Gauss method: examples of solutions and special cases

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Gauss method, also called step method of elimination of unknowns variables, named after the great German scientist KFGauss, while still alive received the unofficial title of "King of mathematics."However, this method has been known long before the birth of European civilization, even in the I century.BC.e.Ancient Chinese scholars have used it in his writings.

Gauss method is a classic way of solving systems of linear algebraic equations (Slough).It is ideal for a quick solution to the limited size matrices.

The method itself consists of two moves: forward and reverse.The direct course is a sequence of linear systems bring to the triangular form, that is, zero values ​​are below the main diagonal.Reversal involves a consistent finding variables, expressing each variable through the previous.

Learning to practice the method of Gauss's just enough to know the basic rules of multiplication, addition and subtraction of numbers.

In order to demonstrate the algorithm for solving linear systems of this method, we explain one example.

So solved using Gauss:

x + 2y + 4z = 3
2x + 6y + 11z = 6
4x-2y-2z = -6

We need the second and third lines to get rid of the variable x.To do this, we add them to the first multiplied by -2 and -4, respectively.We obtain:

x + 2y + 4z = 3
2y + 3z = 0
-10y-18z = -18

now 2-th line multiply by 5 and add it to the third:

x + 2y + 4z= 3
2y + 3z = 0
-3z = -18

We brought our system to a triangular form.Now we carry out the reverse.We start with the last line:
-3z = -18,
z = 6.

second line:
2y + 3z = 0
2y + 18 = 0
2y = -18,
y = -9

first line:
x + 2y + 4z = 3
x-18 + 24 = 3
x = 18-24 + 3
x = -3

Substituting the values ​​of the variables in the original data, we verify the correctness of the decision.

This example can solve a lot of any other substitutions, but the answer is supposed to be the same.

It so happens that on the leading elements of the first row are arranged with too small values.It's not terrible, but rather complicates the calculations.The solution is Gauss method with a choice of the main element of the column.Its essence is as follows: the first line of the maximum sought modulo element, the column in which it is located, change places with the 1st column, that is our maximum element becomes the first element of the main diagonal.The following is a standard process calculations.If necessary, the procedure of swapping the columns can be repeated.

Another modified method of Gauss-Jordan is the method of Gauss.

used for solving linear systems of square, in finding the inverse matrix and the rank of the matrix (the number of non-zero rows).

essence of this method is that the original system is transformed by changes in the identity matrix with a further finding values ​​of variables.

algorithm it is this:

1. The system of equations is, as in the method of Gauss, a triangular form.

2. Each row is divided into a certain number in such a way that the main unit is turned diagonally.

3. The last line is multiplied by some number and is subtracted from the next so as not to get on the main diagonal 0.

4. Step 3 is repeated sequentially for each row until eventually the identity matrix is ​​formed.