ENGR-1100 Introduction to Engineering Analysis Lecture 5

Introduction to System of Linear Equations a1.x+a2.y = b z x a1x+a2y+a3z = b a1x1+a2x2+….+anxn = b

Which of the following are linear equations? x+3y=7 x1-3x2+5x3=cos(10) x1+sin x2=b 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 have no solution x+y=3 x+y=4 If there is at least one solution, it is called consistent.

Infinitely many solutions Three Possibilities y x l1 l2 y x l1and l2 y x l1 l2 One solution No solution Infinitely many solutions

Class Assignment Problem- set 1.1-5 For which value(s) of the constant k does the following system of linear equations have no solutions? Exactly one solution? Infinitely many solutions? 2x-2y=6 2x-2y=k *2 x-y=3 2x-2y=k k=6 infinitely number of solutions k=6 no solutions

Class Assignment Problem- set 1.1-5 For which value(s) of the constant k does the following system of linear equations have no solutions? Exactly one solution? Infinitely many solutions? y x x-y = 3 x+y = k For any k only one solution exists 2x=3+k => x=(3+k)/2 y=x-3=(k-3)/2

Augmented Matrix x1+3x2+4x3=8 2x1+5x2-8x3=1 3x1+7x2-9x3=0 1 3 4 8 x1 x2 x3 RHS 1 3 4 8 2 5 –8 1 3 7 –9 0

Class Assignment Problem- set 1.2-4a Find a system of linear equations corresponding to the following augmented matrix. 1 0 -1 2 2 1 1 3 0 -1 2 4

A 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: Multiple a row through by a nonzero constant, Interchange two rows, Add a multiple of one row to anther. a11 a12 …a1n b1 a21 a22 …a2n b2 a31 a32 …a3n b3 : : : : am1 am2 …amn bm a’11 a’12 …a’1n b’1 a’21 a’22 …a’2n b’2 a’31 a’32 …a’3n b’3 : : : : a’m1 a’m2 …a’mn b’m

Example + + x+y+2z=9 2x+4y-3z=1 3x+6y-5z=0 *-2 1 1 2 9 2 4 -3 1 1 1 2 9 2 4 -3 1 3 6 -5 0 x+y+2z=9 2y-7z=-17 3x+6y-5z=0 1 1 2 9 0 2 -7 -17 3 6 -5 0 *-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

Example - Continue + + x+y+2z=9 y-7/2z=-17/2 3y-11z=-27 1 1 2 9 1 1 2 9 0 1 –7/2 –17/2 0 3 -11 -27 + *-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 *-2 x+y+2z=9 y-7/2z=-17/2 z=3 1 1 2 9 0 1 –7/2 –17/2 0 0 1 3 *-1 +

Example - Continue + x+y+2z=9 1 1 2 9 0 1 –7/2 –17/2 y-7/2z=-17/2 z=3 1 1 2 9 0 1 –7/2 –17/2 0 0 1 3 *-1 + x +11/2z=35/2 y-7/2z=-17/2 z=3 1 0 11/2 35/2 0 1 –7/2 –17/2 0 0 1 3 *-11/2, 7/2 x =1 y =2 z =3 1 0 0 1 0 1 0 2 0 0 1 3

Gauss-Jordan Elimination A systematic procedure for solving a system of linear equations by transforming the augmented matrix to a reduced row-echelon form x =2 y =3 z=1 1 0 0 2 0 1 0 3 0 0 1 1 reduced row-echelon form

Reduced Row - Echelon Form If a row does not consist entirely of zeros, then the first nonzero number in the row is a 1 (leading 1). If there are any rows that consist entirely of zeros, then they are grouped together at the bottom of the matrix. 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. Each column that contains a leading 1 has zeros everywhere else. x =2 y =3 z=1 1 0 0 2 0 1 0 3 0 0 1 1 Reduced row-echelon form

Solve the following reduced-echelon form matrixes 1 0 0 5 0 1 0 -2 0 0 1 4 1 0 0 4 -1 0 1 0 2 6 0 0 1 3 2 a) b) 1 0 0 0 0 1 2 0 0 0 0 1 c)

Class Assignment Problem- set 1.2-7 Solve the following linear equation system by Gauss-Jordan elimination x1 + x2+ 2x3=8 -x1 - 2x2+3x3=1 3x1 - 7x2+4x3=10 Answer: x1=3, x2=1, x3=2