1.1 Mat205 - one day class PDF

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MAT-205 Linear Algebra Course Instructor: Dr. A.S.A.Noor Chapter I

Linear Algebra Book: Elements of linear algebra by Howard Anton & Chris Rorres (9th Edition)

Chapter 1 Section: 1.1 Linear equation: An equation where the power of each variable (unknown) is equal to 1 is called a linear equation. Example1: Example 2: Example 3: Example4:

is non-linear. is non-linear.

Any linear equation with two variables represents a straight line. The term linear comes from the word line. A solution of the linear equation is a sequence of n numbers S1, S2,…., Sn such that the equation is satisfied when we substitute

System of linear equations: A finite number of linear equations in the variables system of linear equations. Ex 5: . . . . . . . .. . .. . .. . . . . .. . .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . .. .

is called a

Example 6:

Solution of the system: Numbers S1, S2,…., Sn is called a solution of the system in Ex5, if satisfy all the equations of the system. Some example: 1.

Solving, we get

So

is a solution.

Here the system has exactly one solution. In fact, two straight lines intersect exactly at one point.

2. =) 𝑥 𝑦

)

𝑥

𝑦

−6

1

Here

1

1 are coincident lines.

and

So, in this system, though we have two equations, actually there is only one equation. So every point of the straight line is a solution. So it has infinitely many solutions.

3.

1

} =)

6

Here two lines are parallel. So it has no solution. Because one point cannot satisfy (sit on) both the parallel lines.

Thus we have three cases: 1. The system has exactly one solution. 2. The system has infinitely many solutions. 3. The system has no solution. Consistent system: If a system has a solution, then it is called a consistent system. If it does not have a solution, then it is called Inconsistent. Thus we have the following diagram.

System of linear equations

Consistent

Exactly one Solution

In consistent

Infinitely many Solutions

No solution

Consider the system:

Here the coefficient matrix So we have, [

][

[ ] [

]

]

This is known as the matrix form of the system of linear equations. Multiplying by −

)

−

)

−

−

−

on both sides, we have

−

−

So the procedure is very simple. First form the coefficient matrix A then find the inverse matrix − . Then the solution is obtained by multiplying − with the column matrix ḇ. Definition of matrix and it’s inverse will be discussed in section: 1.2 Elementary Row operations in matrices: (i)

Multiplying a row by a non-zero constant. [

Example: Let [

1

1 ]

1 ]

Multiplying 1st row by 3 and placing it on new 1st row. (ii)

Interchange of two rows.

Example: Let )[

1

1

[ 1

]

1

]

Interchanging 1st and 3rd rows. (iii)

Adding a row with a multiple of another row and placing it on old row. i.e, multiply ith row by k and add to j th row and place it on new jth row. ]

[

)[

11 ]

Multiplied 1st row by 2 and added to 2nd row and place it on new 2nd row. Row-echelon form of a matrix: A matrix is said to be in row-echelon form if it satisfies the following conditions. 1) If a row is not entirely zero, then its first non-zero entry must be equal to 1 (called leading 1). 2) The rows which are entirely zero must go to the bottom of the matrix. 3) The leading 1 of the lower row will go further to the right of the leading 1 of the higher row. Example: 1

[

1

1

1 1

]

A is in Row-echelon form [

1

1

1

] is of course in Row-echelon form.

[

[

[

1

11

] is not, because it is contradicting condition 3.

1

1

1

] is not, because it is contradicting condition 2.

1 ] is not, because it is contradicting condition 1. 1

4th Condition:

The row-echelon form is called a reduced row-echelon form (or canonical form) if it satisfies the condition 4) All the entries of the columns on which a leading 1 stands must be zero. In above examples, only I is in reduced row-echelon form. Though A is in echelon form, still it is not in reduced form. 1

[

1

1

1 1

]

is in reduced row-echelon form.

Augmented matrix: Consider the system of linear equations

The following matrix obtained from above system [

] is called the Augmented matrix.

Actually, this is a partition matrix [ ] [

]

] [ In solving the system of linear equations by Gaussian Elimination method or Gauss-Jordan Eliminations method we use the Augmented matrix and then the row operations and echelon forms. In Gauss-Jordan elimination form reduced row echelon form is must.

where,

In solving the system of linear equations, we form the augmented matrix and then use row-echelon form and reduced row-echelon form to solve the system. In this chapter, we will use two methods: 1. Gaussian Elimination method. And 2. Gauss-Jordan Elimination method. For Gaussian Elimination method we use Row-echelon form and for Gauss-Jordan Elimination method we use Reduced Row-echelon form. If in Row-echelon form, number of unknowns = n and number of equations = m with Then there are infinitely many solutions, and number of variables are free variables. We are free to choose any value for those free variables. If in echelon form, m = n (i.e, number of equations = number of variables in echelon form), then the system has a unique solution. If in the last equation in Row-echelon form, we have it is inconsistent.

, then the system has no solution, i.e,

Ex : Solve by Gaussian Elimination method (Back substitution method) 1

6

1

1

1

6

6

Solution: Here Augmented matrix for the system is [

1

=) [

)

1

1 1

1

1) => [

=) [

=) [

)

1

1

1

1

) 1

1]

1 1 1 1

)

1] 1

1]

1]

1

1 1

1]

1

1

=) [ 6 Thus we have,

1 1 1

6

1

] ………….. (A)

6

Here in Row-echelon form, number of variables (unknown) = 6 Number of equations = 3 Here Suppose Then

are free variables. 1 1

6

are the solutions where r, s, t are arbitrary real values. Note. If we like to solve the above problem by Gauss-Jordan elimination method, then we use more operations from (A).

)

=) [

=) [

1

1

1

1 ] 1 1

1 )

1 1

]

So we have, 6

Number of unknown = 6, Free variables are Let Then

Number of equations = 3 .

Hence the solution is 6

r, s, t are arbitrary real values. Ex2: Solve the system by Gauss-Jordan Elimination method: 1 Solution: Here Augmented matrix is [

1

1

1]

)

=) [

1

1

11

=) [

=)

[

1

1

1 )

) 1

1] 1

11

]

]

)

[

[

(

1

1

1

)

1 1

1]

]

1 This is the reduced row-echelon form.

Hence,

1

is the (unique) solution. (Ans).

A system of linear equations is said to be homogeneous if the constant terms on right are all equal to zero. That is the system,

---------------------------------------------------------------------------------------------

Every Homogeneous system of linear equations is always consistent. Here, is always a solution which is known as the trivial solution. If there are other solutions, then they are known as non-trivial solutions. Consider the straight lines

Here both lines pass through origin. So the trivial solution point of intersection of the lines.

is the

[graph] Ex: Use Gauss-Jordan elimination to solve the homogeneous linear system: 1

1

6

1

6 6

Solution: Here Augmented matrix is, [

[

1

1

)

1 1

1)

[

1

[

1

1

1

)

]

1 1

)

1 1

]

] 1 )

1

1 1

]

[

[

1

)

1

1 [

6

1

1

)

1

]

1

1 ]

]

This is the reduced row-echelon form.

Here,

6

are free variables.

Let, Then

where

are any real numbers.

6

is a solution

Note: The position of leading 1 in each row is also known as Pivot positions of the matrix. A column that contains a Pivot position is called a Pivot column of the matrix.

Practice problem: Exercise 1.2 (Book; Page:20-21) Solution 6(d). Solve by Gaussian elimination method: 1

Solve: Here Augmented matrix is, [

1

[ [

6

)

1

1

1

[ [

(

1

[

1

1 1

)

1

] 1

] 6

1

6

1 1

1

] 6

1

] 6

]

]

[

1 1

6

1

]

1

From last row we 1 have,

Which means that, the system has no solution. i.e., is inconsistent.

14(b). Solve by any method

Solution: Here Augmented matrix is, 1

[

1 1

[

)

]

1

[

1

1

1

1

]

1

]

[

1 )

[

1

1

]

1

1

]

Here number of equations = 2 number of variables = 4 number of free variables = 4 – 2 = 2 w, x are free variables.

) {

}

17. Determine the values of a for which the system has (i) no solution (ii) exactly one solution (iii) infinitely many solutions. 1)

Solution: Here Augmented matrix is,

[

1

)

(

[

1

1 1 1 ) [

1

)

[

]

1 1] 1

1

1

[

1

1 1

1

]

1 )

)

(i) The system has no solution if

)

This is the row-echelon form.

]

] )

(ii) The system has exactly one solution if

)

, but )

(iii) The system has infinitely many solutions if

)

)

)

)

)

21. Show that, the following non-linear system has 18 solutions if

Solve: Here Augmented matrix is, [

[

1

)

)

1

1

[

1

]

1 ;

1) )

]

[ 1

1

1

[

1

1 1

]

]

This is the reduced row-echelon form.

1

1

]

So, we have

(

) (

) (

)

(

) (

) (

)

(

(

(

(

) ( ) (

) ( ) (

) (

) (

) (

1

) (

) )

)

)

Exercise...