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Gauss – Jordan Elimination Method: Example 1

Gauss – Jordan Elimination Method: Example 1. Solve the following system of linear equations using the Gauss-Jordan elimination method. The system of linear equations. 4x – 3y = 7 3x – 2y = 6. What is the next step?. Convert to a matrix of coefficients. 4x – 3y = 7

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Gauss – Jordan Elimination Method: Example 1

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

  2. The system of linear equations 4x – 3y = 7 3x – 2y = 6 • What is the next step?

  3. Convert to a matrix of coefficients 4x – 3y = 7 3x – 2y = 6 4 – 3 7 3 – 2 6 Now circle the pivot number.

  4. Pivot Number and Pivot Row 4 – 3 7 3 – 2 6 • 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 – 3/4 7/4 3 – 2 6 (1/4) R1 • The notation (1/4)R1 means to multiply all the values in row 1, as signified by the R1, by the value of (1/4) , which is the reciprocal of 4. • 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 3 (in the second row, first column) to a 0, so we take the second row and add it to ( – 3) times the values in the pivot row. • Notation: R2 + (– 3) R1

  8. On a scratch piece of paper, do the following row operation: R2 + (– 3) R1 R2 (– 3 ) R1 3 – 2 6 – 3 9/4 –21/4 0 1/4 3/4 • 1. (– 3) R1 means multiply (– 3) to the values in row 1. So the row • – 3 9/4 – 21/4 is the result of multiplying (– 3) to 1 , – 3/4 and 7/4. • The row of values 0 1/4 3/4 comes from adding the corresponding values in the two rows above, hence the addition symbol in the notation R2 + (– 3) R1 . • Now since R2 is at the beginning of the statement R2 + (– 3) R1 , replace row 2 with the 0 1/4 3/4 values • Thus the new matrix will be the following:

  9. From the matrix in slide 6, the new matrix becomes: 1 – 3/4 7/4 0 1/4 3/4 R2 + (– 3) R1 • Now what is the next step?

  10. Change the pivot number 1 – 3/4 7/4 0 1/4 3/4 • 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 1/4 and the pivot row is the 0 1/4 3/4 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. • Thus the matrix becomes:

  12. From the matrix in slide 10, the new matrix becomes: 1 – 3/4 7/4 0 1 3 ( 4) R2 • ( 4) R2 means that you multiply 4 to the values in row 2 (i.e. multiply 4 to • 0 1/4 and ¾. • The 3 is from multiplying 4 to (3/4). • 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 – 3/4 (in the first row, second column) to a 0, so we take the first row and add it to (3/4) times the values in the pivot row. • Notation: R1 + (3/4) R2

  14. On a scratch piece of paper, do the following row operation: R1 + (3/4) R2 R1 (3/4) R2 1 – 3/4 7/4 0 3/4 9/4 1 0 4 • 1. (3/4) R2 means multiply (3/4) to the values in row 2. So the row • 0 3/4 9/4 is a result of multiplying (3/4) to 0 , 1 and 3 • The row of values 1 0 4 comes from adding the corresponding values in the two rows above. • Now since R1 is at the beginning of the statement R1 + (3/4) R2 , replace row 1 with the 1 0 4 values • Thus the new matrix will be the following:

  15. From the matrix in slide 12, the new matrix becomes: 1 0 4 0 1 3 R1 + (3/4) 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 = 3 0x + 1y = 4 1 0 3 0 1 4 Now simplify the system of equations.

  18. Thus x = 3 y = 4 1x + 0y = 3 0x + 1y = 4 Thus the solution is ( 3 , 4 )

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