dehelloo
  • dehelloo
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?
Mathematics
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katieb
  • katieb
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jim_thompson5910
  • jim_thompson5910
hint: get everything to one side in `x=y`
dehelloo
  • dehelloo
Would I do that by dividing or adding y ?
jim_thompson5910
  • jim_thompson5910
you can think of `x=y` as `x=0+y` the 0 added to anything doesn't change it

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jim_thompson5910
  • jim_thompson5910
then try to undo that `+y`
dehelloo
  • dehelloo
so it turns into x-y=0? 2x+4y=7 which turns into [1 1] [x ] = [0] [2 4 ] [y ] = [7]
jim_thompson5910
  • jim_thompson5910
you made an error on your matrix but you have the correct system
dehelloo
  • dehelloo
-1
dehelloo
  • dehelloo
1 -1 2 4
jim_thompson5910
  • jim_thompson5910
yeah it should be \[\Large \begin{bmatrix} 1 & -1\\ 2 & 4\end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 0 \\ 7\end{bmatrix}\]
dehelloo
  • dehelloo
so then I'd find the inverse of the coefficient matrix? and times that by the solution matrix?
jim_thompson5910
  • jim_thompson5910
what would that inverse be
dehelloo
  • dehelloo
[0.67 0.17] [-0.33 0.17]
jim_thompson5910
  • jim_thompson5910
as fractions you should have \[\Large \begin{bmatrix} 2/3 & 1/6 \\ -1/3 & 1/6\end{bmatrix}\]
dehelloo
  • dehelloo
and then times that by [0] [7] which should equal [1.67] [1.67]
dehelloo
  • dehelloo
Yeah I think I got mate. Thanks.
jim_thompson5910
  • jim_thompson5910
I don't agree with the 1.67 part
jim_thompson5910
  • jim_thompson5910
it's close though
dehelloo
  • dehelloo
1.16?
jim_thompson5910
  • jim_thompson5910
yes, or 1.167
jim_thompson5910
  • jim_thompson5910
you'll find that x = 7/6, y = 7/6
dehelloo
  • dehelloo
Thanks.
dehelloo
  • dehelloo
Misscalc
jim_thompson5910
  • jim_thompson5910
no problem
dehelloo
  • 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?
jim_thompson5910
  • 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
dehelloo
  • dehelloo
If it was 3x=y+1 y=x+2 Does it turn into 3x-y=1 -2=x-y
jim_thompson5910
  • jim_thompson5910
then you can flip `-2=x-y` into `x-y = -2` but yeah, you have it correct
dehelloo
  • dehelloo
So if I got -2=x-y can you just flip it round to x-y=-2 as they are the same equations?
jim_thompson5910
  • jim_thompson5910
yeah because `a = b` is the same as `b = a` (symmetric property of equality)
dehelloo
  • 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.
jim_thompson5910
  • 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
dehelloo
  • 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?
jim_thompson5910
  • 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
dehelloo
  • 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?
jim_thompson5910
  • jim_thompson5910
exactly, the idea is that the two pair up and multiply to 1 5*(1/5) = 1 6*(1/6) = 1
jim_thompson5910
  • jim_thompson5910
with matrices, the original A and its inverse multiply to the identity matrix I (which is a lot like 1) \[\Large A*A^{-1} = I\] \[\Large A^{-1}*A = I\]
dehelloo
  • 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.
jim_thompson5910
  • jim_thompson5910
you're welcome

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