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cutie.patootie
 3 years 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?
cutie.patootie
 3 years 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?

This Question is Closed

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1no...but remember you can check for yourself, plug in those values

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0.... that was my line panlac lol

panlac01
 3 years ago
Best ResponseYou've already chosen the best response.1my bad... I can delete it

cutie.patootie
 3 years ago
Best ResponseYou've already chosen the best response.0Oh 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?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0is this cramer rule? or row operations?

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

cutie.patootie
 3 years ago
Best ResponseYou've already chosen the best response.0Uhh. 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.

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

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0hey cutie,have u been taught matrix inversion?which makes the solution much simpler.....

cutie.patootie
 3 years ago
Best 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.

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

RyanL.
 3 years ago
Best ResponseYou've already chosen the best response.1dw: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

cutie.patootie
 3 years ago
Best ResponseYou've already chosen the best response.0Why would you multiply the first row by two?

cutie.patootie
 3 years ago
Best ResponseYou've already chosen the best response.0Oh wait, nevermind! I gotcha now!

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

cutie.patootie
 3 years ago
Best ResponseYou've already chosen the best response.0so the bottom row would be 0 1 1 ? And would that give you.. 1 2/5 13/5 ?

RyanL.
 3 years ago
Best ResponseYou've already chosen the best response.1dw: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.

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

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

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

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1tough 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 \[\left(\begin{matrix}5 & 3 \\ 2 & 7\end{matrix}\right)^{1} = \left(\begin{matrix}7/41 & 3/41\\ 2/41 & 5/41\end{matrix}\right)\] \[\rightarrow \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}3 \\ 1\end{matrix}\right)\]

RyanL.
 3 years ago
Best ResponseYou've already chosen the best response.1Yeah 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.

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

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

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

RyanL.
 3 years ago
Best ResponseYou've already chosen the best response.1Well 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.

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

cutie.patootie
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, I gotcha. Thanks for the explanation. :) It really helped!

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