Question

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?

Answer 1

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

Answer 2

Would I do that by dividing or adding y ?

Answer 3

you can think of `x=y` as `x=0+y`

the 0 added to anything doesn't change it

Answer 4

then try to undo that `+y`

Answer 5

so it turns into
x-y=0?
2x+4y=7

which turns into
[1 1] [x ] = [0]
[2 4 ] [y ] = [7]

Answer 6

you made an error on your matrix

but you have the correct system

Answer 7

-1

Answer 8

1 -1
2 4

Answer 9

yeah it should be

\[\Large \begin{bmatrix} 1 & -1\\ 2 & 4\end{bmatrix}
\begin{bmatrix} x \\ y\end{bmatrix}
=
\begin{bmatrix} 0 \\ 7\end{bmatrix}\]

Answer 10

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

Answer 11

what would that inverse be

Answer 12

[0.67 0.17]
[-0.33 0.17]

Answer 13

as fractions you should have

\[\Large \begin{bmatrix} 2/3 & 1/6 \\ -1/3 & 1/6\end{bmatrix}\]

Answer 14

and then times that by [0]
[7]

which should equal
[1.67]
[1.67]

Answer 15

Yeah I think I got mate. Thanks.

Answer 16

I don't agree with the 1.67 part

Answer 17

it's close though

Answer 18

1.16?

Answer 19

yes, or 1.167

Answer 20

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

Answer 21

Thanks.

Answer 22

Misscalc

Answer 23

no problem

Answer 24

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?

Answer 25

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

Answer 26

If it was
3x=y+1
y=x+2

Does it turn into
3x-y=1
-2=x-y

Answer 27

then you can flip `-2=x-y` into `x-y = -2`

but yeah, you have it correct

Answer 28

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

Answer 29

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

Answer 30

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.

Answer 31

y-x=2 turns into -x+y = 2

I swapped the x and y terms

you can think of the y as 0+y

Answer 32

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?

Answer 33

`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

Answer 34

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?

Answer 35

exactly, the idea is that the two pair up and multiply to 1

5*(1/5) = 1
6*(1/6) = 1

Answer 36

with matrices, the original A and its inverse multiply to the identity matrix I (which is a lot like 1)

\[\Large A*A^{-1} = I\]
\[\Large A^{-1}*A = I\]

Answer 37

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.

Answer 38

you're welcome