anonymous
  • anonymous
how do i find a solution to a system of equations represented by a matrix?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
this specifically
anonymous
  • anonymous
Well first you need to know how to read a matrix. Do you know how to read a matrix?
anonymous
  • anonymous
no....

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anonymous
  • anonymous
im kind of clueless
anonymous
  • anonymous
and im bad at asking for help
anonymous
  • anonymous
|dw:1438709526900:dw|
anonymous
  • 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
  • anonymous
aw, thank you
anonymous
  • anonymous
You're welcome! :)
anonymous
  • anonymous
itll help alot if i understand this
anonymous
  • anonymous
Alright I'll try the best I can in getting you to understand this. |dw:1438709667710:dw|
anonymous
  • 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
  • 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
  • anonymous
Did you get that so far?
anonymous
  • anonymous
uuhm..
anonymous
  • anonymous
oh, yeah okay i get it.
anonymous
  • anonymous
since one would be just x, sorry takes me awhile to work things out.
anonymous
  • anonymous
It's okay! Take your time! :)
anonymous
  • anonymous
Do you know what the other two equations will be if you used my method that I showed you?
anonymous
  • anonymous
5x+4y+7z=81?
anonymous
  • anonymous
Yes!
anonymous
  • anonymous
-4x+y-6z=-37
anonymous
  • anonymous
Great job! Now those three equations are needed when you are trying to solve equations within a matrix.
anonymous
  • anonymous
\[x+2y-2z=11\]\[5x+4y+7z=81\]\[-4x+y-6z=-37\]
anonymous
  • anonymous
Now do you know how to solve for x's and y's?
anonymous
  • anonymous
replace them with the numbers or..
anonymous
  • 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
  • anonymous
Let's start with the first two equations.
anonymous
  • anonymous
sorry, ah :(
anonymous
  • anonymous
|dw:1438710552563:dw|
anonymous
  • anonymous
And it's okay. That was a rhetorical question.
anonymous
  • anonymous
Oh wait let me rewrite what I wrote on the side.
anonymous
  • 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
  • anonymous
Tell me if you get that part.
anonymous
  • anonymous
hm..
anonymous
  • anonymous
i do, but at the same time i guess i need to know why we are doing that.
anonymous
  • anonymous
what is the reason for finding commons, is that like a big part of the equation?
anonymous
  • anonymous
what is the end result supposed to be? like, is the Y now 1? bleh...
anonymous
  • 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
  • anonymous
why are we doing that?
anonymous
  • 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
  • anonymous
Oh alright. so whenever im doing this i have to find a way to get rid of a variable?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
okay, so, like in this equation for example: |dw:1438711521979:dw|
anonymous
  • anonymous
just as an example
anonymous
  • 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
  • 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
  • anonymous
|dw:1438711724574:dw|
anonymous
  • anonymous
its an example problem of one i did and failed.
anonymous
  • anonymous
totally different problem sorry just trying to see where this applies to other problems too
anonymous
  • 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
  • anonymous
thank you, youre a saint ;-;
anonymous
  • anonymous
okay, so lets finish this problem first
anonymous
  • 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
  • anonymous
did you subtract?
anonymous
  • anonymous
im lost again
anonymous
  • anonymous
-3x-11z=-59 came from the drawing I drew earlier. Take a look at that and tell me if you get that.
anonymous
  • anonymous
where did the other two come from?
anonymous
  • anonymous
It came from the three equations we wrote from the matrix.
anonymous
  • 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
  • anonymous
Oh lol okay.
anonymous
  • anonymous
https://join.me/592-430-542 im sharing my screen with you, is that alright??
anonymous
  • 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
  • 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
  • anonymous
Oh okay. I'm really sorry about that! Are you able to draw here?
anonymous
  • anonymous
We're almost done anyways.
anonymous
  • anonymous
|dw:1438713360885:dw|
anonymous
  • anonymous
alright, sorry i scared you haha
anonymous
  • anonymous
the brushes on here kind of suck there for i kind of suck
anonymous
  • anonymous
It's okay you didn't know! :) And at least you draw better than me XD
anonymous
  • 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
  • anonymous
i changed the background so its less menacing lol
anonymous
  • anonymous
but anyways, MATH.
anonymous
  • anonymous
Lol okay. Now let's use the 1st and 3rd equations so we can get rid of y again.
anonymous
  • anonymous
alright
anonymous
  • anonymous
|dw:1438713734611:dw|
anonymous
  • anonymous
|dw:1438713838855:dw|
anonymous
  • anonymous
|dw:1438713940961:dw|
anonymous
  • anonymous
Did you get that?
anonymous
  • anonymous
noooooooooooooooooo
anonymous
  • anonymous
What part did you get lost on?
anonymous
  • anonymous
|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
anonymous
  • 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
  • anonymous
why did we multiply that row by 2 though?
anonymous
  • 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
  • anonymous
multiply it by two but...
anonymous
  • anonymous
what happened to the row in the middle
anonymous
  • anonymous
anonymous
  • 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
  • anonymous
Did you get that?
anonymous
  • anonymous
so...that whole row was canceled out?
anonymous
  • 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
  • anonymous
i dont get iiiit
anonymous
  • anonymous
Which part?
anonymous
  • anonymous
im a kinesthetic learner, i figure things out differently im sorry this must be frustrating.
anonymous
  • anonymous
everything
anonymous
  • anonymous
|dw:1438715563780:dw|
anonymous
  • anonymous
And it's okay.
anonymous
  • anonymous
Did you understand this picture?
anonymous
  • anonymous
no
anonymous
  • anonymous
Do you know where I got the 1st and 3rd equations from?
anonymous
  • anonymous
lets solve this whole thing and then break it down because i dont know where we are headed
anonymous
  • anonymous
no
anonymous
  • anonymous
Oh okay then hold on.
anonymous
  • anonymous
its really cool of you to help me
anonymous
  • anonymous
thank you
anonymous
  • 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
  • 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
  • anonymous
oooooh
anonymous
  • anonymous
You understand it now?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Okay great! I should have just showed you the steps earlier like that lol. XD
anonymous
  • anonymous
yeah omg
anonymous
  • 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
  • anonymous
oh wow
anonymous
  • anonymous
Tell me if you understand that. This is the final step and it gives the answer.
anonymous
  • anonymous
that is a process
anonymous
  • 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
  • anonymous
omg i actually understand
anonymous
  • anonymous
Yay that's great! :D
anonymous
  • anonymous
see, im taking a summer course and its online, the section imtaking is all about that so im stuck
anonymous
  • anonymous
Oh okay. Well hopefully this was helpful in figuring out how to solve it. :)
anonymous
  • anonymous
it was very helpful!
anonymous
  • anonymous
Lol I'm glad!
anonymous
  • anonymous
i might need help going through another problem because i know there are exceptions
anonymous
  • anonymous
|dw:1438718495088:dw|
anonymous
  • 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
  • anonymous
oh sure
anonymous
  • anonymous
Okay thanks!
anonymous
  • anonymous
so do i get rid of the x?
anonymous
  • anonymous
because 4 can go into the 8 or no
anonymous
  • anonymous
|dw:1438718635151:dw|
anonymous
  • 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
  • anonymous
|dw:1438718776581:dw|
anonymous
  • anonymous
|dw:1438718875081:dw|
anonymous
  • 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
  • anonymous
|dw:1438718996313:dw|
anonymous
  • anonymous
|dw:1438719010410:dw|
anonymous
  • anonymous
Let's use these 2 equations. These are the 2nd and 3rd equations.
anonymous
  • anonymous
Oh wait.
anonymous
  • anonymous
oh nevermind if you wanna use those lets use those
anonymous
  • anonymous
No wait you were fine!
anonymous
  • anonymous
i was?
anonymous
  • anonymous
Yes!!! You got that step right! And yes you have to subtract in order to get rid of the x's.
anonymous
  • anonymous
It doesn't matter which 2 equations you start off with btw.
anonymous
  • anonymous
|dw:1438719205198:dw|
anonymous
  • anonymous
did i do that right?.....
anonymous
  • anonymous
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.
anonymous
  • anonymous
like the second and the third
anonymous
  • anonymous
Ya you can use that. But make sure to cancel out the x again.
anonymous
  • anonymous
|dw:1438719564132:dw|
anonymous
  • anonymous
Make sure to put your variables in there so you won't make any mistakes.
anonymous
  • anonymous
|dw:1438719659795:dw|
anonymous
  • anonymous
|dw:1438719684956:dw|
anonymous
  • anonymous
|dw:1438719764031:dw|
anonymous
  • anonymous
Other than that you got it right :)
anonymous
  • anonymous
oops
anonymous
  • anonymous
then there would be no y?
anonymous
  • anonymous
Now subtract.
anonymous
  • anonymous
And yes sometimes that might happen but we will have z if you can cancel out x and y
anonymous
  • anonymous
|dw:1438719834634:dw|
anonymous
  • anonymous
ooooh
anonymous
  • anonymous
|dw:1438719888536:dw|
anonymous
  • anonymous
Yep! :)
anonymous
  • anonymous
Good job! Now we can substitute the z into the equation that we saved earlier.
anonymous
  • anonymous
ooh
anonymous
  • anonymous
|dw:1438719969259:dw|
anonymous
  • anonymous
|dw:1438719986098:dw|