6.3
Gaussian Elimination: Problem Solving
Gaussian elimination solves a system of m linear equations in n variables by using one equation to eliminate a variable from the other.
Consider factories A, B, and C that produce refrigerators, dishwashers, and stoves.
Taking the factory run days as the variables, the system can be modeled using linear equations E1, E2, and E3. This system is solved using Gaussian elimination.
To begin, eliminate one variable; choose x. Multiply E1 by 2; then subtract it from E2 and solve. Now replace E2 with the result to eliminate the x-term, forming E4.
To remove the x-term from E3, multiply E1 by 5 and E3 by 4, then subtract 5E1 from 4E3 to form E5, without the x-term.
Now, to eliminate the y-term from the E5, multiply the E4 by 11 and the E5 by 4, then solve to get z.
By back-substituting z into the E4, y is obtained.
Similarly, back-substitute y and z into E1 to find x.
The solution shows Factory A runs for 6 days, B for 2, and C for 3.
Systems of linear equations in several variables are pivotal in modeling complex scenarios involving multiple unknowns and constraints. Such systems are widely used in various fields to represent relationships where several conditions must be simultaneously satisfied. Each variable in the system corresponds to an unknown quantity, while each equation imposes a linear constraint, leading to a structured approach for analyzing and solving real-world problems.
A system of three equations with three variables, typically written in the form:
models the relationship among the unknowns. Each equation represents a constraint or condition, and solving the system requires finding values of x, y, and z that simultaneously satisfy all three equations. These systems can represent anything from resource allocation constraints to equilibrium conditions in physical systems.
Solving the system is commonly done using Gaussian elimination, simplifying the system into an equivalent upper triangular form using elementary row operations. This process involves:
- Swapping rows, if needed, to place a non-zero coefficient in the leading position.
- Scaling rows to make the leading coefficients 1.
- Eliminating variables from subsequent rows using linear combinations.
Once in triangular form, back-substitution is employed to determine the values of the variables, starting from the bottom equation and working upward to solve the system completely.
The solution to such a system depends on the geometric configuration of the equations, which can be visualized as planes in three-dimensional space:
- Unique Solution: When the three planes intersect at a single point, the system is consistent and independent.
- Infinitely Many Solutions: If the planes intersect along a line or overlap entirely, the system is dependent.
- No Solution: If the planes are parallel or otherwise do not intersect at a common point, the system is inconsistent.
Understanding the structure and solution of systems of linear equations is essential in fields such as economics, engineering, and data science, where multiple constraints and unknowns frequently arise.
Gaussian elimination solves a system of m linear equations in n variables by using one equation to eliminate a variable from the other.
Consider factories A, B, and C that produce refrigerators, dishwashers, and stoves.
Taking the factory run days as the variables, the system can be modeled using linear equations E1, E2, and E3. This system is solved using Gaussian elimination.
To begin, eliminate one variable; choose x. Multiply E1 by 2; then subtract it from E2 and solve. Now replace E2 with the result to eliminate the x-term, forming E4.
To remove the x-term from E3, multiply E1 by 5 and E3 by 4, then subtract 5E1 from 4E3 to form E5, without the x-term.
Now, to eliminate the y-term from the E5, multiply the E4 by 11 and the E5 by 4, then solve to get z.
By back-substituting z into the E4, y is obtained.
Similarly, back-substitute y and z into E1 to find x.
The solution shows Factory A runs for 6 days, B for 2, and C for 3.
From Chapter 6:
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