## dehelloo one year ago Help with solving systems of equations w/ Matrices x=y 2x+4y=7 I understand what to do if its nicely laid out like 5x+y=10 2x-4y=3 as the solution matrices is on the other side but what if its like the first example?

1. jim_thompson5910

hint: get everything to one side in x=y

2. dehelloo

Would I do that by dividing or adding y ?

3. jim_thompson5910

you can think of x=y as x=0+y the 0 added to anything doesn't change it

4. jim_thompson5910

then try to undo that +y

5. dehelloo

so it turns into x-y=0? 2x+4y=7 which turns into [1 1] [x ] = [0] [2 4 ] [y ] = [7]

6. jim_thompson5910

you made an error on your matrix but you have the correct system

7. dehelloo

-1

8. dehelloo

1 -1 2 4

9. jim_thompson5910

yeah it should be $\Large \begin{bmatrix} 1 & -1\\ 2 & 4\end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 0 \\ 7\end{bmatrix}$

10. dehelloo

so then I'd find the inverse of the coefficient matrix? and times that by the solution matrix?

11. jim_thompson5910

what would that inverse be

12. dehelloo

[0.67 0.17] [-0.33 0.17]

13. jim_thompson5910

as fractions you should have $\Large \begin{bmatrix} 2/3 & 1/6 \\ -1/3 & 1/6\end{bmatrix}$

14. dehelloo

and then times that by [0] [7] which should equal [1.67] [1.67]

15. dehelloo

Yeah I think I got mate. Thanks.

16. jim_thompson5910

I don't agree with the 1.67 part

17. jim_thompson5910

it's close though

18. dehelloo

1.16?

19. jim_thompson5910

yes, or 1.167

20. jim_thompson5910

you'll find that x = 7/6, y = 7/6

21. dehelloo

Thanks.

22. dehelloo

Misscalc

23. jim_thompson5910

no problem

24. dehelloo

So for any equations where there is a variable on both sides can we assume that it is 0+y if its just by itself?

25. jim_thompson5910

yeah you don't need to write 0+y I wrote that to show that you undo the +y by subtracting y from both sides

26. dehelloo

If it was 3x=y+1 y=x+2 Does it turn into 3x-y=1 -2=x-y

27. jim_thompson5910

then you can flip -2=x-y into x-y = -2 but yeah, you have it correct

28. dehelloo

So if I got -2=x-y can you just flip it round to x-y=-2 as they are the same equations?

29. jim_thompson5910

yeah because a = b is the same as b = a (symmetric property of equality)

30. dehelloo

Ok thanks Also when bring variables/coefficients to the other side for y=x+2 If I was to bring x to the other side wouldn't the variable be going after the y to make y-x=2 and then how would you get x-y=2? If that makes any sense.

31. jim_thompson5910

y-x=2 turns into -x+y = 2 I swapped the x and y terms you can think of the y as 0+y

32. dehelloo

So that would equal 3x=y+1 y=x+2 3x-y=1 -x+y=2 [3 1] [x] = 1 [-1 1] [y] = 2 [x]= [3 1]^-1 x [1] [y]= [-1 1] [2] x= -0.25 y= 1.75 Also, what is the point of completing an inverse to cancel out? Is it because we cant divide matrices?

33. jim_thompson5910

Is it because we cant divide matrices? you are correct in general we have Ax = b where A,x,b are matrices A is 2x2 x & b are 2x1 matrices there is no matrix division. So instead, you multiply both sides by the multiplicative inverse to isolate x $\Large Ax = b$ $\Large A^{-1}*Ax = A^{-1}*b$ $\Large x = A^{-1}*b$ it's similar to saying 2x = 5 turns into x = 5/2 after multiplying both sides by 1/2 1/2 is the multiplicative inverse of 2

34. dehelloo

Aaaah that makes sense so is a multiplicative inverse (m.i.) just a fraction of a singular portion like M.i. of 5 is 1/5 M.i. of 6 is 1/6?

35. jim_thompson5910

exactly, the idea is that the two pair up and multiply to 1 5*(1/5) = 1 6*(1/6) = 1

36. jim_thompson5910

with matrices, the original A and its inverse multiply to the identity matrix I (which is a lot like 1) $\Large A*A^{-1} = I$ $\Large A^{-1}*A = I$

37. dehelloo

Great. That makes a lot more sense. Time to hit some more exercises. Thank you for taking the time out of your day to help Jim.

38. jim_thompson5910

you're welcome