Gaussian Elimination

Matrices Solutions

Definition of Augmented Matrix:

x - 2y - 2z = -32x + y - z = 73x - 2y + 5z = 10

Definition of Coefficient Matrix:

x - 2y - 2z = -32x + y - z = 73x - 2y + 5z = 10

523

112

221 These are the coefficients of x, y and z

10

7

3

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221Adding 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.

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Zeroes needed here only.

Gaussian elimination method

Since each row of the matrix represents an equation, it follows that we can interchange rows.

x - 2y - 2z = -32x + y - z = 73x - 2y + 5z = 10

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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 = -32x + y - z = 73x - 2y + 5z = 10

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x - 2y - 2z = -3

2x + y - z = 7

3x - 2y + 5z = 10

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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 = -32x + y - z = 73x - 2y + 5z = 10

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x + 1/2 y - 1/2 z = 7/2

3x - 2y +5z = 10

-3x + 6y + 6z = 9

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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

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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.Also shown is how to get a zero.

x - 2y - 2z = -32x + y - z = 73x - 2y + 5z = 10

2x + y - z = 73x - 2y + 5z = 10x - 2y - 2z = -3

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Now let’s return to our original system and show how to get 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 = 73x - 2y + 5z = 10x - 2y - 2z = -3

To

This x - 2y - 2z = -32x + y - z = 73x - 2y + 5z = 10

Next, write the augmented matrix.

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This is the finalresult we want.For definition click

on link, then click on link there to return to this slide.

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x - 2y - 2z = -32x + y - z = 73x - 2y + 5z = 10

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This is the finalresult 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 2.-2 4 4 6 2 1 -1 7 0 5 3 13

Now multiply row 1 by -3 and add row 3.

-3 6 6 9 3 -2 5 10 0 4 11 19

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Row 1 and 2match thefinal result.

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This is the finalresult 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.

0 -20 -12 -520 20 55 950 0 43 43The sum will replace row 3 in the matrix.

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This is the finalresult we want.

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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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//

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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 = 75x - 2y + 4z = -134x + 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 while guided through the steps:

2x - 3y - z = 75x - 2y + 4z = -134x + 4y + 4z = -24

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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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###

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0

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6111Step 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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Multiply Row 1 by -2 and add row 2 then replace row 2.Let’s short hand this to -2R1 +R2 = NewR2-2 -2 -2 12 2 -3 -1 7 0 -5 -3 19 New R2

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To get the 0 in row 3 do the following:

-5R1 +R3 = NewR3

-5R2 +R3 = NewR3

-5R1 +R3 = NewR3-5 -5 -5 +30 5 -2 4 -13 0 -7 -1 17 = New R3

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0 -5 -3 19 New R2

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0 -7 -1 17 = New R3

Step 1 is complete. Now we move on to step 2which is to get column 2 looking like

##00

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Step 1

Step 2

Since we have coefficients of -5 and -7 we will need to:

R2(-7)= 0 35 21 -133R3( 5)= 0 -35 -5 85 0 0 16 -48Divide the row by 16 0 0 1 -3

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)= 0 35 21 -133R3( 5)= 0 -35 -5 85 0 0 16 -48Divide the row by 16 0 0 1 -3

Now for back substitution.

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

z = -3y = -8x = -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 = -6x -2 - 3 = -6x - 5 = -6x = -1

(-1, -2, -3)