Gaussian Elimination :
The purpose of this article is to describe how the solutions to a linear system are actually found. The fundamental idea is to add multiples of one equation to the others in order to eliminate a variable and to continue this process until only one variable is left. Once this final variable is determined, its value is substituted back into the other equations in order to evaluate the remaining unknowns. This method, characterized by step-by-step elimination of the variables, is called Gaussian elimination.
Example 1: Solve this system:
Multiplying the first equation by −3 and adding the result to the second equation eliminates the variable x:
This final equation, −5 y = −5, immediately implies y = 1. Back-substitution of y = 1 into the original first equation, x + y = 3, yields x = 2. (Back-substitution of y = 1 into the original second equation, 3 x − 2 y = 4, would also yeild x = 2.) The solution of this system is therefore ( x, y) = (2, 1), as noted in Example 1.
Gaussian elimination is usually carried out using matrices. This method reduces the effort in finding the solutions by eliminating the need to explicitly write the variables at each step. The previous example will be redone using matrices.
Hope the above explanation was useful, now let me explain about matrices.
The purpose of this article is to describe how the solutions to a linear system are actually found. The fundamental idea is to add multiples of one equation to the others in order to eliminate a variable and to continue this process until only one variable is left. Once this final variable is determined, its value is substituted back into the other equations in order to evaluate the remaining unknowns. This method, characterized by step-by-step elimination of the variables, is called Gaussian elimination.
Example 1: Solve this system:
Multiplying the first equation by −3 and adding the result to the second equation eliminates the variable x:
This final equation, −5 y = −5, immediately implies y = 1. Back-substitution of y = 1 into the original first equation, x + y = 3, yields x = 2. (Back-substitution of y = 1 into the original second equation, 3 x − 2 y = 4, would also yeild x = 2.) The solution of this system is therefore ( x, y) = (2, 1), as noted in Example 1.
Gaussian elimination is usually carried out using matrices. This method reduces the effort in finding the solutions by eliminating the need to explicitly write the variables at each step. The previous example will be redone using matrices.
Hope the above explanation was useful, now let me explain about matrices.
