Download MATH 1046 Introduction to Linear Algebra

MATH 1046
Introduction to Systems of Linear Equations
(Sections 2.1 and 2.2)
Alex Karassev
Example: Error in scale
• Alex is using a scale that is known to have a
constant error. A can of soup and a can of tuna
are placed on this scale, and it reads 720
grams. Four identical cans of soup and three
identical cans of tuna are placed on an
accurate scale, and a weight of 2.4 kg is
recorded. If two cans of tuna weigh 540 grams
on this bad scale, what is the amount of error
in the scale and what is the correct weight of
each type of can?
Mathematical Model
•
•
•
•
4 cans of soup and 3 cans of tuna on good scale equals 2.4 kg
A can of soup and a can of tuna on bad scale equals 720 g
2 cans of tuna on bad scale equals 540 g
Let x, y, and z be the weights of a can of soup, a can of tuna,
and the error of the scale in grams, respectively. Then we
obtain the following system of linear equations:
 x  y  z  720

 2400
4 x  3 y

2 y  z  540

Example: Logging
• A logging company has a contract to provide 1000 m³ of pine,
800 m³ of spruce, and 600 m³ of fir logs per month. There are
three regions available for logging. The following table gives
the species mix, and timber density for each region.
Region
Volume/hectare % Pine % Spruce
% Fir
West
330 m³
70 %
20 %
10 %
North
390 m³
10 %
60 %
30 %
East
290 m³
5%
20 %
75 %
• How many hectares should one log in each operating region
listed above to deliver exactly the required volume of logs?
Mathematical Model
Region
West
North
East
Volume/hectare
330 m³
390 m³
290 m³
% Pine
70 %
10 %
5%
% Spruce
20 %
60 %
20 %
% Fir
10 %
30 %
75 %
• We need: 1000 m³ pine, 800 m³ spruce, and 600 m³ fir
• Let w, n, e be the number of hectares logged in West,
North, and East regions respectively. Then:
330(.7) w  390(.1)n  290(.05)e  1000

330(.2) w  390(.6)n  290(.2)e  800
330(.1) w  390(.3)n  290(.75)e  600

Modern applications of linear systems
• Linear systems with hundreds (or more!)
unknowns:
 Applications in economics (linear optimization
(programming), transportation problem, linear
models)
 Computer graphics
 Optimization
 Traffic flow
How to solve linear systems?
• Straightforward approach: express one of the
variables form one of the equations and
substitute in the rest of equations
• Continue until you get to one equation with
one variable
Error in scale: solution
• From last equation: z = 540 – 2y
 x  y  z  720
• Substitute in the first, we get

x+ y + (540 – 2y) = 720
 2400
4 x  3 y
• So, we get two equations:

2 y  z  540

x – y = 180
4x + 3y = 2400
• Now from the first equation x = 180 + y and therefore
4(180 + y) + 3y = 2400
• Thus, 7 y = 1680, so y = 240
• Now x = 180 + 240 = 420 and z = 540 – 2y = 60
Problem with straightforward
approach
• It is not efficient (e.g. computationally) when
the number of equations is large
What can we do with systems of equations?
• Example
• Divide all by 10
• Divide 2nd by 2
330(.7) w  390(.1)n  290(.05)e  1000

330(.2) w  390(.6)n  290(.2)e  800
330(.1) w  390(.3)n  290(.75)e  600

33(.7) w  39(.1)n  29(.05)e  100

33(.2) w  39(.6)n  29(.2)e  80
33(.1) w  39(.3)n  29(.75)e  60

33(.7) w  39(.1)n  29(.05)e  100

33(.1) w  39(.3)n  29(.1)e  40
33(.1) w  39(.3)n  29(.75)e  60

What can we do with systems of equations?
• Example
• Subtract
2nd from 3rd
• Change the order
of 1st and 2nd
33(.7) w  39(.1)n  29(.05)e  100

33(.1) w  39(.3)n  29(.1)e  40
33(.1) w  39(.3)n  29(.75)e  60

33(.7) w  39(.1)n  29(.05)e  100

33(.1) w  39(.3)n  29(.1)e  40

29(.65)e  20

33(.1) w  39(.3)n  29(.1)e  40

33(.7) w  39(.1)n  29(.05)e  100

29(.65)e  20

What can we do with systems of equations?
• Example
33(.1) w  39(.3)n  29(.1)e  40

33(.7) w  39(.1)n  29(.05)e  100

29(.65)e  20

• Multiply 1st by 7
and subtract from
2nd
• Add 3rd to 2nd
33(.1) w  39(.3)n  29(.1)e  40

- 39(2)n  29(.65)e  180


29(.65)e  20

33(.1) w  39(.3)n

- 39(2)n



 29(.1)e  40
 160
29(.65)e  20
Matrix form
• Coefficient matrix
 x  y  z  720

 2400
4 x  3 y

2 y  z  540

• Augmented matrix
 1 1 1 720 


 4 3 0 2400 
 0 2 1 540 


1 1 1


 4 3 0
0 2 1


Matrix form
• General 3x3 system
a11 x1  a12 x 2  a13 x 3 b1

a21 x1  a22 x 2  a23 x 3 b 2
a x  a x  a x b
 31 1 32 2 33 3 3
 a11

 a21
a
 31
a12
a22
a32
a13 b1 

a23 b 2 

a33 b 3 
So, we can:
• Change the order of equations
• Multiply an equation by a nonzero number
• Add a multiple of one equation to another
equation
Note: all this operations are invertible, i.e. the
system can be returned to its original form by
applying the same types of operations
Elementary row operations on
augmented matrix
• Switch two rows
• Multiply a row by a nonzero number
• Add a multiple of one row to another row
Note: all this operations are invertible, i.e. the
matrix can be returned to its original form by
applying the same types of operations
Notations
• Switch 1st and 2nd rows: R1 R2
• Multiply 1st row by a nonzero number: aR1
• Add a multiple of 1st row to the 2nd row
What are their inverses?
“Direct” and “Inverse” operations
• Direct
 Switch 1st and 2nd rows: R1 R2
 Multiply 1st row by a nonzero number: aR1
 Add a multiple of 1st row to the 2nd row: R2 → R2 + bR1
• Inverse
 Switch 1st and 2nd rows: R1 R2
 Multiply 1st row by a nonzero number: (1/a)R1
 Add a multiple of 1st row to the second row: R2 → R2 + (-b)R1
Goal of the operations
• If possible, reduce the augmented matrix to
the form:
 1 0 0 r1 


0 1 0 r2 
0 0 1 r 
3 

• Then x1=r1, x2=r2, x3=r3 is the (unique) solution
Linear systems and geometry
• Systems of linear equations can be used to
find intersections of planes and lines
• Exercise
Find the point at which the following planes
intersect
x+y+z=3, x+2y-z=3, 2x+3y+z=1