## anonymous 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?

1. anonymous

no...but remember you can check for yourself, plug in those values

2. amistre64

.... that was my line panlac lol

3. anonymous

my bad... I can delete it

4. amistre64

no need to delete :)

5. anonymous

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?

6. amistre64

is this cramer rule? or row operations?

7. anonymous

I think row of operations is the way to go.

8. anonymous

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.

9. anonymous

yeah you have to divide the whole row by 5 then

10. hartnn

hey cutie,have u been taught matrix inversion?which makes the solution much simpler.....

11. anonymous

@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.

12. anonymous

Well when we row reduce we want every number bellow that 1 to be a zero

13. anonymous

|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

14. anonymous

Why would you multiply the first row by two?

15. anonymous

Oh wait, nevermind! I gotcha now!

16. anonymous

|dw:1344704116600:dw|

17. anonymous

|dw:1344704166215:dw|

18. anonymous

then divide the second row by 8.20 and then subtract the the first row by the second.

19. anonymous

so the bottom row would be 0 1 -1 ? And would that give you.. 1 2/5 13/5 ?

20. anonymous

|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.

21. anonymous

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|

22. anonymous

so I then multiply the second row by$\frac{3}{5}$and then add it to the first row

23. anonymous

|dw:1344704521394:dw|

24. anonymous

|dw:1344704554208:dw|

25. anonymous

|dw:1344704592083:dw|

26. anonymous

multiply the second row by $\frac{5}{3}$now to get them back to 1

27. anonymous

|dw:1344704647344:dw|

28. anonymous

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 $\left(\begin{matrix}5 & -3 \\ 2 & 7\end{matrix}\right)^{-1} = \left(\begin{matrix}7/41 & 3/41\\ -2/41 & 5/41\end{matrix}\right)$ $\rightarrow \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}3 \\ -1\end{matrix}\right)$

29. anonymous

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.

30. anonymous

And I made a mistake I left out a negative sign here|dw:1344704797494:dw|

31. anonymous

Thanks @dumbcow It's always good to show that there is more than one way to solve a problem.

32. anonymous

Oh, so when you multiply and add and what not, does it not apply to the first number in the row?

33. anonymous

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.

34. anonymous

And when you add or subtract a zero from a number it stays the same.

35. anonymous

Okay, I gotcha. Thanks for the explanation. :) It really helped!

36. anonymous

No problem sorry for the messy work hope you could understand it .

37. anonymous

I did :)