Downloaded 52 times
![ So after applying the operation the equation becomes :
𝑎11 𝑥 + 𝑎12 𝑦 + 𝑎13 𝑧 = 𝑏1
0+𝑎′12 𝑦 + 𝑎′13 𝑧 = 𝑏′2 [𝑅′2 = 𝑅2 −
𝑎21
𝑎11
∗ 𝑅1 ]
0+𝑎′12 𝑦 + 𝑎′13 𝑧 = 𝑏′3 [ 𝑅′3 = 𝑅3 −
𝑎31
𝑎11
∗ 𝑅1 ]
then after that we have to apply the same process to make
that pattern to a upper triangular matrix.
𝑎11 𝑥 + 𝑎12 𝑦 + 𝑎13 𝑧 = 𝑏1
0+𝑎′22 𝑦 + 𝑎′23 𝑧 = 𝑏′2
0 + 0 + 𝑎′′33 𝑧 = 𝑏′′3 [ 𝑅′′3 = 𝑅′3 −
𝑎32
𝑎22
∗ 𝑅2 ]
the successively we can find the values Z, Y, X.](https://image.slidesharecdn.com/gausselemination-181103023008/75/GAUSS-ELIMINATION-METHOD-3-2048.jpg)
More Related Content
Similar to GAUSS ELIMINATION METHOD
GAUSS ELIMINATION METHOD
- 1.
- 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 𝑥 = 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒′ 𝑠 𝑚𝑎𝑡𝑟𝑖𝑥 We illustrate the method using the 3 × 3 system 𝑎11 𝑥 + 𝑎12 𝑦 + 𝑎13 𝑧 = 𝑏1 ∶ R1 𝑎21 𝑥 + 𝑎22 𝑦 + 𝑎23 𝑧 = 𝑏2 ∶ 𝑅2 𝑎31 𝑥 + 𝑎32 𝑦 + 𝑎33 𝑧 = 𝑏3 ∶ 𝑅3 Now to make the patter of equation upper triangular matrix we have to use the operation : 𝑅𝑖 = 𝑅𝑖 − 𝑎 𝑖𝑘 𝑎 𝑘𝑘 ∗ 𝑅 𝑘 Where i=(k+1),(k+2),……n ; k= coloumn number
- 3. So afterapplying the operation the equation becomes : 𝑎11 𝑥 + 𝑎12 𝑦 + 𝑎13 𝑧 = 𝑏1 0+𝑎′12 𝑦 + 𝑎′13 𝑧 = 𝑏′2 [𝑅′2 = 𝑅2 − 𝑎21 𝑎11 ∗ 𝑅1 ] 0+𝑎′12 𝑦 + 𝑎′13 𝑧 = 𝑏′3 [ 𝑅′3 = 𝑅3 − 𝑎31 𝑎11 ∗ 𝑅1 ] then after that we have to apply the same process to make that pattern to a upper triangular matrix. 𝑎11 𝑥 + 𝑎12 𝑦 + 𝑎13 𝑧 = 𝑏1 0+𝑎′22 𝑦 + 𝑎′23 𝑧 = 𝑏′2 0 + 0 + 𝑎′′33 𝑧 = 𝑏′′3 [ 𝑅′′3 = 𝑅′3 − 𝑎32 𝑎22 ∗ 𝑅2 ] the successively we can find the values Z, Y, X.
- 4. For examplewe take three equations: 𝑥 − 2𝑦 + 9𝑧 = 8 3𝑥 + 𝑦 − 𝑧 = 3 2x − 8𝑦 + 𝑧 = −5 Now we have to make it a upper triangular matrix using the operation: 𝑅2 = 𝑅2 − 𝑎21 𝑎11 ∗ 𝑅1 𝑅2 = 𝑅2 − 3/1 ∗ 𝑅2 They become : 𝑥 − 2𝑦 + 9𝑧 = 8 0 + 7𝑦 − 28𝑧 = −21 0 − 4𝑦 − 17𝑧 = −21 To get the pattern of upper triangular matrix we again do the operation:
- 5. The operation is: 𝑅3 = 𝑅3 − −4/7 ∗ 𝑅2 Now they become : : 𝑥 − 2𝑦 + 9𝑧 = 8 0 + 7𝑦 − 28𝑧 = −21 0 + 0 − 33𝑧 = −33 So that : 𝑧 = 1; 𝑦 = −21+28𝑧 7 = 1; 𝑥 = 8 − 9𝑧 + 2𝑦 = 1 . The set of solution is : X=1; Y=1; Z=1.
- 6.