Download 11.4: The Earliest Applications of Linear Algebra

11.4: The Earliest Applications of
Linear Algebra
By: Sid & Zoheb
History of Linear Algebra
• Emerged from the study of determinants,
Leibniz
• Gabriel Cramer: Cramer’s Rule
• Gauss: Gaussian elimination
• All were used to solve linear systems
Ancient Civilizations & Linear Systems
• “Linear systems can be found in the earliest
writings of many ancient civilizations”
• Practical problems of early civilizations
included the measurement of land,
distribution of goods, tracks of resources, and
taxation
• Predate the Islamic mathematicians who
created the field of algebra, which eventually
led to the branch of linear algebra
EGYPT
Egyptian Math
• 1650 B.C
• Ahmes (Rhind) Papyrus
-5 meters long
-84 “short” mathematical problems + solutions
-Hieratic
-geometry, notation, mensuration, algebra,
arithmetic
Problem 40
• Divide 100 hekats (Ancient Egyptian volume
unit, equivalent to 4.8 liters today) of barley
among 5 men in arithmetic progression so
that the sum of the smallest is one-seventh
the sum of the three largest.
Solution to Problem 40(Egyptian Style)
• Step 1: Define Variables. a=least amount that
any man obtains, d=common difference of the
terms
• Step 2: set-up equations
- a+(a+d)+(a+2d)+(a+3d)+(a+4d)=100
- (1/7)((a+2d)+(a+3d)+(a+4d))=a+(a+d)
- 5a+10d=100
- 11a-2d=0
Methods to solve for a and d
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Graph
Substitution
Elimination
Gauss-Jordan
Multiply inverse of A by the answer matrix
Cramer’s Rule
False position/assumption(The Egyptian Way)
False Position/Assumption Method
• 5a + 10d=100
11a-2d=0
Step 1: Assume a convenient value of a:(a=1)
Step 2: substitute the value of a to second
equation. (d=11/2)
Step 3: insert a and d values to first
equation(you get 60 which is not equal to 100)
Step 4: 100/60 *1(assumed a)=5/3=a(real a)
False Position/Assumption Method
cont.
• Step 5: substitute a=5/3 into second equation.
d=55/6
• Answer: the quantities of barley by the five
men are : 10/6, 65/6, 120/6,175/6,230/6
Greece
Greek Linear Systems
• 3rd century B.C
• Most famous problem: Archimedes’
celebrated Cattle problem
- Challenge for Eratosthenes by Archimedes.
- No solution from ancient times, not known
how the greeks solved it
Cattle Problem
If thou art diligent and wise, O stranger, compute the number of cattle of the Sun, who
once upon a time grazed on the fields of the Thrinacian isle of Sicily, divided into four
herds of different colours, one milk white, another a glossy black, a third yellow and
the last dappled. In each herd were bulls, mighty in number according to these
proportions: Understand, stranger, that the white bulls were equal to a half and a third
of the black together with the whole of the yellow, while the black were equal to the
fourth part of the dappled and a fifth, together with, once more, the whole of the
yellow. Observe further that the remaining bulls, the dappled, were equal to a sixth
part of the white and a seventh, together with all of the yellow. These were the
proportions of the cows: The white were precisely equal to the third part and a fourth
of the whole herd of the black; while the black were equal to the fourth part once
more of the dappled and with it a fifth part, when all, including the bulls, went to
pasture together. Now the dappled in four parts were equal in number to a fifth part
and a sixth of the yellow herd. Finally the yellow were in number equal to a sixth part
and a seventh of the white herd. If thou canst accurately tell, O stranger, the number
of cattle of the Sun, giving separately the number of well-fed bulls and again the
number of females according to each colour, thou wouldst not be called unskilled or
ignorant of numbers, but not yet shalt thou be numbered among the wise.
Variables for Cattle Problem
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W=number of white bulls
B=number of black bulls
Y=number of yellow bulls
D=number of dappled bulls
w= number of white cows
b=number of black cows
y=number of yellow cows
d=number of dappled cows
Equations
• “The white bulls were equal to a half and a
third of the black bulls together with the
whole of the yellow bulls
W=(.5+1/3)B+Y
“The black bulls were equal to the fourth part of
the dappled bulls and a fifth, together with,
once more, the whole of the yellow bulls
B(.25+.2)D+Y
Equations Cont.
• The remaining bulls, the dappled, were equal
to a sixth part of the white bulls and a
seventh, together with all of the yellow bulls.
• D=(1/6+1/7)W+Y
• The white cows were precisely equal to the
third part and a fourth of the whole herd of
the black
• w=(1/3+1/4)(B+b)
Equations Cont.
• The black cows were equal to the fourth parth
once more of the dappled and with it a 5th
part when all including the bulls, went to
pasture together
• b=(1/4+1/5)(D+d)
• The dappled cows in four parts were equal in
number to a fifth part and a sixth of the
yellow herd
• d=(1/5+1/6)(Y+y)
Equations Cont.
• The yellow cows were in number equal to a
sixth part and a seventh of the white herd
• Y=(1/6 + 1/7)(W+w)
After some math
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W=10,366,482k
B=7,460,514k
Y=4,149,387k
D=7,358,060k
w=7,206,360k
b=4,893,246k
y=5439213k
d=3,515,820k
India
Indian Math
• Fourth century A.D
• Bakhsali Manuscript
-70 leaves or sheets of birch bark
-mathematical problems and solutions
-equalization problems that leads to systems of
linear equations
Indian Problem
• Once merchant has seven asava horses, a
second has 9 haya horses, and a third has then
camels. They are equally well off in the value
of their animals if each gives two animals, one
to each of the others. Find the price of each
animals possessed by each merchant
Variables & Equations
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x=price of an asava horse
y=price of a haya horse
z= price of a camel
K= total value of animals
5x +y +z=K
x + 7y +z=K
x + y +8z= K
Solution Method
subtract (x+y+z) from both sides of each
equation
4x=6y=7z=K-(x+y+z)
K-(x+y+z) must be an integer that is divisible by
4,6,7. so it equals 4*6*7=168
Plugging it into the equation above. x=42, y=28,
z=24
China
Chinese Problems
• There are three classes of corn, of which three
bundles of the first class, two of the second,
and one of the third make 39. Two of the first,
three of the second and one of the third make
34. And one of the first, two of the second,
and three of the third make 26. How many
measures of corn are contained in one bundle
of each class?
Game #1
• A passenger jet took three hours to fly 1800
miles in the direction of the jetstream. The
return trip against the jetstream took four
hours. What was the jet's speed in still air
and the jetstream's speed?
Problem 63 of the Ahmes
• 700 loaves are to be divided among recipients
in the following proportions:
2/3; ½;1/3;1/4
How much does each person receive?
Problem 24
• Find the heep if the heap and a seventh of the
heep is 19. Use False position
Berlin Papyrus #1
• You are told the area of a square of 100
square cubits is equal to that of two smaller
squares, the side of one square is 1/2 + 1/4 of
the other. What are the sides of the two
unknown squares.