Question

Can someone help me figure out how to solve this matrix using row operations step by step? I don't know where to start.

2 6 15 -12
4 7 13 -10
3 6 12 -9

Answer 1

I can help, but it's a long process.
Do you have the patience?

The first step is to multiply the first row by something to make the number in the first position (the 2) into a 1.

Answer 2

So multiply row 1 by 1/2?

Answer 3

yes

Answer 4

I helped another user with this a little while ago, you might want to try following that example.

http://openstudy.com/updates/5089d4e7e4b077c2ef2e02b0

I also posted some links to some useful websites at the end. Check those out.

Answer 5

Next make the numbers beneath the 1, 0's

1 3 15/2 -6
4 7 13 -10 add (-4 times row 1) to row 2
3 6 12 -9 add (-3 times row 1) to row 3

Answer 6

@ChmE , I'll let you take this. I have a class to teach in the morning and should get to bed. Thanks.

Answer 7

Okay so as of now I have:
1 3 15/2 -6
0 3 9 6
0 3 9 -6

Answer 8

Oops for row two, last should be -6*

Answer 9

I have this
1 3 15/2 -6
0 -5 -17 14
0 -3 -21/2 9

Answer 10

you are not subtracting 4 and 3, you are adding the product of -4 times each term in row 1, to row 2. And adding the product of -3 times row 1, to row 3

Answer 11

oh! okay I understand now

Answer 12

-4 -12 -30 24 (This is -14 times row 1) Add this to row 2
4 7 13 -10

Answer 13

Yup, try it again and see if you get the same matrix as me.

Answer 14

Got it. But why would you do that, because the first column has its leading 1 and 0s already, wouldn't this mess with that?

Answer 15

you can add rows together to manipulate a matrix. but you can't add a number to only one row. That's why we do this. Technically, after we multiply row 1 by -4 and do the addition, we would divide row 1 by -4 to return it to its original state.

This is just understood that after each operation you do this. So we just leave the first row be.

Understand that explanation?

Answer 16

Yup! Thanks for clarifying.

Answer 17

So we have this now, we want row 2 column 2 to be a 1 and row 3 column 2 to be a 0.
1 3 15/2 -6
0 -5 -17 14 divide by -5
0 -3 -21/2 9

1 3 15/2 -6
0 1 17/5 -14/5
0 -3 -21/2 9 add ( -3 times row 2) to row 3

Answer 18

Can we remove the fractions? I hate them
1 3 7.5 -6
0 1 3.4 -2.8
0 0 -20.7 17.4

Answer 19

How are you doing?

Answer 20

Good! Got it so far!

Answer 21

Yay or nay on fractions? I'll do whatever you prefer

Answer 22

I think fractions would be a bit easier for me but I understand either way!

Answer 23

1 3 15/2 -6
0 1 17/5 -14/5
0 0 -207/10 87/5 times row 3 by -10/207

Answer 24

1 3 15/2 -6
0 1 17/5 -14/5
0 0 1 -174/207

Answer 25

Did you get the same thing?

Answer 26

Indeed!

Answer 27

Now we work up
1 3 15/2 -6 add (-15/2 times row 3) to row 1
0 1 17/5 -14/5 add (-17/5 times row 3) to row 2
0 0 1 -174/207

Answer 28

got it!

Answer 29

I just used a matrix solver online and they got a different answer for our c value. Our method is correct, but I need to check for a mistake.

You can keep going, you seem like you got the hang of it. last operation is add (-3 times row 2) to row 1

Answer 30

There is a mistake somewhere. OMG

Answer 31

I want to rearrange the matrix

4 7 13 -10 divide by 4
3 6 12 -9 add (-3/4 times row 1) to row 2
2 6 15 -12 add (-1/2 times row 1) to row 3

1 7/4 13/4 -10/4
3 6 12 -9 add (-3 times row 1) to row 2
2 6 15 -12 add (-2 times row 1) to row 3

1 7/4 13/4 -10/4 Multiply everything by 4 to get rid of the fractions
0 3/4 9/4 -6/4
0 10/4 34/4 -28/4

4 7 13 -10
0 3 9 -6
0 10 34 -28

Answer 32

4 7 13 -10
0 3 9 -6
0 10 34 -28 add (-10/3 times row 2) to row 3

4 7 13 -10
0 3 9 -6
0 0 4 -8 divide by 4

4 7 13 -10
0 3 9 -6
0 0 1 -2

Answer 33

(as you work this out I just want you to know that you are fantastic and I sincerely appreciate this! hahaha)

Answer 34

4 7 13 -10 add (-13 times row 3) to row 1
0 3 9 -6 add (-9 times row 3) row 2
0 0 1 -2

4 7 0 16
0 3 0 12 divide by 3
0 0 1 -2

4 7 0 16 add (-7 times row 2) to row 1
0 1 0 4
0 0 1 -2

4 0 0 -12 Divide by 4
0 1 0 4
0 0 1 -2

1 0 0 -3 Divide by 4
0 1 0 4
0 0 1 -2

Success !!!!

Answer 35

ignore that last divide by 4. Copy and paste error

Answer 36

Okay what I did differently was I rearranged the matrix so that the highest first column value was at the top.

After making this one, I kept everything in this denominator. This allowed me to get rid of them afterwards.

Everything just kinda fell into place after that.

Sorry for the wrong answer at first. But that was good practice none the less I guess. :/

This is the site I used to check my answer http://www.bluebit.gr/matrix-calculator/linear_equations.aspx

Answer 37

Practice indeed! Thanks so much.

Answer 38

Your welcome. That was a bunch of fun ?!

Answer 39

"Fun" is a relative term. :p However it is nice to have someone who is patient and helpful so thanks again!!

Answer 40

@kamuela710