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?

- dehelloo

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- jim_thompson5910

hint: get everything to one side in `x=y`

- dehelloo

Would I do that by dividing or adding y ?

- 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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## More answers

- jim_thompson5910

then try to undo that `+y`

- dehelloo

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

- jim_thompson5910

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

- dehelloo

-1

- dehelloo

1 -1
2 4

- 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

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

- jim_thompson5910

what would that inverse be

- dehelloo

[0.67 0.17]
[-0.33 0.17]

- jim_thompson5910

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

- dehelloo

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

- dehelloo

Yeah I think I got mate. Thanks.

- jim_thompson5910

I don't agree with the 1.67 part

- jim_thompson5910

it's close though

- dehelloo

1.16?

- jim_thompson5910

yes, or 1.167

- jim_thompson5910

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

- dehelloo

Thanks.

- dehelloo

Misscalc

- jim_thompson5910

no problem

- 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

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

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

- jim_thompson5910

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

- 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

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

- 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

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

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

`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

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

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

- 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

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

you're welcome

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