Похожие презентации:

# Linear Algebra. Chapter 1. Linear Equations in Linear Algebra

## 1. Chapter 1 Linear Equations in Linear Algebra

Linear AlgebraChapter 1

Linear Equations in Linear

Algebra

## 2. 1.1 Matrices and Systems of Linear Equations

Definition• An equation such as x+3y=9 is called a linear equation

(in two variables or unknowns).

• The graph of this equation is a straight line in the xy-plane.

• A pair of values of x and y that satisfy the equation is called

a solution.

Ch1_2

## 3.

DefinitionA linear equation in n variables x1, x2, x3, …, xn has the

form a1 x1 + a2 x2 + a3 x3 + … + an xn = b

where the coefficients a1, a2, a3, …, an and b are real

numbers.

Ch1_3

## 4. Solutions for system of linear equations

Figure 1.1Unique solution

x + 3y = 9

–2x + y = –4

Lines intersect at (3, 2)

Unique solution:

x = 3, y = 2.

Figure 1.2

No solution

–2x + y = 3

–4x + 2y = 2

Lines are parallel.

No point of intersection.

No solutions.

Figure 1.3

Many solution

4x – 2y = 6

6x – 3y = 9

Both equations have the

same graph. Any point on

the graph is a solution.

Many solutions.

Ch1_4

## 5.

A linear equation in three variables corresponds toa plane in three-dimensional space.

※ Systems of three linear equations in three variables:

Unique solution

Ch1_5

## 6.

No solutionsMany solutions

Ch1_6

## 7.

A solution to a system of a three linear equations will bepoints that lie on all three planes.

The following is an example of a system of three linear

equations:

x1

x2

x3 2

2 x1 3 x2

x3 3

x1 x2 2 x3 6

How to solve a system of linear equations? For this we

introduce a method called Gauss-Jordan elimination.

(Section1.2)

Ch1_7

## 8.

Definition• A matrix is a rectangular array of numbers.

• The numbers in the array are called the elements of the

matrix.

Matrices

2 3 4

A

7

5

1

7 1

B 0 5

8 3

6

3 5

C 0 2 5

8 9 12

Ch1_8

## 9.

2 3 4A

7

5

1

Row and Column

2 3 4

7 5 1

row 1

row 2

2

7

column 1

3

5

column 2

4

1

column 3

Submatrix

1 7 4

A 2 3 0

5 1 2

matrix A

1 7

P 2 3

5 1

7

4

1

Q 3

R

5

2

1

submatrice s of A

Ch1_9

## 10.

Size and Type1 0

2 4

2 5 7

9 0 1

3 5 8

3 3 matrix

3

5

Size : 2 3

1 4 matrix

8

3

2

3 1 matrix

a row matrix

a column matrix

4

a square matrix

3 8

5

Location

2 3 4

A

7 5 1

a13 4, a21 7

The element aij is in row i , column j

The element in location (1,3) is 4

Identity Matrices

diagonal size

I 2 1 0

0 1

1 0 0

I 3 0 1 0

0 0 1

Ch1_10

## 11.

Relations between system of linear equationsand matrices

matrix of coefficients and augmented matrix

x1 x2 x3 2

2 x1 3x2 x3 3

x1 x2 2 x3 6

1

1 1

2

3

1

1 1 2

matrix of coefficien ts

1

2

1 1

2

3

1

3

1 1 2 6

augmented matrix

Ch1_11

## 12.

Elementary Row Operations of MatricesElementary Transformation

1. Interchange two equations.

2. Multiply both sides of an

equation by a nonzero

constant.

3. Add a multiple of one

equation to another equation.

Elementary Row

Operation

1. Interchange two rows of a

matrix.

2.

Multiply the elements of a

row by a nonzero constant.

3.

Add a multiple of the

elements of one row to the

corresponding elements of

another row.

Ch1_12

## 13. Example 1

Solving the following system of linear equation.x1 x2 x3 2

2 x1 3 x2 x3 3

x1 x2 2 x3 6

row equivalent

Solution

Equation Method

Initial system:

x1 x2 x3 2

Eq2+(–2)Eq1

2 x1 3 x2 x3 3

x1 x2 2 x3 6

Eq3+(–1)Eq1

x1 x2 x3 2

Analogous Matrix Method

Augmented matrix:

1

2

1 1

3

1

3

2

1 1 2 6

x2 x3 1 R2+(–2)R1

2 x2 3x3 8 R3+(–1)R1

1

1

2

1

1 1 1

0

0 2 3 8

Ch1_13

## 14.

Eq1+(–1)Eq2Eq3+(2)Eq2

x1 x2 x3 2

x2 x3 1

2 x2 3 x3 8

(–1/5)Eq3

x1 2 x3 3

x2 x3 1

5 x3 10

R1+(–1)R2

R3+(2)R2

2

3

1 0

0 1 1 1

0 0 5 10

x1 2 x3 3

x2 x3 1

x3 2

(–1/5)R3

2

3

1 0

0 1 1 1

1

2

0 0

R1+(–2)R3

R2+R3

1

0

0

Eq1+(–2)Eq3

Eq2+Eq3

The solution is

x1 1, x2 1, x3 2.

x1 1

x2 1

x3 2

1

1

2

1

1 1 1

0

0 2 3 8

0

0

1

0

0

1

The solution is

x1 1, x2 1, x3 2.

1

1

2

Ch1_14

## 15. Example 2

Solving the following system of linear equation.x1 2 x2 4 x3 12

2 x1 x2 5 x3 18

x1 3 x2 3 x3 8

Solution

4 12

4 12

1 2

1 2

R2 ( 2)R1

5 18

3 3 6

2 1

0

R3 R1

3 3 8

1

1

4

1

0

1

R2

3

1 2 4 12

1 1 2

0

1 1

4

0

R1 (2)R2

R3 ( 1) R2

8

1 0 2

0 1 1 2

6

0 0 2

8

1 0 2

1 0 0 2

0 1 1 2 R1 ( 2)R3

0

1

0

1

1

R2 R3

R3

0 0

1

3

0 0 1 3

2

x1 2

solution x2 1.

x 3

3

Ch1_15

## 16. Example 3

Solve the system4 x1 8 x2 12 x3 44

3 x1 6 x2 8 x3 32

2 x1 x2

7

Solution

1 2 3 11

4 8 12 44

1 2 3 11

3 6 8 32 R2 ( 3)R1

3 6 8 32

0

0

1

1

1 R1

R3

2

R1

2 1

0 7

0 7 4

0 3 6 15

2 1

1 2 3 11

1 2 3 11

0 1 2

0 3 6 15

5

R2 R3

R1 ( 2)R2

1 R2

0 0

0 0

1 1

1 1 3

1 0 0

R1 ( 1)R3

R2 2R3

2

0 1 0

.

3

0 0 1 1

1 1

1 0

0 1 2

5

0 0

1 1

The solution is x1 2, x2 3, x3 1.

Ch1_16

## 17. Summary

8 12 444

[ A : B ] 3 6 8 32

2 1

0 7

4 x1 8 x2 12 x3 44

3 x1 6 x2 8 x3 32

2 x1 x2

7

A

Use row operations to [A: B] :

1

8 12 44

4

3 6 8 32 0

2 1

0 7

0

0

0

1

0

0

1

2

3 .

1

B

i.e., [ A : B] [ I n : X ]

Def. [In : X] is called the reduced echelon form of [A : B].

Note. 1. If A is the matrix of coefficients of a system of n equations

in n variables that has a unique solution,

then A is row equivalent to In (A In).

2. If A In, then the system has unique solution.

Ch1_17

## 18. Example 4 Many Systems

Solving the following three systems of linear equation, all ofwhich have the same matrix of coefficients.

x1 x2 3 x3 b1

b1 8 0 3

2 x1 x2 4 x3 b2 for b2 11 , 1 , 3 in turn

b 11 2 4

x1 2 x2 4 x3 b3

3

Solution

3

8 0

3

3

8 0

3

1 1

1 1

4

11 1

3 R2+(–2)R1 0

1 2 5 1 3

2 1

R3+R1

1 1 3 2 1

1 2 4 11 2 4

0

1 0

1

3 1

0

1 0 0 1 0 2

0 1 2 5 1 3 R1 ( 1) R3

R1 R2

0 1 0 1 3

1

R3 ( 1)R2

0 0

R2 2 R3

1

2

1

2

2 1

2

0 0 1

x1 1 x1 0 x1 2

The solutions to

x

1

,

x

3

,

2

x2 1 .

the three systems are 2

x 2 x 1 x 2

3

3

3

Ch1_18

## 19. Homework

Exercises will be given by the teachers ofthe practical classes.

Ch1_19

## 20. 1.2 Gauss-Jordan Elimination

DefinitionA matrix is in reduced echelon form if

1. Any rows consisting entirely of zeros are grouped at the

bottom of the matrix.

2. The first nonzero element of each other row is 1. This

element is called a leading 1.

3. The leading 1 of each row after the first is positioned to

the right of the leading 1 of the previous row.

4. All other elements in a column that contains a leading 1

are zero.

Ch1_20

## 21.

Examples for reduced echelon form1 0 8

0 1 2

0 0 0

( )

1 2 0 4

0 0 0 0

0 0 1 3

( )

1 0 0 7

0 1 0 3

0 0 1 9

( )

1 2 0 3 0

0 0 3 4 0

0 0 0 0 1

( )

elementary row operations reduced echelon form

The reduced echelon form of a matrix is unique.

Ch1_21

## 22. Gauss-Jordan Elimination

System of linear equationsaugmented matrix

reduced echelon form

solution

Ch1_22

## 23. Example 1

Use the method of Gauss-Jordan elimination to find reducedechelon form of the following matrix.

2 2 2

0 0

9 12

3 3 3

4 4 2 11 12

pivot (leading 1)

Solution

3

3 3

9 12 1 1 1

3 4

2 2 2

0

2 2

2 1 R1 0 0

0

R1 R2

4 4 2 11 12

3

4

4

2

11

12

1 1 1

2

0 0

R3 ( 4)R1

0 0

pivot 2

3

2

1

1

1

2 R2 0

2

4

0

4

1

1

3

0

1

1

0

2

1

4

1

4

5

1 1 0 2

1 1 0 0 17

R1 R2 0 0 1 1

1 R1 ( 2)R3 0 0 1 0 5

R3 ( 2)R2 0 0 0

1 6 R2 R3 0 0 0 1 6

The matrix is the reduced echelon form of the given matrix.

Ch1_23

## 24. Example 2

Solve, if possible, the system of equations3 x1 3 x2 3 x3 9

2 x1 x2 4 x3 7

3 x1 5 x2 x3 7

Solution

1

3

3 3 3 9 1 1 1 3 1 1

1

2

1

2 1 4 7 1 R1 2 1 4 7 R2 ( 2)R1 0

3 5 1 7 3 3 5 1 7 R3 ( 3)R1 0 2 4 2

1 0 3 4

R1 R2 0

1 2 1

0 0 0 0

x 3 x3 4 x1 3 x3 4

1

x2 2 x3 1 x2 2 x3 1

R3 2R2

The general solution to the system is

x1 3r 4

x2 2r 1

x3 r

, where r is real number (called a parameter) .

Ch1_24

## 25. Example 3

Solve the system of equations2 x1 4 x2 12 x3 10 x4 58

x1 2 x2 3 x3 2 x4 14

2 x1 4 x2 9 x3 6 x4 44

Solution

2 4 12

2 3

1

9

2 4

1 2

R2 R1 0

0

R3 ( 2)R1

0

0

1 2

R1 ( 6)R2 0

0

R3 3R2

0

0

many

sol.

10

58 1 2

6 5

29

2 14 1 R1 1

2 3

2 14

6

44 2 2 4

9 6

44

6 5

29 1 2

6 5

29

3 3

15 1 R2 0

0

1 1

5

3

4 14 3 0

0 3

4 14

0

1 1 1 2 0 0 2

1 1 5 R1 ( 1)R3 0

0 1 0

6

0

1 1 R2 R3 0

0 0 1

1

x1 2 x2 2

x3 6

x4 1

x1 2r 2

x2 r

, for some r.

x3 6

x4 1

Ch1_25

## 26. Example 4

Solve the system of equationsx1 2 x2 x3 3 x4 x5 2

2 x1 4 x2 2 x3 6 x4 3 x5 6

x1 2 x2 x3 x4 3 x5 4

Solution

2 1 3 1 2 1 2 1 3 1 2

1

4 2 6 3 6 R2 ( 2)R1 0 0 0 0 1 2

2

1 1 3 4 R3 R1 0 0 0 2 4 6

1 2

1 2 1 3 1 2 1 2 1 3 1 2

0 0 0 2 4 6

0 0 0 1 2 3

1

R

2

R2 R3

0 0 0 0 1 2 2 0 0 0 0 1 2

1 2 1 0 5 7

1 2 1 0 0

3

0 0 0 1

2

3 R1 5R3 0 0 0 1 0 1

R1 ( 3)R2

1

2 R2 ( 2)R3 0 0 0 0 1 2

0 0 0 0

x1 2r s 3

x1 2 x2 x3 3

x4 1

x2 r , x3 s, x4 1, , for some r and s.

x5 2

x5 2

Ch1_26

## 27. Example 5

This example illustrates a system that has no solution. Let us tryto solve the system

x1 x2 2 x3 3

2 x1 2 x2 5 x3 4

x1 2 x2 x3 3

2 x2 2 x3 1

Solution

1 1 2 3 1 1 2 3

1 1 2 3

2 2 5 4 R2 ( 2)R1 0 0 1 2 0 3 3 6

1 2 1 3 R3 ( 1)R1 0 3 3 6 R2 R3 0 0 1 2

0 2 2 1

0 2 2 1

0 2 2 1

1

0 0

3

1 1 2 3 1 0 1 1

0

1

0

4

0 1 1 2

0 1 1 2 R1 ( 1)R3

R1 R2

1

R2

R3

3 R2 0 0 1 2 R4 ( 2)R2 0 0 1 2 R4 ( 4)R3 0 0 1 2

0 2 2 1

0 0 4 5

0

0 0 13

1 0 0 3

0x1+0x2+0x3=1

0 1 0 4

131 R4 0 0 1 2 The system has no solution.

0 0 0 1

Ch1_27

## 28. Homogeneous System of linear Equations

DefinitionA system of linear equations is said to be homogeneous if all

the constant terms are zeros.

Example:

x1 2 x2 5 x3 0

2 x1 3 x2 6 x3 0

Observe that x1 0, x2 0, x3 0 is a solution.

Theorem 1.1

A system of homogeneous linear equations in n variables always

has the solution x1 = 0, x2 = 0. …, xn = 0. This solution is called

the trivial solution.

Ch1_28

## 29. Homogeneous System of linear Equations

Note. Non trivial solutionExample: x1 2 x2 5 x3 0

2 x1 3 x2 6 x3 0

The system has other nontrivial solutions.

2 5 0

1

1 0 3 0

2 3 6 0 0 1 4 0

x1 3r , x2 4r , x3 r

Theorem 1.2

A system of homogeneous linear equations that has more

variables than equations has many solutions.

Ch1_29

## 30. Homework

Exercise will be given by the teachers of thepractical classes.

Ch1_30

## 31. 1.3 Gaussian Elimination

DefinitionA matrix is in echelon form if

1. Any rows consisting entirely of zeros are

grouped at the bottom of the matrix.

2. The first nonzero element of each row is 1. This

element is called a leading 1.

3. The leading 1 of each row after the first is

positioned to the right of the leading 1 of the

previous row.

(This implies that all the elements below a leading 1

are zero.)

Ch1_31

## 32.

Example 6Solving the following system of linear equations using the

method of Gaussian elimination.

x1 2 x2 3 x3 2 x4 1

x1 2 x2 2 x3 x4 2

2 x1 4 x2 8 x3 12 x4 4

Solution

Starting with the augmented matrix, create zeros below the pivot

in the first column.

2

3 2 1

1

1 2 3 2 1

1 2 2 1 2 R 2 R1 0 0 1 3

1

2

4

8 12 4 R3 ( 2) R1 0 0 2 8 6

At this stage, we create a zero only below the pivot.

1 2 3 2 1 1 2 3 2 1

0 0 1 3

1 1 0 0 1 3

1

R3

R3 ( 2) R 2

2

0 0 0 2 4

0 0 0 1 2

Echelon form

We have arrived at the echelon form.

Ch1_32

## 33.

The corresponding system of equation isx1 2 x2 3x3 2 x4 1

x3 3x4 1

x4 2

We get

x3 3(2) 1

x3 5

Substituting x4 = 2 and x3 = 5 into the first equation,

x1 2 x2 3( 5) 2(2) 1

x1 2 x2 10

x1 2 x2 10

Let x2 = r. The system has many solutions. The solutions are

x1 2r 10, x2 r , x3 5, x4 2

Ch1_33

## 34.

Example 7Solving the following system of linear equations using the

method of Gaussian elimination, performing back substitution

using matrices.

x1 2 x2 3 x3 2 x4 1

x1 2 x2 2 x3 x4 2

2 x1 4 x2 8 x3 12 x4 4

Solution

We arrive at the echelon form as in the previous example.

2

3 2 1

1

1 2 3 2 1

1 2 2 1 2

0 0 1 3

1

2

0 0 0 1 2

4

8 12 4

Echelon form

This marks the end of the forward elimination of variables from

equations. We now commence the back substitution using

matrices.

Ch1_34

## 35.

1 2 3 2 10 0 1 3

1

0 0 0 1 2

1 2 3 0 5

R1 ( 2) R 3 0 0 1 0 5

R 2 ( 3) R3 0 0 0 1

2

1 2 0 0 10

R1 ( 3) R 2 0 0 1 0 5

0 0 0 1

2

This matrix is the reduced echelon form of the original

augmented matrix. The corresponding system of equations is

x1 2 x2 10

x3 5

x4 2

Let x2 = r. We get same solution as previously,

x1 2r 10, x2 r , x3 5, x4 2

Ch1_35