Gaussian Elimination Matrices Solutions By Dr. Julia Arnold

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.

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.

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

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.

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

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.

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.

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.

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.

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).

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

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.

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.

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

-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

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:

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.

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

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)

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