Need help with matrices 2x2 transformations

[terse orchid]

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[hard crag]

hey im the guy who asked what u meant, sorry i dont know about using matrices to transform graphs

im sure theres someone else who can help you out

good luck

i know about matrices but i didnt know u could use it to show graphical transformations

sorry

[lavish vale]

You'd need to write down the points

Say for point C you have [0 1] vertical

And, when multiplied by something, we gwt C'

Well, how would you solve the matrix equation

When you have XA=A'

We'd want to move the A to the other side. How would we do that?

We dont have division defined correct

So what you do is you multiply with the inverse on the right

$XAA^{-1}$

[livid aspen]

Ephesians 2:8-9

[lavish vale]

That way we get XI

And anything multiplied with the identity matrix is itself so we just get X

$XAA^{-1}=A'A^{-1}$

[livid aspen]

Ephesians 2:8-9

[lavish vale]

And what we do to one side, we also do to the other

$X=A'A^{-1}$ is what you're looking for

[livid aspen]

Ephesians 2:8-9

[lavish vale]

Then you'd need to find the inverse and multiply them

To find X

@cerulean spear did I lose you somewhere?

So

Hehe

Wdym

A is just the coordinate

Which can be represented as a 2x1 matrix

And A' is the resulting point, also a 2x1 matrix

I think that's the way to work out this problem. Let me know if it works

Yeah

Yeah, I overlooked that

Give me a moment

$\begin{bmatrix}
a & b\
c & d
\end{bmatrix}\begin{bmatrix}
1\
0
\end{bmatrix} = \begin{bmatrix}
2\
0
\end{bmatrix}$

[livid aspen]

Ephesians 2:8-9

[lavish vale]

So a×1+b×0=2 right?

@cerulean spear

What else can you get from there

a is 2 that's correct

c is not 2

c×1+d×0=0

Where are you from č

Lol

Its a letter in my language, that's why I ask

Anyways č=0 correct

We have a and c

Lets try to find the other two using another point

Cuz you cant figure it out from this one. The other two could be anything cuz theyre multiplied by 0

Try point C for example

Yeah

Looks good

And you can check if you multiply it by B whether you get B'

Looks good

So it's simpler than what I initally imagined

But hey you learned an additional technique when working with square matrices

Worked out all in all

Yep

Yeah the transform goes first

Yw