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Gaussian Elimination Matrices Solutions By Dr. Julia Arnold

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Definition of Augmented Matrix:Augmented Matrix x - 2y - 2z = -3 2x + y - z = 7 3x - 2y + 5z = 10 Definition of Coefficient Matrix: x - 2y - 2z = -3 2x + y - z = 7 3x - 2y + 5z = 10 These are the coefficients of x, y and z Adding the constant column creates the augmented matrix. Do not click on the link. This will take you to slide 8.

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The Gaussian elimination method is manipulating the matrix so that we have zeros below the main diagonal as explained in the last lesson. Zeroes needed here only.

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Since each row of the matrix represents an equation, it follows that we can interchange rows. x - 2y - 2z = -3 2x + y - z = 7 3x - 2y + 5z = 10 x - 2y - 2z = -3 2x + y - z = 7 Interchanging rows does not disrupt the solution of the system. 3x - 2y + 5z = 10

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Since each row of the matrix represents an equation, it follows that we can interchange rows. x - 2y - 2z = -3 2x + y - z = 7 3x - 2y + 5z = 10 x - 2y - 2z = -3 2x + y - z = 7 3x - 2y + 5z = 10 Interchanging rows does not disrupt the solution of the system.

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We can also multiply any row by any number (not 0 of course) that we wish. x - 2y - 2z = -3 2x + y - z = 7 3x - 2y + 5z = 10 x + 1/2 y - 1/2 z = 7/2 3x - 2y +5z = 10 -3x + 6y + 6z = 9 Row 1 multiplied by -3 Row 2 multiplied by 1/2 Row 3 multiplied by 1

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After multiplying by a number we can add two equations together or two rows of a matrix and replace the added row. x + 1/2 y - 1/2 z = 7/2 3x - 2y + 5z = 10 -3x + 6y + 6z = 9 Row 1 multiplied by -3 Row 2 multiplied by 1/2 Row 3 multiplied by 1 Let’s add equation 1 and equation 3 together: -3x + 6y + 6z = 9 3x - 2y + 5z = 10 0 x +4y +11z = 19 The sum replaces one of the rows in the system. I also showed you how we can get a zero.

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2x + y - z = 7 3x - 2y + 5z = 10 x - 2y - 2z = -3 Now let’s return to our original system and show how I got the new system. It’s helpful to have a coefficient of 1 for the first element in the matrix. So look in the x column of your system and see if there is a coefficient of 1. If there is, make that equation #1. Change this 2x + y - z = 7 3x - 2y + 5z = 10 x - 2y - 2z = -3 To This x - 2y - 2z = -3 2x + y - z = 7 3x - 2y + 5z = 10 Next, write the augmented matrix.augmented matrix This is the final result we want. For definition click on link, then click on link there to return to this slide.

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x - 2y - 2z = -3 2x + y - z = 7 3x - 2y + 5z = 10 This is the final result we want. The first task is to make column 1 match the final result. Row 1 matches already. Since we have a 1 in row 1 we can multiply by any number that appears below in row 2 or row 3 to create a sum of 0. Multiply row 1 by -2 and add row Now multiply row 1 by -3 and add row The result: Row 1 and 2 match the final result.

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This is the final result we want. Now we move to column 2. Notice, to make it match the final result we only need to change the 4 to a 0. However, to do that will be a little more complicated. Since we are working on column 2 we can only use row 2 to help us get the job done. Above the 4 in row 2 is a 5. What do you think we could do? How about multiply row 2 by -4 and multiply row 3 by 5 and then add them The sum will replace row 3 in the matrix.

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This is the final result we want. How can we make them match? How about multiplying row 3 by 1/43 (or just say divide the row by 43).

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First, look in the x column for a coefficient of 1. (None) Is their anyway to get a 1 without creating fractions? (Yes) Divide equation 3 by 4 and make it equation 1. Write the augmented matrix. Which is the correct augmented matrix? Now you should try this problem and let me guide you through the steps: 2x - 3y - z = 7 5x - 2y + 4z = -13 4x + 4y + 4z = -24

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First, look in the x column for a coefficient of 1. (None) Is their anyway to get a 1 without creating fractions? (Yes) Divide equation 3 by 4 and make it equation 1. Write the augmented matrix. Which is the correct augmented matrix? Now you should try this problem and let me guide you through the steps: 2x - 3y - z = 7 5x - 2y + 4z = -13 4x + 4y + 4z = -24 While this is also correct, it did not reflect the directions given above. This is what you want to start with.

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Step 1: Get column 1 to look like this Row 1 will (now and forever) be) the same throughout. What action will get you a 0 in the 2nd row, 1st column? Add Row 1 to Row 2 and replace Row 2. Multiply Row 1 by -2 and add row 2 then replace row 2.

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Let’s short hand this to -2R1 +R2 = NewR New R2 To get the 0 in row 3 do the following: -5R1 +R3 = NewR3 -5R2 +R3 = NewR3

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-5R1 +R3 = NewR = New R New R = New R3 Step 1 is complete. Now we move on to step 2 which is to get column 2 looking like

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Step 1 Step 2 Since we have coefficients of -5 and -7 we will need to: R2(-7)= R3( 5)= Divide the row by multiply the -5 by -7 and the -7 by 5 multiply the -5 by 7 and the -7 by -5 Correct: either will work. However, doing the first suggestion:

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Step 1 Step 2 Since we have coefficients of -5 and -7 we will need to multiply the -5 by -7 and the -7 by 5 (or the equivalent) which will create 0 when added. R2(-7)= R3( 5)= Divide the row by Now for back substitution.

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z = -3 y = --2 x = -1 z = -3 y = -8 x = -3

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1z = -3 or z = -3 -5y -3z = -5y - (3)(-3) = 19 -5y + 9 = 19, -5y = 10 y = -2 x + y + z = -6 x = -6 x - 5 = -6 x = -1 (-1, -2, -3)

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Practice Problems will be found on a separate power point presentation.

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