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The simplest method for solving a system of linear equations is to repeatedly eliminate variables. This method can be described as follows: In the first equation, solve for one of the variables in terms of the others. Substitute this expression into the remaining equations. This yields a system of equations with one fewer equation and unknown.
Another way of solving the same system of linear equations is by substitution. {+ = = An equivalent for y can be deduced by using one of the two equations. Using the second equation: = Subtracting from each side of the equation:
For example, to solve a system of n equations for n unknowns by performing row operations on the matrix until it is in echelon form, and then solving for each unknown in reverse order, requires n(n + 1)/2 divisions, (2n 3 + 3n 2 − 5n)/6 multiplications, and (2n 3 + 3n 2 − 5n)/6 subtractions, [9] for a total of approximately 2n 3 /3 operations.
are equivalent to Newton's equations for the function =, where T is the kinetic, and V the potential energy. In fact, when the substitution is chosen well (exploiting for example symmetries and constraints of the system) these equations are much easier to solve than Newton's equations in Cartesian coordinates.
An early iterative method for solving a linear system appeared in a letter of Gauss to a student of his. He proposed solving a 4-by-4 system of equations by repeatedly solving the component in which the residual was the largest [ citation needed ] .
To solve the equations, we choose a relaxation factor = and an initial guess vector = (,,,). According to the successive over-relaxation algorithm, the following table is obtained, representing an exemplary iteration with approximations, which ideally, but not necessarily, finds the exact solution, (3, −2, 2, 1) , in 38 steps.
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