Gaussian Elimination

Introduction of Gaussian Elimination:
In linear algebra, Gaussian elimination is an algorithm for solving systems of linear equations, finding the rank of a matrix, and calculating the inverse of an invertible square matrix. Gaussian elimination is named after German mathematician and scientist Carl Friedrich Gauss. Elementary row operations are used to reduce a matrix to row echelon form. Gauss–Jordan elimination, an extension of this algorithm, reduces the matrix further to reduced row echelon form. Gaussian elimination alone is sufficient for many applications.


Solving Gaussian Elimination:

•    Solve three-variable, three-equation linear systems is more complex, at least in the beginning, than solving the two-variable systems, because the computations concerned are messier.

•    You will require being very efficient in your functioning, and you should outline to use lots of scratch paper.

•    The method for explain these systems is an extension of the two-variable solving-by-addition method, so make sure you identify this process well and can use it consistently properly.

•    Though the way of explanation is based on addition/elimination, demanding to do real addition tends to get very messy, so there is a systematized process for solving the three-or-more-variables systems. This method is called "Gaussian elimination".

Example problems of Gaussian Elimination:

1) Solve: 3x - 4y = 20 …(i)

5x + 6y = 8 …(ii)

Solution:

Multiply (i) by 3 and (ii) by 2:

9x – 12y =  60
10x + 12y = 16
-------------------
19x         = 76

x = 76 / 19

= 4

Substituting x = 4 in (ii), we get

5(4) + 6y = 8

6y = 8 - 20

6y = -12

y = -2

The solution is x = 4 and y = -2.

2) solve the following system using Gaussian Elimination method.

3x + y = 9

3x – y = 15

Solution:

If add down, the y determination cancel out. So sketch an "equals" bar below the system, and add down:

3x + y =  9
3x – y = 15
--------------
6x = 24
x = 24 / 6

x = 4

At the present divide from side to side to solve for x = 4, and then back-solve, using either of the original equations, to find the value of y. The first equation have lesser facts, so back - explain in that one:

2(4) + y = 10

8 + y = 10

y = 2

Then the solution is (x, y) = (4, 2)

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Example of Gaussian Elimination calculator:


Consider the system of equations,

`[[x,-3y,z],[2x,-8y,8z],[-6x,3y,-15z]]`  = `[[4],[-2],[9]]`

• To solve for x, y, and z we should eliminate some of the unknowns from some of the equations. regard as adding -2 times the first equation to the second equation and also adding 6 times the first equation to the third equation. The result is

`[[x,-3y,z],[0x,-2y,6z],[0x,-15y,-9z]]`  = `[[4],[-10],[33]]`

• To remove the y term in the last equation, multiplication the second equation by -5 and add it to the third equation:

`[[x,-3y,z],[0x,-y,2z],[0x,0y,-18z]]`  = `[[4],[-5],[36]]`

-18z = 36

z = 36/-18

z = -2

Then plug the value z=-2, in equation 2,

-y+2z = -5

-y+2(-2) = -5

-y = -5+4

-y = -1

Y = 1

Now plug the values (y=1, z=-2), in equation 1,

x-3y+z = 4

x-3(1)-2 = 4

x-5 = 4

x = 9

• The third equation says z=-2. Substituting this into the second equation yields y=1. Using both of these result in the first equation gives x=9. The method of gradually solving for the unknowns is called back-substitution.

• This is the essence of Gaussian elimination. However, we may clean up the notation in our worked by using matrices.