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Precalculus4E1
Gaussian Elimination
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
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
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
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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
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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
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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
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Precalculus51b
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:
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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
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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:
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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
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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:
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