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Gauss – Jordan Elimination Method: Example 2 Solve the following system of linear equations using the Gauss - Jordan elimination method Slide 1

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The system of linear equations – 3x + 2y = 6 2x + 4y = 3 What is the next step? Slide 2

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Convert to a matrix of coefficients – 3x + 2y = 6 2x + 4y = 3 – 3 2 6 2 4 3 Now circle the pivot number. Slide 3

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Pivot Number and Pivot Row – 3 2 6 2 4 3 1.Recall that the row with the pivot number (circled number) is called the pivot row. 2.What is the next step? Slide 4

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1.Change the pivot number to a 1 by multiplying the pivot number, and all the other numbers in the pivot row, by the reciprocal of the circled number. 2.Remember that when you change a pivot number to a 1, you use the second elementary row operation; Multiply an equation by a nonzero value. 3.Thus the matrix becomes: Slide 5

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From the matrix in slide 4, the new matrix becomes: 1 – 2/3 – 2 2 4 3 1.The notation (– 1/3) R 1 means to multiply all the values in row 1, as signified by the R 1, by the value (– 1/3), which is the reciprocal of – 3. 2.Now what is the next step? (– 1/3) R 1 Slide 6

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1.Change any values above and or below the pivot value to a 0. 2.Do this by multiplying the pivot row by the opposite number (i.e. change the sign of the number) that you want to change to a 0. 3.In this case we want to change the 2 (in the second row, first column) to a 0, so we take the second row and add it to ( – 2) times the values in the pivot row. 4.Notation: R 2 + (– 2) R 1 Slide 7

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On a scratch piece of paper, do the following row operation: R 2 + (– 2) R 1 R 2 (– 2 ) R 1 2 4 3 – 2 4/3 4 0 16/3 7 1.(– 2) R 1 means multiply (– 2) to the values in row 1. So the row – 2 – 4/3 4 is a result of multiplying (– 2) to 1 – 2/3 – 2 2.The row of values 0 16/3 7 comes from adding the corresponding values in the two rows above, hence the addition symbol in the notation [2] + (– 2) [ 1 ]. 3.Now since R 2 is at the beginning of the statement R 2 + (– 2) R 1, replace row 2 with the 0 16/3 7 values 4.Thus the new matrix will be the following: Slide 8

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From the matrix in slide 6, the new matrix becomes: 1 – 2/3 – 2 0 16/3 7 1.Now what is the next step? [2] + (– 2)[ 1] Slide 9

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Change the pivot number 1 – 2/3 – 2 0 16/3 7 1.Since all the values below the pivot value of 1 are now zeros, the pivot value moves down the diagonal. 2.The pivot value is now 16/3 and the pivot row is the 0 16/3 7 row (i.e. row 2, or R 2 ). 3.What is the next step? Slide 10

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1.Change the pivot number to a 1 by multiplying the pivot number, and all the other numbers in the pivot row, by the reciprocal of the circled number. 2.Remember that when you change a pivot number to a 1, you use the second elementary row operation; Multiply an equation by a nonzero value. 3.Thus the matrix becomes: Slide 11

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From the matrix in slide 10, the new matrix becomes: 1 – 2/3 – 2 0 1 21/16 1.( 3/16) R 2 means that you multiply 3/16 to the values in row 2 (i.e. multiply 3/16 to 0, 16/3, and 7. 2.The 21/16 is from multiplying (3/16) to 7. 3.Now what is the next step? ( 3/16) R 2 Slide 12

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1.Change any values above and or below the pivot value to a 0. 2.Do this by multiplying the pivot row by the opposite number (i.e. change the sign of the number) that you want to change to a 0. 3.In this case we want to change the – 2/3 (in the first row, second column) to a 0, so we take the first row and add it to (2/3) times the values in the pivot row. 4.Notation: R 1 + (2/3) R 2 Slide 13

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On a scratch piece of paper, do the following row operation: R 1 + (2/3) R 2 R 1 (2/3) R 2 1 – 2/3 – 2 0 2/3 7/8 1 0 – 9/8 1.(2/3) R 2 means multiply (2/3) to the values in row 2. So the row 0 2/3 7/8 is a result of multiplying (2/3) to 0, 1 and 21/16 2.The row of values 1 0 – 9/8 comes from adding the corresponding values in the two rows above. 3.Now since R 1 is at the beginning of the statement R 1 + (2/3) R 2, replace row 1 with the 1 0 – 9/8 values 4.Thus the new matrix will be the following: Slide 14

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From the matrix in slide 12, the new matrix becomes: 1 0 – 9/8 0 1 21/16 1.Now what is the next step? R 1 + (2/3) R 2 Slide 15

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Convert the matrix back to a system of equations Now that there are 1’s on the diagonals (from top left corner to the bottom right corner) and 0’s above and/or below the 1’s, then convert the matrix back to the system of linear equations. Slide 16

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Convert back to a system of equations 1x + 0y = – 9/8 0x + 1y = 21/16 1 0 – 9/8 0 1 21/16 Now simplify the system of equations. Slide 17

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1x + 0y = – 9/8 0x + 1y = 21/16 Thus the solution is ( – 9/8, 21/16 ) Slide 18 Thus x = – 9/8 y = 21/16

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