how do i find a solution to a system of equations represented by a matrix?

- anonymous

how do i find a solution to a system of equations represented by a matrix?

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- anonymous

this specifically

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- anonymous

Well first you need to know how to read a matrix. Do you know how to read a matrix?

- anonymous

no....

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## More answers

- anonymous

im kind of clueless

- anonymous

and im bad at asking for help

- anonymous

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- anonymous

Okay I'm just writing the problem here so I can see it more easier. And it's okay. I'll write how to solve the problem step by step. Just tell me when you are lost.

- anonymous

aw, thank you

- anonymous

You're welcome! :)

- anonymous

itll help alot if i understand this

- anonymous

Alright I'll try the best I can in getting you to understand this. |dw:1438709667710:dw|

- anonymous

First you need to know how to read the matrix. As you can see there is a column name called x y and z. Those numbers below the variables are going to be part of the equation.

- anonymous

For example in the first row you see the numbers 1 2 and -2. When you put this into an equation you will use the variables above the first row to combine it as shown below:\[x+2y-2z\]The 11 will be after the equals sign since that line separating -2 and 11 technically means equals. So now the equation will look like this:\[x+2y-2z=11\]

- anonymous

Did you get that so far?

- anonymous

uuhm..

- anonymous

oh, yeah okay i get it.

- anonymous

since one would be just x, sorry takes me awhile to work things out.

- anonymous

It's okay! Take your time! :)

- anonymous

Do you know what the other two equations will be if you used my method that I showed you?

- anonymous

5x+4y+7z=81?

- anonymous

Yes!

- anonymous

-4x+y-6z=-37

- anonymous

Great job! Now those three equations are needed when you are trying to solve equations within a matrix.

- anonymous

\[x+2y-2z=11\]\[5x+4y+7z=81\]\[-4x+y-6z=-37\]

- anonymous

Now do you know how to solve for x's and y's?

- anonymous

replace them with the numbers or..

- anonymous

Not in this case since they didn't give us any numbers to put in for x y or z. So the first step is to pick any two sets of equations out of the three.

- anonymous

Let's start with the first two equations.

- anonymous

sorry, ah :(

- anonymous

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- anonymous

And it's okay. That was a rhetorical question.

- anonymous

Oh wait let me rewrite what I wrote on the side.

- anonymous

Pick which variable to make common. For example the y's can be made common by multiplying the 2y by 2 and leaving the 4y. When you multiply 2y by 2 you can get 4y. 4y and 4y equal each other so now you can subtract the whole equation to get rid of y.

- anonymous

Tell me if you get that part.

- anonymous

hm..

- anonymous

i do, but at the same time i guess i need to know why we are doing that.

- anonymous

what is the reason for finding commons, is that like a big part of the equation?

- anonymous

what is the end result supposed to be? like, is the Y now 1? bleh...

- anonymous

We are doing that so that we can get rid of one term and just concentrate on finding x and z. We will find y at the end.

- anonymous

why are we doing that?

- anonymous

In order to find the solution to this matrix we need to find what x y and z equals. So we need to break it down by finding one variable at a time.

- anonymous

Oh alright. so whenever im doing this i have to find a way to get rid of a variable?

- anonymous

Yes.

- anonymous

okay, so, like in this equation for example: |dw:1438711521979:dw|

- anonymous

just as an example

- anonymous

they have to be in the same row to be able to get rid of them right? the 9 and -9 cant cancel each other out

- anonymous

I'm sorry but where did 31 -25 and 28 come from? It's the last column. And no you can't cancel out the 9 and -9 because one is a y and the other is a z. Remember the 2nd column are y's and the 3rd column are z's.

- anonymous

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- anonymous

its an example problem of one i did and failed.

- anonymous

totally different problem sorry just trying to see where this applies to other problems too

- anonymous

You understood more than before at least :) And oh ok. I'll help you out with those if you don't understand still.

- anonymous

thank you, youre a saint ;-;

- anonymous

okay, so lets finish this problem first

- anonymous

Lol you're welcome! And okay. We have one equation: -3x-11z=-59 after canceling the y. Now when you look at the 3 equations from earlier, you can see that we need an equation with just an x and z and no y but we don't have that. So we need to repeat the step in finding another 2 equations to cancel out y.

- anonymous

did you subtract?

- anonymous

im lost again

- anonymous

-3x-11z=-59 came from the drawing I drew earlier. Take a look at that and tell me if you get that.

- anonymous

where did the other two come from?

- anonymous

It came from the three equations we wrote from the matrix.

- anonymous

|dw:1438712602958:dw| gosh im so sorry, im not sure where the two came from. i wrote down the equation again, to see if i can do what youre doing

- anonymous

Oh lol okay.

- anonymous

https://join.me/592-430-542 im sharing my screen with you, is that alright??

- anonymous

I'm pretty freaked out by that screen. Lol sorry but I don't feel like it's safe. No offense to you or anything. I just feel like it's a hack since that kind of screen popped up when my uncle's computer got hacked.

- anonymous

pfft, im not trying to hack anyone and learn math at the same time but thats understandable. i usually use joinme with my friends when im drawin and stuff

- anonymous

Oh okay. I'm really sorry about that! Are you able to draw here?

- anonymous

We're almost done anyways.

- anonymous

|dw:1438713360885:dw|

- anonymous

alright, sorry i scared you haha

- anonymous

the brushes on here kind of suck there for i kind of suck

- anonymous

It's okay you didn't know! :) And at least you draw better than me XD

- anonymous

\[x+2y-2z=11\]\[5x+4y+7z=81\]\[-4x+y-6z=-37\]*Just rewriting this so I don't have to keep scrolling up.*

- anonymous

i changed the background so its less menacing lol

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- anonymous

but anyways, MATH.

- anonymous

Lol okay. Now let's use the 1st and 3rd equations so we can get rid of y again.

- anonymous

alright

- anonymous

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- anonymous

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- anonymous

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- anonymous

Did you get that?

- anonymous

noooooooooooooooooo

- anonymous

What part did you get lost on?

- anonymous

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i dont understand any of it, how do we get to the answers that it wants, why am i so BAD at UNDERSTANDING THINGS

- anonymous

Oh we didn't get to finding the answers yet. We needed to find two equations to find x and z. We did those last two steps to get rid of y.

- anonymous

why did we multiply that row by 2 though?

- anonymous

To get rid of the y. You need to multiply the equation by a certain number to cancel out the y. In that case the y in the 1st equation is 2 while the 3rd equation's y is 1. So if you were to just look at\[2y\]&\[y\]How would you be able to get that y to equal the 2y?

- anonymous

multiply it by two but...

- anonymous

what happened to the row in the middle

- anonymous

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- anonymous

Oh it was cancelled because we need to cancel out variables to find x or z. In order to find what x y and z equals we need to have two variables in two equations in order to solve it. That's why we got rid of y.

- anonymous

Did you get that?

- anonymous

so...that whole row was canceled out?

- anonymous

No just y was cancelled out. We subtracted the two equations to get rid of y and have only x and z left in a new equation.

- anonymous

i dont get iiiit

- anonymous

Which part?

- anonymous

im a kinesthetic learner, i figure things out differently im sorry this must be frustrating.

- anonymous

everything

- anonymous

|dw:1438715563780:dw|

- anonymous

And it's okay.

- anonymous

Did you understand this picture?

- anonymous

no

- anonymous

Do you know where I got the 1st and 3rd equations from?

- anonymous

lets solve this whole thing and then break it down because i dont know where we are headed

- anonymous

no

- anonymous

Oh okay then hold on.

- anonymous

its really cool of you to help me

- anonymous

thank you

- anonymous

\[x+2y-2z=11\]\[5x+4y+7z=81\]\[-4x+y-6z=-37\]-----------------------
1st step:\[x+2y-2z=11\]\[5x+4y+7z=81\]-----------------------
\[2(x+2y-2z=11)\]\[5x+4y+7z=81\]-----------------------
\[2x+4y-4z=22\]\[5x+4y+7z=81\]-----------------------
\[(2x+4y-4z=22)\]minus \[(5x+4y+7z=81)\]equals\[(-3x+0y-11z=-59)\]OR\[-3x-11z=-59\](1st equation with no y's)
-----------------------
2nd step: (Pick another two sets of equations to cancel out y.)\[x+2y-2z=11\]\[-4x+y-6z=-37\]-----------------------\[x+2y-2z=11\]\[2(-4x+y-6z=-37)\]-----------------------\[x+2y-2z=11\]\[-8x+2y-12z=-74\]-----------------------\[x+2y-2z=11\]minus \[-8x+2y-12z=-74\]equals\[9x+0y+10z=85\]OR\[9x+10z=85\](2nd equation with no y's)
-----------------------

- anonymous

And you're welcome! :) This doesn't give you the answers yet but I'm just going to show you this part to see if you understand it.

- anonymous

oooooh

- anonymous

You understand it now?

- anonymous

yes

- anonymous

Okay great! I should have just showed you the steps earlier like that lol. XD

- anonymous

yeah omg

- anonymous

1st equation with no y's:\[-3x-11z=-59\]2nd equation with no y's:\[9x+10z=85\]Now we need to get rid of x's in this step since that's the easiest to multiply with.
We need to get -3x equal to 9x. We have to multiply the 1st equation with no y's by 3.\[3(-3x-11z=-59)\]\[9x+10z=85\]-----------------------\[-9x-33z=-177\]\[9x+10z=85\]-----------------------\[-9x-33z=-177\]plus \[9x+10z=85\]equals\[0x-23z=-92\]OR\[-23z=-92\]-----------------------
Now that we have one variable left, we can finally find what z equals. Divide -92 by -23.\[z=\frac{ -92 }{ -23 }\]\[z=4\]-----------------------
After finding one variable, you can now substitute the z in the equation with no y's. You can do either the 1st or 2nd equation with no y's. It doesn't matter.\[9x+10z=85\]\[9x+10(4)=85\]\[9x+40=85\]\[9x=85-40\]\[9x=45\]\[x=5\]-----------------------
Lastly we can solve for y by using the equations givin to us in the beginning of the question (from the matrices). Either 3 of the equations can work. Make sure to substitute the x and z.\[x+2y-2z=11\]\[(5)+2y-2(4)=11\]\[5+2y-8=11\]\[2y-3=11\]\[2y=11+3\]\[2y=14\]\[y=7\]

- anonymous

oh wow

- anonymous

Tell me if you understand that. This is the final step and it gives the answer.

- anonymous

that is a process

- anonymous

Ya it might look really long but don't let that scare you. Once you keep practicing it, it will be really easy! :)

- anonymous

omg i actually understand

- anonymous

Yay that's great! :D

- anonymous

see, im taking a summer course and its online, the section imtaking is all about that so im stuck

- anonymous

Oh okay. Well hopefully this was helpful in figuring out how to solve it. :)

- anonymous

it was very helpful!

- anonymous

Lol I'm glad!

- anonymous

i might need help going through another problem because i know there are exceptions

- anonymous

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- anonymous

Okay. If it's another problem can you write it in the Ask a question box? That way I can scroll through the steps that belong to that problem?

- anonymous

oh sure

- anonymous

Okay thanks!

- anonymous

so do i get rid of the x?

- anonymous

because 4 can go into the 8 or no

- anonymous

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- anonymous

Ya you can get rid of the x or y since you can just multiply the 3rd equation with 4 so it can be the same x and y in the 2nd equation.

- anonymous

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- anonymous

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- anonymous

Okay remember the 1st step. Just pick only 2 equations. Don't write all 3 down or else it will confuse you when you try to get rid of a variable like x y or z.

- anonymous

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