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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]
For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.
Most but not all overdetermined systems, when constructed with random coefficients, are inconsistent. For example, the system x 3 – 1 = 0, x 2 – 1 = 0 is overdetermined (having two equations but only one unknown), but it is not inconsistent since it has the solution x = 1.
When the denominator b(x) is monic and linear, that is, b(x) = x − c for some constant c, then the polynomial remainder theorem asserts that the remainder of the division of a(x) by b(x) is the evaluation a(c). [18] In this case, the quotient may be computed by Ruffini's rule, a special case of synthetic division. [20]
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.
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.
x 3 has been divided leaving no remainder, and can therefore be marked as used by crossing it out. The result x 2 is then multiplied by the second term in the divisor −3 = −3x 2. Determine the partial remainder by subtracting −2x 2 − (−3x 2) = x 2. Mark −2x 2 as used and place the new remainder x 2 above it.
Input: initial guess x (0) to the solution, (diagonal dominant) matrix A, right-hand side vector b, convergence criterion Output: solution when convergence is reached Comments: pseudocode based on the element-based formula above k = 0 while convergence not reached do for i := 1 step until n do σ = 0 for j := 1 step until n do if j ≠ i then ...