Does anyone know how to solve this?

Factual Questions
1

I’m not sure if this is the right forum but i was wondering if anyone knows how to solve this system of equations using augmented matrices.

2x+6y+8z=5
-2x+9y-12z=-1
4x+6y-4z=3

I came up with this:

z= -1
y= -1 1/8
x= -1 15/16

We did it in class today and everyone was getting different answers I don’t think it is right. Any help would be appreciated, thanks.

2

Every one of your answers is negative, so when you substitute them back in the first equation, every term on the left hand side is going to be negative–and their sum is going to be negative. But the right hand side is positive, so there is no chance that those are the right answers.

I assume you know the method of augmented matrices. Perhaps show your work, and we might be able to isolate where your error occurred.

3

I’m not sure I know the method of augmented matrices, but I do know how to solve it.

Solve the first equation for x.

Substitute that into the second equation and solve for z. (seemed easier than solving for y.)

Substitute both the x and z values into the third equation and solve for y.

Use the y value to calculate z.

Use the y and z values to calculate x.

The main thing is to take the time to do each step carefully, particularly on the substitutions. Hint: all three are positive fractions.

4

Whoa. I guessed the answer with just that hint.

5

If you understand augmented matrices then this is a very straight-forward problem.

First, make your matrix:


 2   6   8 |   5
-2   9 -12 |  -1
 4   6  -4 |   3

The object is to use addition and subtraction to get one row and one column to contain two zeroes and one row and one column to contain one zero (not including the last column). For instance, these are the final zeroes i had:


 ?   ?   ? |   ?
 0   ?   ? |   ?
 0   ?   0 |   ?

I’ll do the first step (this was a fairly simple problem and only required 3 easy steps to get the 3 zeroes).

Subtract twice the first row from the third row, to get:


 2   6   8 |   5
-2   9 -12 |  -1
 0  -6 -20 |  -7

Finding the other zeroes should be easy once you know what you’re doing … just be careful not to make any addition/subtraction errors. Then, of course, solve the equations, starting with the row with 2 zeroes.

6

here are my steps

[ 2 6 8 5]
|-2 9 -12 -1|
[ 4 6 -4 3]

[ 2 6 8 5]x2
|-2 9 -12 -1|x2
[ 4 6 -4 3]

First I multiplied the top two rows by 2.
I got:
[ 4 12 16 10]
|-4 18 -24 -2|
[ 4 6 -4 3]

Next I added the first two rows and also added the bottom two rows:
[ 0 30 -8 8]
| 0 24 -28 1|
[ 4 6 -4 3]

Then I multiplied the first row by 4 and the second row by
-5:
[ 0 30 -8 8] by 4
| 0 24 -28 1| by -5
[ 4 6 -4 3]

I came up with:
[ 0 120 -32 32]
| 0 -120 140 -5|
[ 4 6 -4 3]

Then I added the first two rows to get:
[ 0 0 108 27]
| 0 -120 140 -5|
[ 4 6 -4 3]

Does that look correct? Then I would just figure out x, y, and z from the matrix. Thanks for the help. As you can see JEFFB that I did my homework I just made an error previously. Thanks for all your help

7

The first of your final equations should be 108z = 27, right? So how can z = 1? Shouldn’t it be 0.25? Maybe that was the problem.

8

You’re on the right track, or should I say you’re on a right track. Your next step would be to divide the top row by 108 to get:
0 0 1 1/4
which yields the result z = 1/4 (which is correct).

The way you seem to be going about it looks like it will have bigger and bigger numbers. It might eventually get the right answers, but the chances of error grow.

The classic way (which is also error prone) is to set up the augmented matrix:

2 6 8 5
-2 9 -12 -1
4 6 -4 3

Now employ a process called “row reduction”.
The first step is to divide the top row by 2:
1 3 4 5/2
-2 9 -12 -1
4 6 -4 3

With a 1 in the first row of the first column, the idea is to put zeros in every other position in the first column.
To put a zero in the second row, first column; multiply the first row by 2 and add it to the second row:

1 3 4 5/2
0 15 -4 4
4 6 -4 3

To put a zero in the third row, first column; multiply the first row by -4 and add it to the third row:

1 3 4 5/2
0 15 -4 4
0 -6 -20 -7

That takes care of the first column. To get to work on the second column, divide the second row by 15:

1 3 4 5/2
0 1 -4/15 4/15
0 -6 -20 -7

Then put zeros in all the other positions in the second column (I’ll spare you the details):

1 0 24/5 17/5
0 1 -4/15 4/15
0 0 108/5 27/5

which takes care of the second column.
Finally, take care of the third column. Start by putting a 1 in the third row, third column by multiplying the third row by 5/108 to get:

1 0 24/5 17/5
0 1 -4/15 4/15
0 0 1 1/4

Add appropriate multiples of the third row to the second and first rows to put zeros in the third column to get:

1 0 0 1/2
0 1 0 1/3
0 0 1 1/4
and you’re done.

I’ve found that the best way to keep from making errors is to not be afraid to completely re-write the whole matrix for each step along the way. If you try to do too many steps in your head involving things like (-24/15)-20 you’re bound to make a mistake.

9

Thanks for all your help I see where I messed up now. Thank you to all that replied.