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Gauss – Jordan Elimination Method: Example 2

Gauss – Jordan Elimination Method: Example 2. Solve the following system of linear equations using the Gauss - Jordan elimination method. The system of linear equations. – 3x + 2y = 6 2x + 4y = 3. What is the next step?. Convert to a matrix of coefficients. – 3x + 2y = 6 2x + 4y = 3.

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Gauss – Jordan Elimination Method: Example 2

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

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

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

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

  5. 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. • 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. • Thus the matrix becomes:

  6. From the matrix in slide 4, the new matrix becomes: 1 – 2/3 – 2 2 4 3 (– 1/3) R1 • The notation (– 1/3) R1 means to multiply all the values in row 1, as signified by the R1 , by the value (– 1/3) , which is the reciprocal of – 3. • Now what is the next step?

  7. Change any values above and or below the pivot value to a 0. • 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. • 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. • Notation: R2 + (– 2) R1

  8. On a scratch piece of paper, do the following row operation: R2 + (– 2) R1 R2 (– 2 ) R1 2 4 3 – 2 4/3 4 0 16/3 7 • 1. (– 2) R1 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 • 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 ]. • Now since R2 is at the beginning of the statement R2 + (– 2) R1 , replace row 2 with the 0 16/3 7 values • Thus the new matrix will be the following:

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

  10. Change the pivot number 1 – 2/3 – 2 0 16/3 7 • Since all the values below the pivot value of 1 are now zeros, the pivot value moves down the diagonal . • The pivot value is now 16/3 and the pivot row is the 0 16/3 7 row (i.e. row 2, or R2 ). • What is the next step?

  11. 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. • 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. • Thus the matrix becomes:

  12. From the matrix in slide 10, the new matrix becomes: 1 – 2/3 – 2 0 1 21/16 ( 3/16) R2 • ( 3/16) R2 means that you multiply 3/16 to the values in row 2 (i.e. multiply 3/16 to 0, 16/3, and 7. • The 21/16 is from multiplying (3/16) to 7. • Now what is the next step?

  13. Change any values above and or below the pivot value to a 0. • 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. • 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. • Notation: R1 + (2/3) R2

  14. On a scratch piece of paper, do the following row operation: R1 + (2/3) R2 R1 (2/3) R2 1 – 2/3 – 2 0 2/3 7/8 1 0 – 9/8 • 1. (2/3) R2 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 • The row of values 1 0 – 9/8 comes from adding the corresponding values in the two rows above. • Now since R1 is at the beginning of the statement R1 + (2/3) R2 , replace row 1 with the 1 0 – 9/8 values • Thus the new matrix will be the following:

  15. From the matrix in slide 12, the new matrix becomes: 1 0 – 9/8 0 1 21/16 R1 + (2/3) R2 • Now what is the next step?

  16. 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.

  17. 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.

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

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