gaussian elimination matrices solutions definition of augmented matrix:augmented matrix x - 2y - 2z...
TRANSCRIPT
![Page 1: Gaussian Elimination Matrices Solutions Definition of Augmented Matrix:Augmented Matrix x - 2y - 2z = -3 2x + y - z = 7 3x - 2y + 5z = 10 Definition](https://reader033.vdocuments.net/reader033/viewer/2022061402/5697bfd21a28abf838cab99e/html5/thumbnails/1.jpg)
Gaussian Elimination
Matrices Solutions
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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
523
112
221Adding 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.
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Zeroes needed here only.
Gaussian elimination method
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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
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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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3221
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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 = -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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21
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1
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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
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1
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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
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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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1100
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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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###
###
0
0
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
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-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
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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)