Gaussian Elimination
Matrices Solutions
ByDr. Julia Arnold
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.
The Gaussian elimination method is manipulating the matrix so that we have zeros below the main diagonal as explained in the last lesson.
8200
9410
5132
Zeroes needed here only.
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
10523
7112
3221
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
10523
7112
3221
x - 2y - 2z = -3
2x + y - z = 7
3x - 2y + 5z = 10
7112
3221
10523
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
10523
7112
3221
x + 1/2 y - 1/2 z = 7/2
3x - 2y +5z = 10
-3x + 6y + 6z = 9
1052327
21
21
1
9663
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
19114027
21
21
1
9663
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 = 73x - 2y + 5z = 10x - 2y - 2z = -3
10523
7112
3221
Now let’s return to our original system and show how I got thenew 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.
1100
13350
3221
This is the finalresult we want.
For definition click on link, then click on link there to return to this slide.
10523
7112
3221
x - 2y - 2z = -32x + y - z = 73x - 2y + 5z = 10
1100
13350
3221
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
191140
13350
3221The result:
Row 1 and 2match thefinal result.
191140
13350
3221
1100
13350
3221
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.
434300
13350
3221
1100
13350
3221
This is the finalresult we want.
434300
13350
3221
How can we make them match?
How about multiplying row 3 by 1/43 (or just say divide the row by 43).
4343434300
13350
3221
//
1100
13350
3221
13425
7132
6111
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
24444
13425
7132
13425
7132
6111
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
24444
13425
7132
While this is also correct, it did not reflect the directions given above.
This is what you want to start with.
13425
7132
6111
###
###
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.
13425
7132
6111
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
###
###
0
0
6111
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
13425
7132
6111
13425
19350
6111
0 -5 -3 19 New R2
17170
19350
6111
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
19350
6111
13425
7132
6111
13425
19350
6111
17170
19350
6111
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:
13425
7132
6111
13425
19350
6111
17170
19350
6111
3100
19350
6111
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.
3100
19350
6111
z = -3y = --2x = -1
z = -3y = -8x = -3
3100
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6111
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)
Practice Problems will be found on a separate power point presentation.