1. systemsofequations handout

8/13/2019 1. SystemsofEquations Handout

1/44

A FIRST COURSE

IN LINEAR ALGEBRAAn Open Text by Ken Kuttler

Systems of Linear Equations

Lecture Notes by Karen Seyffarth

Adapted byLYRYX SERVICE COURSE SOLUTION

Attribution-NonCommercial-ShareAlike (CC BY-NC-SA)This license lets others remix, tweak, and build upon your work

non-commercially, as long as they credit you and license their new creationsunder the identical terms.

Systems of Linear Equations Page 1/44


2/44


Example

A system oflinear equations:

x1 2x2 7x3 = 1x1 + 3x2 + 6x3 = 0

variables: x1, x2, x3.

coefficients:1x1 2x2 7x3 = 1

1x1 + 3x2 + 6x3 = 0

constant terms:

x1 2x2 7x3 = 1x1 + 3x2 + 6x3 = 0

Systems of Linear Equations Systems of Linear Equations Page 2/44


3/44

A Graphical Solution

Example

Use a graph to find the solution to the following system of equations

x+y= 3y x= 5

y=x+5

y=-x+3

(-1,4)



4/44

An Algebraic Solution

Example

x1 =3, x2 =1, x3 = 0 is a solutionto the system

x1 2x2 7x3 = 1x1 + 3x2 + 6x3 = 0

as is x1 = 6, x2 = 0, x3 = 1.

However, x1 =1, x2 = 0, x3 = 0is nota solution to the system.

A solution to the system must be a solution to every equation in the

system.

The system above is consistent, meaning that the system has at leastone solution.



5/44

Example (continued)x1 + x2 + x3 = 0x1 + x2 + x3 = 8

is an example of an inconsistent system, meaning that it has no solutions.

Why are there no solutions?



6/44


Example

For any sa real number(written s R), then

x1 =3 + 9s, x2 =1 +s, x3 =s

is a solution to the system

x1 2x2 7x3 = 1x1 + 3x2 + 6x3 = 0

Verify this by substituting the expressions for x1, x2, and x3 into the twoequations.

s is called aparameter, and the solution is written in terms ofsx1 =3 + 9s, x2 =1 +s, x3 =s, where s R



7/44


Problem

Find all solutions to a system of m linear equations in n variables, i.e.,

solve a system of linear equations.

Definition

Two systems of linear equations areequivalentif they have exactly the

same solutions.

Example

The two systems of linear equations

2x + y = 23x = 3

and x + y = 1y = 0

areequivalent because both systems have the unique solution x= 1,y= 0.



8/44

Elementary Operations

We will solve systems of linear equations by usingElementary Operationsto transform the system into an equivalent but much simpler system fromwhich we can read the solution.

The Elementary Operations are of 3 kinds:Type I: Interchange two equations.

Type II: Multiply an equation by a nonzero number.

Type III: Add a (nonzero) multiple of one equation to a different

equation.

Systems of Linear Equations Elementary Operations Page 8/44


9/44

Theorem (Elementary Operations and Solutions)

Suppose you have a system of two linear equations

E1 =b1E2 =b2 (1)

Then the following systems have the same solution set as (1):

1

E2 =b2E1 =b1

2

E1 =b1

kE2 =kb2

for any scalar k, provided k= 0.

3

E1 =b1

E2+kE1 =b2+kb1=Systems of Linear Equations Elementary Operations Page 9/44


10/44

Solving a System with Elementary Operations

Example

Use Elementary Operations to solve the system

2x+y= 4x 3y= 1

Solution: Add (2) times the second row to the first row.

2x+y+ (2)x (2)(3)y= 4+ (2)1x 3y= 1

The result is an equivalent system

7y= 2x 3y= 1

Systems of Linear Equations Elementary Operations Page 10/44


11/44

Example (continued)

The first equation of the system,

7y= 2

tells us that

y=2

7

Now, use this value in the second equation:

x 3y=x 3

2

7

= 1

Simplifying,x= 1 +

6

7=

13

7

The solution is x= 137 , y= 27 .

The method used here is calledback substitution.Systems of Linear Equations Elementary Operations Page 11/44


12/44

The Augmented MatrixRepresent a system of linear equations with its augmented matrix.

Example

The system of linear equations

x1 2x2 7x3 = 1x1 + 3x2 + 6x3 = 0

is represented by theaugmented matrix 1 2 7 11 3 6 0

(Amatrix is a rectangular array of numbers.)

Note. Two othermatricesassociated with a system of linear equations are

thecoefficient matrixand theconstant matrix. 1 2 71 3 6

,

1

0

Systems of Linear Equations The Augmented Matrix Page 12/44


13/44


14/44

Elementary Row Operations

Type II: Multiply a row by a nonzero number.

Example

Multiply row 4 by 2.

2 1 0 5 3

2 0 3 3 10 5 6 1 01 4 2 2 2

2 1 0 5 32 0 3 3 1

0 5 6 1 02 8 4 4 4

Systems of Linear Equations Elementary Row Operations Page 14/44


15/44

Elementary Row Operations

Type III: Add a (nonzero) multiple of one row to a different row.

Example

Add 2 times row 4 to row 2.

2 1 0 5 3

2 0 3 3 10 5 6 1 01 4 2 2 2

2 1 0 5 30 8 7 7 30 5 6 1 01 4 2 2 2

Systems of Linear Equations Elementary Row Operations Page 15/44


16/44

Row-Echelon Matrix

All rows consisting entirely of zeros are at the bottom.

The first nonzero entry in each nonzero row is a 1(called the leading 1 for that row).

Each leading 1 is to the right of all leading 1s in rows above it.

Example

0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1

0 0 0 0 0 0 0 00 0 0 0 0 0 0 0

where can be any number.

Systems of Linear Equations Row-Echelon Matrix Page 16/44


17/44

Reduced Row-Echelon Matrix

Row-echelon matrix.

Each leading 1 is the only nonzero entry in its column.

Example

0 1 0 0 0

0 0 0 1 0 00 0 0 0 1 00 0 0 0 0 0 0 10 0 0 0 0 0 0 00 0 0 0 0 0 0 0

where can be any number.

We use elementary row operations to carry a matrix to either row-echelonor reduced row-echelon form.

Systems of Linear Equations Reduced Row-Echelon Matrix Page 17/44

S S f


18/44

Solving Systems of Linear Equations

Solving a system of linear equations means finding all solutions to thesystem.

Method I: Gauss-Jordan Elimination

1 Use elementary row operations to transform the augmented matrix toan equivalent (not equal) reduced row-echelon matrix. Theprocedure for doing this is called theGaussian Algorithm, also calledtheReduced Row-Echelon Form Algorithm.

2 If a row of the form [0 0 0|1] occurs, then there is no solution

3 Otherwise assign parameters to the nonleading variables (if any), andsolve for the leading variables in terms of the parameters.

Systems of Linear Equations Gauss-Jordan Elimination Page 18/44

G J d Eli i i


19/44

Gauss-Jordan EliminationExample

Solve the system2x + y + 3z = 12y z + x = 09z + x 4y = 2

Solution:

2 1 3 11 2 1 01 4 9 2

1 2 1 0

2 1 3 11 4 9 2

1 2 1 0

0 3 5 10 6 10 2

1 2 1 00 3 5 1

0 0 0 0

1 2 1 00 1 53

13

0 0 0 0

1 0 7323

0 1 53 13

0 0 0 0

Systems of Linear Equations Gauss-Jordan Elimination Page 19/44


20/44

S l i S t f Li E ti


21/44

Solving Systems of Linear Equations

Method II: Gaussian Elimination with Back-Substitution

1 Use elementary row operations to transform the augmented matrix toan equivalent row-echelon matrix.

2 The solutions (if they exist) can be determined usingback-substitution.

Systems of Linear Equations Gaussian Elimination Page 21/44

G i Eli i ti ith B k S b tit ti


22/44

Gaussian Elimination with Back Substitution

Example

Solve the system2x + y + 3z = 12y z + x = 09z + x 4y = 2

Solution: 2 1 3 11 2 1 0

1 4 9 2

1 2 1 02 1 3 1

1 4 9 2

1 2 1 00 3 5 1

0 6 10 2

1 2 1 00 3 5 1

0 0 0 0

1 2 1 00 1 53

13

0 0 0 0


Example (continued)


23/44

Example (continued)

This row-echelon matrix corresponds to the system

x + 2y z = 0

y 53 z = 13 , so

x = 2y+z

y = 13 + 53 z ,

and thusx = 2( 13 +

53 z) +z =

23

73 z

y = 13 + 53 z

Setting z=s, where s R, gives us (as before):

x = 23 73 s

y = 13 + 5

3 s

z = s


Example


24/44

Example

Solve the systemx + y + 2z = 1y + 2x + 3z = 0z 2y = 2

Solution:

1 1 2 12 1 3 0

0 2 1 2

1 1 2 10 1 1 2

0 2 1 2

1 1 2 10 1 1 2

0 2 1 2

1 0 1 10 1 1 20 0 3 2

1 0 1 10 1 1 20 0 1 23

1 0 0 530 1 0 430 0 1 2

3

The unique solution is x= 53 , y=43 , z=

23 .

Check your answer!


Example


25/44

Example

Solve the system

3x1 9x2 + x3 = 9

2x1 + 6x2 x3 = 6x1 + 3x2 x3 = 2

Solution:

1 3 1 22 6 1 63 9 1 9

1 3 1 20 0 1 20 0 2 3

1 3 0 40 0 1 20 0 0 1

The last row of the final matrix corresponds to the equation

0x1+ 0x2+ 0x3 = 1

which is impossible!Therefore, this system is inconsistent, i.e., it has no solutions.



26/44


27/44

Example (continued)

1 3 2 10 1 2 c2 3 a b

1 3 2 1

0 1 2 c0 3 a+ 4 b 2

1 3 2 10 1 2 c

0 3 a+ 4 b 2

1 0 4 1 + 3c0 1 2 c

0 0 a 2 b 2 3c

Case 1. a 2= 0, i.e., a= 2. In this case,

1 0 4 1 + 3c0 1 2 c

0 0 1 b23ca2

1 0 0 1 + 3c 4

b23c

a2

0 1 0 c+ 2 b23ca2 0 0 1 b23c

a2

Systems of Linear Equations General Patterns Page 27/44


28/44

Example (continued)

1 0 0 1 + 3c 4b23ca2

0 1 0 c+ 2

b23ca2

0 0 1 b23c

a2

(i) When a= 2, the unique solution is

x= 1 + 3c 4

b 2 3c

a 2

, y=c+ 2

b 2 3c

a 2

,

z=b 2 3c

a 2 .



29/44

Example (continued)

Case 2. Ifa= 2, then the augmented matrix becomes

1 0 4 1 + 3c0 1 2 c

0 0 a 2 b 2 3c

1 0 4 1 + 3c0 1 2 c

0 0 0 b 2 3c

From this we see that the system has no solutions when b 2 3c= 0.

(ii) When a= 2 and b 3c= 2, the system has no solutions.



30/44

Example (continued)

Finally when a= 2 and b 3c= 2, the augmented matrix becomes

1 0 4 1 + 3c0 1 2 c

0 0 0 b 2 3c

1 0 4 1 + 3c0 1 2 c

0 0 0 0

and the system has infinitely many solutions.(iii) When a= 2 and b 3c= 2, the system has infinitely many solutions,given by

x = 1 + 3c 4s

y = c + 2sz = s

where s R.


Uniqueness of the Reduced Row-Echelon Form


31/44

Uniqueness of the Reduced Row Echelon Form

Theorem

The two linear systems of equations corresponding to two equivalentaugmented matrices have exactly the same solutions.

Theorem

Every matrix A is equivalent to a unique matrix in reduced row-echelonform.

Systems of Linear Equations Uniqueness Page 31/44

Homogeneous Systems of Equations


32/44

Homogeneous Systems of Equations

DefinitionA system of equations is called homogeneous if each equation in thesystem is equal to 0. A homogeneous system has the form

a11x1+a12x2+ +a1nxn = 0a21x1+a22x2+ +a2nxn = 0

...am1x1+am2x2+ +amnxn = 0

where aijare scalars and xi are variables.

Systems of Linear Equations Homogeneous Systems Page 32/44

The Trivial Solution


33/44

Notice that x1 = 0, x2 = 0, , xn = 0 is always a solution to ahomogeneous system of equations. We call this thetrivial solution.

We are interested in finding thenon trivial solutionsto a homogeneoussystem, if they exist.


Homogeneous Equations


34/44

g q

Example

Solve the system

x1 + x2 x3 + 3x4 = 0

x1 + 4x2 + 5x3 2x4 = 0x1 + 6x2 + 3x3 + 4x4 = 0

1 1 1 3 0

1 4 5 2 0

1 6 3 4 0

1 0 95

145 0

0 1 4515 0

0 0 0 0 0

The system has infinitely many solutions, and the general solution is

x1 = 9

5 s 14

5t

x2 = 45 s 15 t

x3 = sx4 = t

or

x1x2x3x4

=

95 s

145t

45 s 15 t

st

, where s, t R.


Definition


35/44

IfX1, X2, . . . , Xpare columns with the same number of entries, and ifa1, a2, . . . ar R (are scalars) then a1X1+a2X2+ +apXp is a linearcombinationof columns X1, X2, . . . , Xp.

Example (continued)

In the previous example,

x1x2x3x4

=

9

5

s 14

5

t

45 s 15 t

st

=

9

5 s 45 s

s0

+

14

5t 15 t

0t

= s

95

4510

+t

145 15

01


Example (continued)


36/44

This gives us

x1x2x3x4

=s95

4510

+t

145

1501

=sX1+tX2,

where X1 =

9

5

4510

and X2 =

14

5

1501

.

The columns X1 and X2 are calledbasic solutionsto the originalhomogeneous system.

The general solution to a homogeneous system can be expressed as alinear combinationofbasic solutions.



37/44

Example (continued)

Notice that

x1x2x3x4

=s

95

4510

+t

145 15

01

= s

5

94

50

+ t5

141

05

= r

94

50

+q

141

05

= r(5X1) +q(5X2)

where r, q R.



38/44

Example (continued)

The columns 5X1 =

94

50

and 5X2 =

141

05

are also basic solutions

to the original homogeneous system.

In general, any nonzero multiple of a basic solution (to a homogeneoussystem of linear equations) is also a basic solution.



39/44

Example

Find all values ofa for which the system

x + y = 0ay + z = 0

x + y + az = 0

has nontrivial solutions, and determine the solutions.Solution: Non trivial solutions occur when a= 0. Therefore the solutions

when a= 0 are given by

xy

z

=s

1

1

0

, s R.


Rank


40/44

Definition

Therankof a matrix A, denoted rank A, is the number of leading 1s inany row-echelon matrix obtained from A by performing elementary rowoperations.

Example

Find the rank ofA= a b 5

1 2 1

.

Solution:

a b 51 2 1

1 2 1a b 5

1 2 10 b+ 2a 5 a

Ifb+ 2a= 0 and 5 a= 0, i.e., a= 5 and b=10, then rank A= 1.Otherwise, rank A= 2.

Systems of Linear Equations Rank Page 40/44

What does the rank of an augmented matrix tell us?


41/44

SupposeA is the augmented matrix of a consistent system ofm linearequations in n variables, and rank A= r.

m

1 0 0 1 0 0 0 1 0 0 0 0 0

0 0 0 0 0

n

r leading 1s

Then the set of solutions to the system has n rparameters, so

ifr


42/44

Thus for any system of linear equations, exactly one of the followingholds:

1 the system is inconsistent;

2 the system has a unique solution, i.e., exactly one solution;

3 the system has infinitely many solutions.



43/44

( )


44/44

Example (continued)

From the reduced row-echelon matrix

1 2 0 0 13 90 0 1 0 0 20 0 0 1 4 20 0 0 0 0 0

x1 = 9 + 2r+ 13sx2 = rx3 = 2x4 = 2 4s

x5 = s

r, s R

The solution has two parameters (r and s) as we expected.


1. systemsofequations handout

Documents