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Gauss-Jordan Method.

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andie
20 diapositives · 907 Views · Aug 23, 2012

Gauss-Jordan Method. . How To complete Problem 2.2 # 29 Produced by E. Gretchen Gascon. The problem. Plan to solve. Step 1 – write a matrix with the coefficients of the terms and as the last column the constant equivalents.

#newton polynomial #elimination theory #lagrange polynomial #least squares #the crystal method #gauss-legendre algorithm #jordan river #jordan normal form
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Gauss-Jordan Method.
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  1. Gauss-Jordan Method. How To complete Problem 2.2 # 29 Produced by E. Gretchen Gascon

  2. The problem

  3. Plan to solve • Step 1 – write a matrix with the coefficients of the terms and as the last column the constant equivalents. • Step 2 – use the Gauss-Jordan method to manipulate the matrix so that the solution will become evident

  4. The matrix:

  5. 2. Keep Row1 fixed Focus on the first row-first column element 4. We want to make the rest of the column #1 into 0’s. To accomplish that we will perform the various row operation. Start with row 2. What would be a LCM of 4 and 2? (This helps to define the formula to be used. Ans: 4. Therefore you multiply 1 times 4 = 4, and 2 times 2 = 4. This is the formula 2*Elements of R2 Plus -1 times the Elements of R1

  6. 3. Replace Each Element of Row 2 6 2+4 -18-24 -42 4-4 0 -6 -2-4 Add 2*R2C1 -1*R1C1  R2C1, this would be 2(2) -1(4)  0 Replace all the cells in Row 2 with this formula, just change the column you are working in. 2*R2C2 -1*R1C2  R2C2, 2*R2C3 -1*R1C3  R2C3, 2(-1) - 1(4)  -2+ -4 -6 2(1) -1(-4)  2+46 2*R2C4 -1*R1C4  R2C4, 2(-9) -1(24)  -18-24 -42

  7. 4. Replace Each Element of Row 3 8+4 12 -16 -4+24 -4+4 -12-4 20 0 Add -4*R3C1 +*R1C1  R3C1, this would be -4(1) +(4)  0 Replace all the cells in Row 2 with this formula, just change the column you are working in. -4*R3C2 +*R1C1  R3C2, -4*R3C3 +*R1C3  R3C3, -4*R3C4 +*R1C4  R3C4,

  8. Column 1 Complete Column 1 – row 1 has a value and the rest of the column values are 0.

  9. Column 2 -6 in column 2 is the pivot element. We want to make the rest of the column 0’s

  10. 6. Replace Each Element of Row 1 12 12-12 0 -12+12 0 12+0 72-84 -12 Add 3*R1C1 +2*R2C1  R1C1, this would be 3(4) +0 12 Replace all the cells in Row 1 with this formula, just change the column you are working in. Add 3*R1C2 +2R2C2  R1C2, Add 3*R1C3 +2R2C3  R1C3, Add 3*R1C4 +2R2C4  R1C4,

  11. 7. Replace Each Element of Row 3 0+0 0 0 12-12 -4 -16+12 -64 20-84 Add R3C1 +2*R2C1  R3C1, this would be 0 + 3(0)  0 Replace all the cells in Row 1 with this formula, just change the column you are working in. Add R3C2 +2R2C2  R3C2, Add R3C3 +2R2C3  R3C3, Add R3C4 +2R2C4  R3C4,

  12. Column 2 Complete Column 2 – row 2 has a value and the rest of the column values are 0.

  13. Work on column 3 -4 in column 3 remains, we want all the other elements in column 3 0’s. We will accomplish this by using row operations, but the first row already has a 0 in the third column so we can skip that row operation.

  14. 9 Replace Each Element of Row 2 0+0 0 -12 -12+12 0 -276 -12+0 -84-192 Add 2*R2C1 +3*R3C1  R1C1, this would be (0) +3(0) 0 Replace all the cells in Row 1 with this formula, just change the column you are working in. Add 2*R2C2 +3*R3C2  R1C2, Add 2*R2C3 +3*R3C3  R1C3, Add 2*R2C4 +3*R3C4  R1C4,

  15. Column 3 Complete Column 3 – row 3 has a value and the rest of the column values are 0.

  16. Done with + - rows You have the diagonal elements with values and 0’s in the rest of those columns

  17. Two options:Option # 1 – covert the matrix back to an algebraic equation and solve 12x = -12 (from row #1) x = -1 -12y = -276 (from row # 2) y = 23 -4z = -64 (from row # 3) z = 16 We can write these equations because if you remember col#1 was the coefficients of the x, col#2 the coefficients of the y and col#3 the coefficients of the z.

  18. Option 2: Use the rules for Gauss-Jordan - Divide through by the left most element in each row. 12/12 1 -1 -12/12 1 -6/-6 -276/(-6) 23 1 -64/(-4) 16 -4/-4 R1/r1c1 r1/12 R2/r2c2  r2/(-6) R3/r3c3 r3/(-4)

  19. You are finished At this point you can read the answer from the matrix. x = -1 y = 23 z = 16

  20. Solutions • X = 1 • Y = 23 • Z = 16