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Precalculus4E1
Gaussian Elimination
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Resolvability of a system of linear equationsResolvability of a system of linear equations
a1 x b1 y = c1
a2 x b2 y = c2
Case A (regular):
● the system is inconsistent, there is no solution
● the equations are not independent, there is an infinite set of solutions
Case B (singular):
These characteristics of the resolvability of a system of 2 linear equationsstay when considering systems of linear equations with an arbitrary numberof equations and unknowns.
There is exactly one solution
There is no uniquely defined solution
Precalculus4E2
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Determine analytically and graphically the solutions of thesystems of linear equations given below
−0.5 x y = 1
2 x − y = 2
−0.5 x y = 1
−0.5 x y = 0
x y = 2
2 x 2 y = 4
Resolvability of a system of linear equations: Resolvability of a system of linear equations: Exercise 1Exercise 1
system 1:
system 2:
system 3:
Precalculus4A
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Fig. 31: Linear functions f (x) = 1 + 0.5x and g (x) = 2 x 2
−0.5 x y = 1
2 x − y = 2The system 1 has one solution x = 2, y = 2 :
Resolvability of a system of linear equations: Resolvability of a system of linear equations: Solution 1Solution 1
Precalculus41
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−0.5 x y = 1
−0.5 x y = 0
Fig. 32: Linear functions f (x) = 1 + 0.5 x and g (x) = 0.5 x
The straight lines y = 1 + 0.5 x and y = 0.5 x are parallel. Therefore theyhave no common point. This system of equations has no solution.
system 2:
Resolvability of a system of linear equations: Resolvability of a system of linear equations: Solution 1Solution 1
Precalculus42
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Fig. 33: linear functions y = 2 x and 2 y = 4 2 x
x y = 2
2 x 2 y = 4
The two equations are equivalent, describing the same function.The number of solutions is infinite.
system 3:
Resolvability of a system of linear equations: Resolvability of a system of linear equations: Solution 1Solution 1
Precalculus43
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Gaussian Elimination
Below we demonstrate the algorithm Gaussian Elimination by solvinga system of three linear equations with three unknowns.
Carl Friedrich Gauß (17771855), brilliant German mathematician
Precalculus51a
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Precalculus51b
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Gaussian Elimination:Gaussian Elimination: ExampleExample
We call the first equation elimination row. It stays unchangedin the subsequent transformations. Below it is multiplied by afactor.
E1 : − x y z = 0
E2 : x − 3 y − 2 z = 5
E3 : 5 x y 4 z = 3
Step 1: elimination of x
E1 E2 = E1 : −2 y − z = 5
5 E1 E3 = E 2 : 6 y 9 z = 3⇔
−2 y − z = 5
2 y 3 z = 1
Step 2: elimination of y
E1 13
E2 = E* : z = 3
system:
Precalculus52
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Gaussian EliminationGaussian Elimination
−x y z = 0
−2 y − z = 5
z = 3
Triangular system of equations:
The equation without x and the equation without x and y form together with the first equation a triangular system ofequations from which the three unknowns can be calculatedone by one from bottom to top.
Unique solution: x = −1, y = −4, z = 3
or as triple of numbers: −1, − 4, 3
or as column vector : v = xyz =
−1−4 3
Precalculus53
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Gaussian Elimination: Gaussian Elimination: Exercises 25Exercises 25
Solve the systems of equations given below:
2 x − y z = 8
x 2 y 2 z = 6
4 x − 2 y − 3 z = 1
Exercise 2:
x − z = 2
2 x − y − 3 z = −9
−3 x y 5 z = 4
Exercise 3:
2 x − y − z = 4
3 x 4 y − 2 z = 11
3 x − 2 y 4 z = 11
Exercise 4:
x y 2 z = −1
2 x − y 2 z = −4
4 x y 4 z = −2
Exercise 5:
Precalculus61
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x1 x2 2 x3 3 x4 = 1
Exercise 6:3 x1 − x2 − x3 − 2 x4 = − 4
2 x1 3 x2 − x3 − x4 = −6
x1 2 x2 3 x3 − x4 = − 4
x1 2 x2 3 x3 − 2 x4 = 6
Exercise 7:2 x1 − x2 − 2 x3 − 3 x4 = 8
3 x1 2 x2 − x3 2 x4 = 4
2 x1 − 3 x2 2 x3 x4 = −8
Solve the systems of equations given below:
Gaussian Elimination: Gaussian Elimination: Exercises 6, 7Exercises 6, 7
Precalculus62
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Solution 2: x = 2 , y = −1 , z = 3
x = −1 , y = 16 , z = −3
x = 3 , y = 1 , z = 1
x = 1 , y = 2 , z = −2
x 1 = x 2 = −1 , x 3 = 0, x 4 = 1
x 1 = 1, x 2 = 2 , x 3 = −1, x 4 = −2
Gaussian Elimination: Gaussian Elimination: Solutions 27Solutions 27
Solution 3:
Solution 4:
Solution 5:
Solution 6:
Solution 7:
Precalculus63