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In mathematics, a system of linear equations (or linear system) is a collection of two or more linear equations involving the same variables. [ 1 ] [ 2 ] For example, { 3 x + 2 y − z = 1 2 x − 2 y + 4 z = − 2 − x + 1 2 y − z = 0 {\displaystyle {\begin{cases}3x+2y-z=1\\2x-2y+4z=-2\\-x+{\frac {1}{2}}y-z=0\end{cases}}}
For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1.
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 ...
Hilbert's tenth problem is the tenth on the list of mathematical problems that the German mathematician David Hilbert posed in 1900. It is the challenge to provide a general algorithm that, for any given Diophantine equation (a polynomial equation with integer coefficients and a finite number of unknowns), can decide whether the equation has a solution with all unknowns taking integer values.
has no real number solution since no real number squared equals −1. Sometimes a quadratic equation has a root of multiplicity 2, such as: (+) = For this equation, −1 is a root of multiplicity 2. This means −1 appears twice, since the equation can be rewritten in factored form as
Because of this, often, the only simple effective way to deal with multiplication by expressions involving variables is to substitute each of the solutions obtained into the original equation and confirm that this yields a valid equation. After discarding solutions that yield an invalid equation, we will have the correct set of solutions.
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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.