6.3
Gaussian Elimination: Problem Solving
Systems of linear equations in several variables are pivotal in modeling complex scenarios involving multiple unknowns and constraints. Such systems a…
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.
View the full transcript and gain access to JoVE Core videos
Q1: What is Gaussian elimination and how does it solve systems of equations?
Gaussian elimination solves systems of linear equations by using one equation to eliminate a variable from the others. The method simplifies the system into an upper triangular form through elementary row operations like multiplying and subtracting equations. Once in triangular form, back-substitution determines variable values starting from the bottom equation and working upward to find the complete solution.
Q2: How do you eliminate variables in the first step of Gaussian elimination?
To eliminate a variable, choose which variable to remove first. Multiply one equation by a constant, then subtract it from another equation containing that variable. This creates a new equation without the chosen variable. Repeat this process for each variable across all equations to progressively build the triangular form.
Q3: What is back-substitution and when is it used?
Back-substitution is used after the system is converted to upper triangular form. Starting with the bottom equation, solve for the last variable. Substitute that value into the equation above it to find the next variable. Continue this process upward through all equations until all variable values are determined.
Q4: How many solutions can a system of three linear equations have?
A system of three equations can have three solution types: a unique solution when the three planes intersect at a single point; infinitely many solutions when planes intersect along a line or overlap entirely; or no solution when planes are parallel or do not intersect at a common point. The geometric configuration determines which outcome occurs.
Q5: What are elementary row operations in Gaussian elimination?
Elementary row operations are manipulations used to simplify a system without changing its solution. These include swapping rows to place non-zero coefficients in leading positions, scaling rows by multiplying by constants, and eliminating variables by subtracting linear combinations of rows. These operations transform the system into upper triangular form.
Q6: How can Gaussian elimination be applied to real-world problems?
Gaussian elimination models complex scenarios with multiple unknowns and constraints. For example, factory production problems use variables for run days and equations for output constraints. By solving the resulting system, you find how many days each factory must operate to meet production goals, demonstrating how linear equations represent real resource allocation and planning decisions.
Q7: Why is upper triangular form important in solving systems?
Upper triangular form simplifies solving because each equation contains progressively fewer variables. The bottom equation has only one variable, making it solvable directly. Each equation above contains one additional variable, allowing systematic back-substitution. This structured arrangement makes finding all variable values efficient and organized.