ENGR-1100 Introduction to Engineering Analysis

Lecture 5Notes courtesy of: Prof. Yoav Peles

Important Information!

• Exam No. 1 covers Lectures 1-5 in Syllabus !

Lecture outline(Linear Algebra today)

• Introduction to linear equations• Gauss-Jordan elimination method

Introduction to system of linear equations

a1x + a2y = b

• a1x1 + a2x2 +….+anxn = b• a1x + a2y + a3z =b

x

y

z

Which of the following are linear equations?

a) x + 3y = 7b) x1 - 3x2 + 5x3 = cos(10)

c) x1 + sin x2 = b

d) a1x12 + a2x2 +….+ anxn = b

Answer: a, b

System of linear equations• A finite set of linear equations in the

variables x1, x2,.. xn is called a system of linear equations. A sequence of numbers s1, s2,.. sn is called a solution of the system if x1=s1, x2=s2,.. xn=sn , is a solution of every equation in the system. For example:

4x1 - x2 + 3x3 = -1

3x1 + x2 + 9x3 = -4

has the solution x1 = 1, x2 = 2, x3 = -1.

No solution- inconsistent

• The following set of linear equations has no solution

x + y = 3x + y = 4

If there is at least one solution, it is called consistent.

Three possibilities

No solution 1 solution Infinite solutions

y

x

l1 l2y

x

l1and l2y

x

l1

l2

Augmented matrix

x1 +3 x2 + 4x3 = 8

2x1 + 5x2 - 8x3 = 1

3x1 + 7x2 - 9x3 = 0

1 3 4 8

2 5 –8 1

3 7 –9 0

Class assignment problem

Find a system of linear equations corresponding to the following augmented matrix, assuming the variables are x, y, and z

1 0 -1 2

2 1 1 3

0 -1 2 4

Class assignment solution

Find a system of linear equations corresponding to the following augmented matrix, assuming the variables are x, y, and z

4 2z y -

3 z y 2x

2 z -x

4

3

2

210

112

101

Basic method for solving a system of linear equations• Replace the given system by a new

system that has the same solution set, but is easier to solve:

1) Multiply a row through by a nonzero constant.

2) Interchange two rows.3) Add a multiple of one row to another.a11 a12 …a1n b1

a21 a22 …a2n b2

a31 a32 …a3n b3

: : : :

an1 an2 …ann bn

1 0 … 0 b’1

0 1 … 0 b’2

0 0 … 0 b’3

: : : :

0 0 … 1 b’n

Examplex+y+2z = 9

2x+4y-3z = 1

3x+6y-5z = 0

1 1 2 9

2 4 -3 1

3 6 -5 0

*-2*-2++

x+y+2z=9

2y-7z=-17

3x+6y-5z=0

1 1 2 9

0 2 -7 -17

3 6 -5 0

*-3*-3

+ +

x+y+2z=9

2y-7z=-17

3y-11z=-27

1 1 2 9

0 2 -7 -17

0 3 -11 -27

*1/2 *1/2

Example-continued

1 1 2 9

0 1 –7/2 –17/2

0 3 -11 -27

x+y+2z=9

y-7/2z=-17/2

3y-11z=-27+

*-3*-3+

x+y+2z=9

y-7/2z=-17/2

-1/2z=-3/2

1 1 2 9

0 1 –7/2 –17/2

0 0 –1/2 –3/2

x +11/2z=35/2

y-7/2z=-17/2

-1/2z=-3/2

1 0 11/2 35/2

0 1 –7/2 –17/2

0 0 –1/2 –3/2

*-1*-1++

*-2 *-2

Example-continuex+y+2z=9

y-7/2z=-17/2

z=3

x =1

y-7/2z=-17/2

z=3

1 0 0 1

0 1 –7/2 –17/2

0 0 1 3

*-11/2

+ +*-11/2

x = 1

y = 2

z = 3

1 0 0 1

0 1 0 2

0 0 1 3

1 0 11/2 35/2

0 1 –7/2 –17/2

0 0 1 3

* 7/2 * 7/2 ++

Gauss-Jordan elimination

• A systematic procedure for solving system of linear equations by transforming the augmented matrix to a reduced row-echelon form

1 0 0 2

0 1 0 3

0 0 1 1

x = 2

y = 3

z = 1

Reduced row-echelon form

Reduced row-echelon form1) If a row does not consist entirely of zeros,

then the first nonzero number in the row is a 1 (leading 1).

2) If there are any rows that consist entirely of zeros, then they are grouped together at the bottom of the matrix.

3) In any two successive rows that do not consist entirely of zeros, the leading 1 in the lower row occurs farther to the right than the leading 1 in the higher row.

4) Each column that contains a leading 1 has zeros everywhere else.

1 0 0 2

0 1 0 3

0 0 1 1

x = 2

y = 3

z= 1

Reduced row-echelon form

Solve the following reduced-echelon form matrices

1 0 0 5

0 1 0 -2

0 0 1 4

a) 1 0 0 4 -1

0 1 0 2 6

0 0 1 3 2

b)

c) 1 0 0 0

0 1 2 0

0 0 0 1

Gauss-Jordan elimination

• A systematic procedure for solving system of linear equations by transforming the augmented matrix to a reduced row-echelon form

Class assignment problemFor the following system of linear equations:

Write down the equations in the augmented formUse G-J elimination method to determine the RREFDetermine the solution for that system

Class assignment solution

1 z

1- y

2 x

1

1

2

100

010

001

R

2R

1

2

0

100

110

201

7/

7

2

0

700

110

201

8R

2R-

23

2

4

1580

110

421

3R-

2R-

11

10

4

323

952

421

3

3

2

2

1

1