# Questions on Vectors and Matrices

#### Phavonic

Hi,

I have a few questions in this area which I am still stuck on. Hopefully you will give me some help....

1) How do I find the matrix of a transformation if only the coordinates are given?

For example, a triangle with coordinates (0,1) (3, 5) (1, 5) is reflected in the line x = 3, such that

[FONT=&quot]
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[FONT=&quot](? ?)x(0 3 1) = (6 3 5)[/FONT]
[FONT=&quot](? ?)x(1 5 5) = (1 5 5)

But how do I find the matrix that transforms this, given the new coordinates?

2) I also don't fully understand this question....

Write the statements x - 4y = 2m and 2y - x = n as vector equations....

3) A triangle ABC has the coordinates A = (0, 1), B = (3, 5), C = (1, 5)

M is a reflection in the line x = 0, mapping it onto A1B1C1, and T is a translation determined by the vector

(6)
(0)

Mapping the triangle A1B1C1 onto A2B2C2
Find the transformation K such that K = TM

So how to I multiply a matrix (M) with a vector (T)? If this is indeed what it is asking me.

(-1 0)x(6)
( 0 1)x(0)

Thanks
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#### Idea

(1)
the transformation is given by

$$T(x,y)=(-x+6,y)$$

this is not a linear transformation since it does not map $0$ to $0$

you could write it as a linear transformation followed by a translation

1 person

#### Phavonic

So if I've got this correctly, it means that only reflections in the lines y = 0 and x = 0 can be written as matrices.

And regarding rotations, I've only ever seen matrices denoting rotations of 90 or 180 degrees, never 60 or 45 (and always about the origin)

Are matrices generally limited in what they can or cannot represent?

I hope you can see where I'm coming from.

#### romsek

MHF Helper
So if I've got this correctly, it means that only reflections in the lines y = 0 and x = 0 can be written as matrices.

And regarding rotations, I've only ever seen matrices denoting rotations of 90 or 180 degrees, never 60 or 45 (and always about the origin)

Are matrices generally limited in what they can or cannot represent?

I hope you can see where I'm coming from.
You don't have it correctly.

A linear transformation is one of the type $x \to Ax$

$\forall A, ~A\begin{pmatrix}0\\0\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$

what isn't a linear transformation is

$x \to Ax + B,~\text{where$B \neq 0$. I believe the term for this sort of transformation is affine}$

a general rotation matrix counter clockwise is $R(\theta) = \begin{pmatrix}\cos(\theta)&-\sin(\theta) \\ \sin(\theta)&\cos(\theta)\end{pmatrix},~\forall \theta \in \mathbb{R}$

If it's a linear transformation then a matrix can represent it.

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1 person

#### Phavonic

Thanks romsek,

but I have to let you know that my maths ability is only at UK GCSE level (highschool?), so much of what you said is wasted on me. I have only just been introduced to the basics of matrices and vectors, hence the easy questions.

I'm looking it from the point of the reflection in question 3, with the same triangle reflected in the line x = 0

(-1 0)x(0 3 1)=(0 -3 -1)
( 0 1)x(1 5 5)=(1 5 5)

I am obviously naive in thinking that all reflections are linear transformations.

#### romsek

MHF Helper
ok let's look at the questions

First off it should be clear that a reflection about any line that doesn't pass through (0,0) cannot be a linear transformation.

what I think you are after is a sneaky little trick.

Suppose we consider the transformation $x \to A \begin{pmatrix}x\\y\\1\end{pmatrix}$

$\begin{pmatrix}6-x\\y\end{pmatrix}=A \begin{pmatrix}x\\y\\1\end{pmatrix}$

By inspection $A = \begin{pmatrix}-1 &0&6\\0&1&0\end{pmatrix}$

Now we can apply this to each of the triangle vertices.

For example $A\begin{pmatrix}0\\1\\1\end{pmatrix} = \begin{pmatrix}6\\1\end{pmatrix}$

(2) Is straightforward enough

$\begin{pmatrix}1&-4\\-1&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}2m\\n\end{pmatrix}$

(3) in next post

#### romsek

MHF Helper
(3) is confusing but I think we can apply the results of (1)

The reflection about the line $x=0$ is linear and has a matrix representation $M=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$

Using the same idea as in (1) we absorb the translation by augmenting the input vector with a 1.

I think we are after $y = Mx + \begin{pmatrix}6\\0\end{pmatrix}$

$y = \begin{pmatrix}-x\\y\end{pmatrix}+\begin{pmatrix}6\\0\end{pmatrix} =\begin{pmatrix}6-x\\y\end{pmatrix}$

we showed in (1) the matrix for this transformation on the augmented input vector is

$K=\begin{pmatrix}-1&0&6\\0&1&0\end{pmatrix}$

This could all be totally incorrect for what you're doing. I don't know. I don't have your textbook in front of me.

1 person