GAUSS ELIMINATION METHOD
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The Gauss-Elemination method is used to solve systems of linear equations by reducing the system to upper triangular form using elementary row operations. It works by first making the coefficients of the variables above the main diagonal equal to zero one by one, then back-substituting the solutions. The method is illustrated using a 3x3 system that is reduced to upper triangular form by subtracting appropriate multiples of rows from each other. The unique solution can then be found by back-substituting the values of z, y, and x.
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![So after applying the operation the equation becomes :
11 + 12 + 13 = 1
0+12 + 13 = 2 [2 = 2
21
11
1 ]
0+12 + 13 = 3 [ 3 = 3
31
11
1 ]
then after that we have to apply the same process to make
that pattern to a upper triangular matrix.
11 + 12 + 13 = 1
0+22 + 23 = 2
0 + 0 + 霞33 = 霞3 [ 霞3 = 3
32
22
2 ]
the successively we can find the values Z, Y, X.](https://image.slidesharecdn.com/gausselemination-181103023008/85/GAUSS-ELIMINATION-METHOD-3-320.jpg)
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3. The same method was used to find the roots of x3-2x2-5x+6=0, resulting in roots of 3, -2, and 1.
4. The method can also determine complex roots, using properties of how the coefficients fluctuate under squsystem of algebraic equation by Iteration method
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The document discusses iterative methods for solving systems of linear equations, specifically Jacobi's method and Gauss-Seidel method. It provides examples of using these methods to solve several systems of 3 equations with 3 unknowns. For each system, it shows rewriting the equations in a form suitable for the methods, choosing initial approximations, iterating to obtain better approximations, and concluding when subsequent iterations yield identical results to 3 significant digits.Solving Linear system of equations Mathematics.pdf.pptx
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Class: B.Sc CS.
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GAUSS ELIMINATION METHOD
- 2. GAUSS ELEMINATION METHOD : The method is based on the idea of reducing the given system of equations Ax = B; to an upper triangular system of equations Ux = z, using elementary row operations .That is, the solutions of both the systems are identical. Here = ゐ ｐ ＄ We illustrate the method using the 3 3 system 11 + 12 + 13 = 1 R1 21 + 22 + 23 = 2 2 31 + 32 + 33 = 3 3 Now to make the patter of equation upper triangular matrix we have to use the operation : = Where i=(k+1),(k+2),n ; k= coloumn number
- 3. So after applying the operation the equation becomes : 11 + 12 + 13 = 1 0+12 + 13 = 2 [2 = 2 21 11 1 ] 0+12 + 13 = 3 [ 3 = 3 31 11 1 ] then after that we have to apply the same process to make that pattern to a upper triangular matrix. 11 + 12 + 13 = 1 0+22 + 23 = 2 0 + 0 + 霞33 = 霞3 [ 霞3 = 3 32 22 2 ] the successively we can find the values Z, Y, X.
- 4. For example we take three equations: 2 + 9 = 8 3 + = 3 2x 8 + = 5 Now we have to make it a upper triangular matrix using the operation: 2 = 2 21 11 1 2 = 2 3/1 2 They become : 2 + 9 = 8 0 + 7 28 = 21 0 4 17 = 21 To get the pattern of upper triangular matrix we again do the operation:
- 5. The operation is : 3 = 3 4/7 2 Now they become : : 2 + 9 = 8 0 + 7 28 = 21 0 + 0 33 = 33 So that : = 1; = 21+28 7 = 1; = 8 9 + 2 = 1 . The set of solution is : X=1; Y=1; Z=1.
- 6. THANK YOU