 1 For Equation 2, divide Equation 1 by and multiply by .

)...( 11313212111

11

21 bxaxaxaxaa

ann

1

11

211

11

21212

11

21121 ... b

a

axa

a

axa

a

axa nn

11a

21a

FORWARD ELIMINATION

FORWARD ELIMINATION

1

11

211

11

21212

11

21121 ... b

a

axa

a

axa

a

axa nn

22323222121 ... bxaxaxaxa nn

1

11

2121

11

212212

11

2122 ... b

a

abxa

a

aaxa

a

aa nnn

'

2

'

22

'

22 ... bxaxa nn

Subtract the result from Equation 2.

−_________________________________________________

or

Repeat this procedure for the remaining equations to reduce the set of equations as

11313212111 ... bxaxaxaxa nn

'

2

'

23

'

232

'

22 ... bxaxaxa nn

'

3

'

33

'

332

'

32 ... bxaxaxa nn

''

3

'

32

'

2 ... nnnnnn bxaxaxa

. . .

. . .

. . .

End of Step 1

FORWARD ELIMINATION

Step 2Repeat the same procedure for the 3rd term of Equation 3.

11313212111 ... bxaxaxaxa nn

'

2

'

23

'

232

'

22 ... bxaxaxa nn

"

3

"

33

"

33 ... bxaxa nn

""

3

"

3 ... nnnnn bxaxa

. .

. .

. .

End of Step 2

FORWARD ELIMINATION

At the end of (n-1) Forward Elimination steps, the system of equations will look like

'

2

'

23

'

232

'

22 ... bxaxaxa nn

"

3

"

33

"

33 ... bxaxa nn

11 n

nn

n

nn bxa

. .

. .

. .

11313212111 ... bxaxaxaxa nn

End of Step (n-1)

FORWARD ELIMINATION

MATRIX FORM AT END OF

FORWARD ELIMINATION

)(n-

n

"

'

n

)(n

nn

"

n

"

'

n

''

n

b

b

b

b

x

x

x

x

a

aa

aaa

aaaa

1

3

2

1

3

2

1

1

333

22322

1131211

0000

00

0

BACK SUBSTITUTION

735.0

21.96

8.106

7.000

56.18.40

1525

3

2

1

x

x

x

Solve each equation starting from the last equation

Example of a system of 3 equations

BACK SUBSTITUTION STARTING EQNS

'

2

'

23

'

232

'

22 ... bxaxaxa nn

"

3

"

3

"

33 ... bxaxa nn

11 n

nn

n

nn bxa

. .

. .

. .

11313212111 ... bxaxaxaxa nn

Start with the last equation because it has only one unknown

)1(

)1(

n

nn

n

n

na

bx

BACK SUBSTITUTION

1,...,1for...

1

1

,2

1

2,1

1

1,

1

nia

xaxaxabx

i

ii

n

i

nii

i

iii

i

ii

i

i

i

1,...,1for1

1

11

nia

xab

xi

ii

n

ijj

i

ij

i

i

i

)1(

)1(

n

nn

n

n

na

bx

BACK SUBSTITUTION

FORWARD ELIMINATION

NUMBER OF STEPS OF FORWARD

ELIMINATION

Number of steps of forward elimination is

(n1)(31)2

Divide Equation 1 by 25 and

multiply it by 64, .

.

408.27356.28.126456.28.1061525

208.96 56.18.4 0

408.273 56.2 8.1264

177.2 1 8 64

2.279112144

2.1771864

8.1061525

2.279112144

208.9656.18.40

8.1061525

56.225

64

Subtract the result from Equation 2

Substitute new equation for Equation 2

FORWARD ELIMINATION

FORWARD ELIMINATION: STEP 1 (CONT.)

.

168.61576.58.2814476.58.1061525

2.279112144

208.9656.18.40

8.1061525

968.335 76.48.16 0

168.615 76.5 8.28 144

279.2 1 12 144

968.33576.48.160

208.9656.18.40

8.1061525

Divide Equation 1 by 25 and

multiply it by 144, .76.525

144



FORWARD ELIMINATION: STEP 2

728.33646.58.1605.3208.9656.18.40

968.33576.48.160

208.9656.18.40

8.1061525

.760 7.0 0 0

728.33646.516.80

335.968 76.416.80

76.07.000

208.9656.18.40

8.1061525

Divide Equation 2 by −4.8

and multiply it by −16.8,

.5.38.4

8.16



BACK SUBSTITUTION

BACK SUBSTITUTION

76.0

208.96

8.106

7.000

56.18.40

1525

7.07.000

2.9656.18.40

8.1061525

3

2

1

a

a

a

08571.1

7.0

76.0

76.07.0

3

3

3

a

a

a

Solving for a3

BACK SUBSTITUTION (CONT.)

Solving for a2

690519.

4.8

1.085711.5696.208

8.4

56.1208.96

208.9656.18.4

2

2

32

32

a

a

aa

aa

76.0

208.96

8.106

7.000

56.18.40

1525

3

2

1

a

a

a

BACK SUBSTITUTION (CONT.)

Solving for a1

290472.0

25

08571.16905.1958.106

25

58.106

8.106525

321

321

aaa

aaa

76.0

2.96

8.106

7.000

56.18.40

1525

3

2

1

a

a

a

GAUSSIAN ELIMINATION

SOLUTION

2279

2177

8106

112144

1864

1525

3

2

1

.

.

.

a

a

a

08571.1

6905.19

290472.0

3

2

1

a

a

a

THANK YOU

Matrix presentation By DHEERAJ KATARIA

Education

Transcript of Matrix presentation By DHEERAJ KATARIA