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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, [10] for a total of approximately 2n 3 /3 operations.
Indeed, multiplying each equation of the second auxiliary system by , adding with the corresponding equation of the first auxiliary system and using the representation = +, we immediately see that equations number through of the original system are satisfied; it only remains to satisfy equation number .
The solution set for the equations x − y = −1 and 3x + y = 9 is the single point (2, 3). A solution of a linear system is an assignment of values to the variables ,, …, such that each of the equations is satisfied. The set of all possible solutions is called the solution set. [5]
In linear algebra, Cramer's rule is an explicit formula for the solution of a system of linear equations with as many equations as unknowns, valid whenever the system has a unique solution. It expresses the solution in terms of the determinants of the (square) coefficient matrix and of matrices obtained from it by replacing one column by the ...
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): = =.
In numerical linear algebra, the Gauss–Seidel method, also known as the Liebmann method or the method of successive displacement, is an iterative method used to solve a system of linear equations. It is named after the German mathematicians Carl Friedrich Gauss and Philipp Ludwig von Seidel .
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.
The system Q(Rx) = b is solved by Rx = Q T b = c, and the system Rx = c is solved by 'back substitution'. The number of additions and multiplications required is about twice that of using the LU solver, but no more digits are required in inexact arithmetic because the QR decomposition is numerically stable .