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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?

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no...but remember you can check for yourself, plug in those values
.... that was my line panlac lol
my bad... I can delete it

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no need to delete :)
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?
is this cramer rule? or row operations?
I think row of operations is the way to go.
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.
yeah you have to divide the whole row by 5 then
hey cutie,have u been taught matrix inversion?which makes the solution much simpler.....
@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.
Well when we row reduce we want every number bellow that 1 to be a zero
|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
Why would you multiply the first row by two?
Oh wait, nevermind! I gotcha now!
then divide the second row by 8.20 and then subtract the the first row by the second.
so the bottom row would be 0 1 -1 ? And would that give you.. 1 2/5 13/5 ?
|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.
Well no I see that I want to make the first row second column zero right. I'm talking about this number|dw:1344704459156:dw|
so I then multiply the second row by\[\frac{3}{5}\]and then add it to the first row
multiply the second row by \[\frac{5}{3}\]now to get them back to 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 \[\left(\begin{matrix}5 & -3 \\ 2 & 7\end{matrix}\right)^{-1} = \left(\begin{matrix}7/41 & 3/41\\ -2/41 & 5/41\end{matrix}\right)\] \[\rightarrow \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}3 \\ -1\end{matrix}\right)\]
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.
And I made a mistake I left out a negative sign here|dw:1344704797494:dw|
Thanks @dumbcow It's always good to show that there is more than one way to solve a problem.
Oh, so when you multiply and add and what not, does it not apply to the first number in the row?
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.
And when you add or subtract a zero from a number it stays the same.
Okay, I gotcha. Thanks for the explanation. :) It really helped!
No problem sorry for the messy work hope you could understand it .
I did :)

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