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1 Gauss-Jordan Method. How To complete Problem 2.2 # 29 Produced by E. Gretchen Gascon.

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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: 44-424 21-9 1-231

5 2. Keep Row1 fixed 44-424 21-9 1-231 Focus on the first row-first column element 4. We want to make the rest of the column #1 into 0s. 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 44-424 21-9 1-231 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 4-40-2-4-6 2+46-18-24-42

7 4. Replace Each Element of Row 3 44-424 0-6+6-42 1-231 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, -4+4 0 8+4 -12-4 -4+2412-16 20

8 Column 1 Complete 44-424 0-6+6-42 012-1620 Column 1 – row 1 has a value and the rest of the column values are 0.

9 Column 2 44-424 0-6+6-42 012-1620 -6 in column 2 is the pivot element. We want to make the rest of the column 0s

10 6. Replace Each Element of Row 1 44-424 0-66-42 012-1620 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, 12+0 12-12-12+12 72-84 1200 -12

11 7. Replace Each Element of Row 3 1200-12 0-66-42 012-1620 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, 0+0 12-12 - 16+12 20-84 00-4-64

12 Column 2 Complete 1200-12 0-66-42 00-4-64 Column 2 – row 2 has a value and the rest of the column values are 0.

13 Work on column 3 1200-12 0-66-42 00-4-64 -4 in column 3 remains, we want all the other elements in column 3 0s. 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 1200-12 0-66-42 00-4-64 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, 0+0 0 -12+0 -12 -12+12 0 -84-192 -276

15 Column 3 Complete 1200-12 0 0-276 00-4-64 Column 3 – row 3 has a value and the rest of the column values are 0.

16 Done with + - rows 1200-12 0 0-276 00-4-64 You have the diagonal elements with values and 0s in the rest of those columns

17 Two options: Option # 1 – covert the matrix back to an algebraic equation and solve 1200-12 0 0-276 00-4-64 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. 1200-12 0-60-276 00-4-64 R1/r1c1 r1/12 R2/r2c2 r2/(-6) R3/r3c3 r3/(-4) 12/12 1 -12/12 -6/-6 1 -276/(-6) 23 -4/-4 1 -64/(-4) 16

19 You are finished 100 01023 00116 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

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