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Check my answer
Solve with matrices.
5x 3y = 18 and 2x + 7y = 1
I got ....
x=31/18
y=3/5
Am I right?
 one year ago
 one year ago
Check my answer Solve with matrices. 5x 3y = 18 and 2x + 7y = 1 I got .... x=31/18 y=3/5 Am I right?
 one year ago
 one year ago

This Question is Closed

dumbcowBest ResponseYou've already chosen the best response.1
no...but remember you can check for yourself, plug in those values
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
.... that was my line panlac lol
 one year ago

panlac01Best ResponseYou've already chosen the best response.1
my bad... I can delete it
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
no need to delete :)
 one year ago

cutie.patootieBest ResponseYou've already chosen the best response.0
Oh okay, well I did & it wasn't.. Crud. Well I am confused on part of it, so when I have to do one row of the matrices that is 5 13 18 to get it to 1 0 c1 How do I do that?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
is this cramer rule? or row operations?
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
I think row of operations is the way to go.
 one year ago

cutie.patootieBest ResponseYou've already chosen the best response.0
Uhh. I don't know.. But I have to get the row to 1 0 c1, like 5 has to change to 1 and 13 has to change to 0 somehow.
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
yeah you have to divide the whole row by 5 then
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
hey cutie,have u been taught matrix inversion?which makes the solution much simpler.....
 one year ago

cutie.patootieBest ResponseYou've already chosen the best response.0
@RyanL. Well that would give me 1 13/5 18/5. How would I get 13/5 to zero with out changing the 1? @hartnn Maybe, math isn't my strong point to I have no idea the fancy names... Sorry.
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
Well when we row reduce we want every number bellow that 1 to be a zero
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
dw:1344704018436:dw well you need to subtract the second row by the first row but before that you need to multiply the first row by 2
 one year ago

cutie.patootieBest ResponseYou've already chosen the best response.0
Why would you multiply the first row by two?
 one year ago

cutie.patootieBest ResponseYou've already chosen the best response.0
Oh wait, nevermind! I gotcha now!
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
then divide the second row by 8.20 and then subtract the the first row by the second.
 one year ago

cutie.patootieBest ResponseYou've already chosen the best response.0
so the bottom row would be 0 1 1 ? And would that give you.. 1 2/5 13/5 ?
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
dw:1344704298500:dw I like to make them 1s so if I need to divide to multiply it it's going to be easier to do it later.
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
Well no I see that I want to make the first row second column zero right. I'm talking about this numberdw:1344704459156:dw
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
so I then multiply the second row by\[\frac{3}{5}\]and then add it to the first row
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
multiply the second row by \[\frac{5}{3}\]now to get them back to 1
 one year ago

dumbcowBest ResponseYou've already chosen the best response.1
tough to teach row operations online...there are a lot of steps use matrix algebra \[Ax = b\] \[x=A^{1}b\] where A is coefficient matrix and b is right side constants \[\left(\begin{matrix}5 & 3 \\ 2 & 7\end{matrix}\right)\left(\begin{matrix}x \\ y\end{matrix}\right)= \left(\begin{matrix}18 \\ 1\end{matrix}\right)\] \[\left(\begin{matrix}x \\ y\end{matrix}\right)= \left(\begin{matrix}5 & 3 \\ 2 & 7\end{matrix}\right)^{1}\left(\begin{matrix}18 \\ 1\end{matrix}\right)\] det(A) = 5*7  2*3 = 41 \[\left(\begin{matrix}5 & 3 \\ 2 & 7\end{matrix}\right)^{1} = \left(\begin{matrix}7/41 & 3/41\\ 2/41 & 5/41\end{matrix}\right)\] \[\rightarrow \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}3 \\ 1\end{matrix}\right)\]
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
Yeah but row operations is essential to everything that you will learn ahead so you need to be able to do them at ease. After a while it takes seconds.
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
And I made a mistake I left out a negative sign heredw:1344704797494:dw
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
Thanks @dumbcow It's always good to show that there is more than one way to solve a problem.
 one year ago

cutie.patootieBest ResponseYou've already chosen the best response.0
Oh, so when you multiply and add and what not, does it not apply to the first number in the row?
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
Well I think you mean when we add the second row to the first. The thing is that the first number is a zero so if you multiply it a zero by a number or divide it by another number it will remain zero so when we add it to the first row we will be adding or subtracting a zero.
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
And when you add or subtract a zero from a number it stays the same.
 one year ago

cutie.patootieBest ResponseYou've already chosen the best response.0
Okay, I gotcha. Thanks for the explanation. :) It really helped!
 one year ago

RyanL.Best ResponseYou've already chosen the best response.1
No problem sorry for the messy work hope you could understand it .
 one year ago
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