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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
2x4y=3
as the solution matrices is on the other side but what if its like the first example?
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 2x4y=3 as the solution matrices is on the other side but what if its like the first example?

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1hint: get everything to one side in `x=y`

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1Would I do that by dividing or adding y ?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you can think of `x=y` as `x=0+y` the 0 added to anything doesn't change it

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1then try to undo that `+y`

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1so it turns into xy=0? 2x+4y=7 which turns into [1 1] [x ] = [0] [2 4 ] [y ] = [7]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you made an error on your matrix but you have the correct system

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah it should be \[\Large \begin{bmatrix} 1 & 1\\ 2 & 4\end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 0 \\ 7\end{bmatrix}\]

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1so then I'd find the inverse of the coefficient matrix? and times that by the solution matrix?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1what would that inverse be

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1[0.67 0.17] [0.33 0.17]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1as fractions you should have \[\Large \begin{bmatrix} 2/3 & 1/6 \\ 1/3 & 1/6\end{bmatrix}\]

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1and then times that by [0] [7] which should equal [1.67] [1.67]

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1Yeah I think I got mate. Thanks.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I don't agree with the 1.67 part

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1it's close though

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yes, or 1.167

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you'll find that x = 7/6, y = 7/6

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1So for any equations where there is a variable on both sides can we assume that it is 0+y if its just by itself?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah you don't need to write 0+y I wrote that to show that you undo the +y by subtracting y from both sides

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1If it was 3x=y+1 y=x+2 Does it turn into 3xy=1 2=xy

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1then you can flip `2=xy` into `xy = 2` but yeah, you have it correct

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1So if I got 2=xy can you just flip it round to xy=2 as they are the same equations?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah because `a = b` is the same as `b = a` (symmetric property of equality)

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1Ok 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 yx=2 and then how would you get xy=2? If that makes any sense.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yx=2 turns into x+y = 2 I swapped the x and y terms you can think of the y as 0+y

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1So that would equal 3x=y+1 y=x+2 3xy=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?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1`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

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1Aaaah 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?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1exactly, the idea is that the two pair up and multiply to 1 5*(1/5) = 1 6*(1/6) = 1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1with matrices, the original A and its inverse multiply to the identity matrix I (which is a lot like 1) \[\Large A*A^{1} = I\] \[\Large A^{1}*A = I\]

dehelloo
 one year ago
Best ResponseYou've already chosen the best response.1Great. 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.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you're welcome
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