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

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

Mathematics
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this specifically
Well first you need to know how to read a matrix. Do you know how to read a matrix?
no....

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im kind of clueless
and im bad at asking for help
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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.
aw, thank you
You're welcome! :)
itll help alot if i understand this
Alright I'll try the best I can in getting you to understand this. |dw:1438709667710:dw|
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.
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:\[x+2y-2z=11\]
Did you get that so far?
uuhm..
oh, yeah okay i get it.
since one would be just x, sorry takes me awhile to work things out.
It's okay! Take your time! :)
Do you know what the other two equations will be if you used my method that I showed you?
5x+4y+7z=81?
Yes!
-4x+y-6z=-37
Great job! Now those three equations are needed when you are trying to solve equations within a matrix.
\[x+2y-2z=11\]\[5x+4y+7z=81\]\[-4x+y-6z=-37\]
Now do you know how to solve for x's and y's?
replace them with the numbers or..
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.
Let's start with the first two equations.
sorry, ah :(
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And it's okay. That was a rhetorical question.
Oh wait let me rewrite what I wrote on the side.
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.
Tell me if you get that part.
hm..
i do, but at the same time i guess i need to know why we are doing that.
what is the reason for finding commons, is that like a big part of the equation?
what is the end result supposed to be? like, is the Y now 1? bleh...
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.
why are we doing that?
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.
Oh alright. so whenever im doing this i have to find a way to get rid of a variable?
Yes.
okay, so, like in this equation for example: |dw:1438711521979:dw|
just as an example
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
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.
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its an example problem of one i did and failed.
totally different problem sorry just trying to see where this applies to other problems too
You understood more than before at least :) And oh ok. I'll help you out with those if you don't understand still.
thank you, youre a saint ;-;
okay, so lets finish this problem first
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.
did you subtract?
im lost again
-3x-11z=-59 came from the drawing I drew earlier. Take a look at that and tell me if you get that.
where did the other two come from?
It came from the three equations we wrote from the matrix.
|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
Oh lol okay.
https://join.me/592-430-542 im sharing my screen with you, is that alright??
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.
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
Oh okay. I'm really sorry about that! Are you able to draw here?
We're almost done anyways.
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alright, sorry i scared you haha
the brushes on here kind of suck there for i kind of suck
It's okay you didn't know! :) And at least you draw better than me XD
\[x+2y-2z=11\]\[5x+4y+7z=81\]\[-4x+y-6z=-37\]*Just rewriting this so I don't have to keep scrolling up.*
i changed the background so its less menacing lol
but anyways, MATH.
Lol okay. Now let's use the 1st and 3rd equations so we can get rid of y again.
alright
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Did you get that?
noooooooooooooooooo
What part did you get lost on?
|dw:1438714161470:dw| i dont understand any of it, how do we get to the answers that it wants, why am i so BAD at UNDERSTANDING THINGS
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.
why did we multiply that row by 2 though?
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?
multiply it by two but...
what happened to the row in the middle
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.
Did you get that?
so...that whole row was canceled out?
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.
i dont get iiiit
Which part?
im a kinesthetic learner, i figure things out differently im sorry this must be frustrating.
everything
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And it's okay.
Did you understand this picture?
no
Do you know where I got the 1st and 3rd equations from?
lets solve this whole thing and then break it down because i dont know where we are headed
no
Oh okay then hold on.
its really cool of you to help me
thank you
\[x+2y-2z=11\]\[5x+4y+7z=81\]\[-4x+y-6z=-37\]----------------------- 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) -----------------------
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.
oooooh
You understand it now?
yes
Okay great! I should have just showed you the steps earlier like that lol. XD
yeah omg
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.\[x+2y-2z=11\]\[(5)+2y-2(4)=11\]\[5+2y-8=11\]\[2y-3=11\]\[2y=11+3\]\[2y=14\]\[y=7\]
oh wow
Tell me if you understand that. This is the final step and it gives the answer.
that is a process
Ya it might look really long but don't let that scare you. Once you keep practicing it, it will be really easy! :)
omg i actually understand
Yay that's great! :D
see, im taking a summer course and its online, the section imtaking is all about that so im stuck
Oh okay. Well hopefully this was helpful in figuring out how to solve it. :)
it was very helpful!
Lol I'm glad!
i might need help going through another problem because i know there are exceptions
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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?
oh sure
Okay thanks!
so do i get rid of the x?
because 4 can go into the 8 or no
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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.
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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.
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Let's use these 2 equations. These are the 2nd and 3rd equations.
Oh wait.
oh nevermind if you wanna use those lets use those
No wait you were fine!
i was?
Yes!!! You got that step right! And yes you have to subtract in order to get rid of the x's.
It doesn't matter which 2 equations you start off with btw.
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did i do that right?.....
Yes you were right again! :) Save this equation. Now we need to go back to the 3 equations and try to do the same process except use different equations.
like the second and the third
Ya you can use that. But make sure to cancel out the x again.
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Make sure to put your variables in there so you won't make any mistakes.
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