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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.
An example of using Newton–Raphson method to solve numerically the equation f(x) = 0. In mathematics, to solve an equation is to find its solutions, which are the values (numbers, functions, sets, etc.) that fulfill the condition stated by the equation, consisting generally of two expressions related by an equals sign.
A system of polynomial equations which has fewer equations than unknowns is said to be underdetermined. It has either infinitely many complex solutions (or, more generally, solutions in an algebraically closed field) or is inconsistent. It is inconsistent if and only if 0 = 1 is a linear combination (with polynomial coefficients) of the ...
Consider a system of n linear equations for n unknowns, represented in matrix multiplication form as follows: = where the n × n matrix A has a nonzero determinant, and the vector = (, …,) is the column vector of the variables. Then the theorem states that in this case the system has a unique solution, whose individual values for the unknowns ...
When solving systems of equations, b is usually treated as a vector with a length equal to the height of matrix A. In matrix inversion however, instead of vector b , we have matrix B , where B is an n -by- p matrix, so that we are trying to find a matrix X (also a n -by- p matrix):
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.
For each solution (c 0, s 0) of this system, there is a unique solution x of the equation such that 0 ≤ x < 2 π. In the case of this simple example, it may be unclear whether the system is, or not, easier to solve than the equation.
At any step in a Gauss-Seidel iteration, solve the first equation for in terms of , …,; then solve the second equation for in terms of just found and the remaining , …,; and continue to . Then, repeat iterations until convergence is achieved, or break if the divergence in the solutions start to diverge beyond a predefined level.