MHB Constructing a matrix version of the transformation algorithm?

[eleventhxhour]

Algorithms like the transformation algorithm: $(x, y)$ --> $(\frac{x}{k} + p, ay + d)$ are not generally used in mathematics. Instead, we use matrices.

Multiplying matrixes: you multiply a row of the first matrix by a column of the second. Use the following example:

$ \begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}a & b \\ c & d \end{bmatrix} = \begin{bmatrix}ax + cy & bx + dy \end{bmatrix} $

Use this information to construct a matrix version of the transformation algorithm for the transformation $y=af[k(x−p)]+d$

 

[I like Serena]

Algorithms like the transformation algorithm: $(x, y)$ --> $(\frac{x}{k} + p, ay + d)$ are not generally used in mathematics. Instead, we use matrices.

Multiplying matrixes: you multiply a row of the first matrix by a column of the second. Use the following example:

$ \begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}a & b \\ c & d \end{bmatrix} = \begin{bmatrix}ax + cy & bx + dy \end{bmatrix} $

Use this information to construct a matrix version of the transformation algorithm for the transformation $y=af[k(x−p)]+d$

Welcome to MHB, eleventhxhour!

Usually matrices are used if you have at least 2 variables (typically x and y) that are transformed in 2 new variables.

That is not the case for your transformation, so it makes little sense to write it in matrix notation.

Anyway, if you really want to, it could be written with 1x1 matrices as:
$$\begin{bmatrix} y \end{bmatrix}
= \begin{bmatrix} x−p \end{bmatrix} \begin{bmatrix} afk \end{bmatrix} + \begin{bmatrix} d \end{bmatrix}$$
or
$$\begin{bmatrix} y \end{bmatrix}
= \begin{bmatrix} x \end{bmatrix} \begin{bmatrix} afk \end{bmatrix} + \begin{bmatrix} d -afkp \end{bmatrix}
$$

Or a more advanced form:
$$\begin{bmatrix} y \end{bmatrix} = \begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} afk \\ d - afkp \end{bmatrix}$$

 

[Evgeny.Makarov]

Use this information to construct a matrix version of the transformation algorithm for the transformation $y=af[k(x−p)]+d$

I suspect that $f$ is a function here. Let $g(x)=af\big(k(x-p)\big)+d$. Then the required transformation probably converts the graph of $f(x)$ into the graph of $g(x)$. OP, please confirm if this is the correct interpretation.

 

[eleventhxhour]

I suspect that $f$ is a function here. Let $g(x)=af\big(k(x-p)\big)+d$. Then the required transformation probably converts the graph of $f(x)$ into the graph of $g(x)$. OP, please confirm if this is the correct interpretation.

Yes, I'm pretty sure that's the correct interpretation.

 

[I like Serena]

Yes, I'm pretty sure that's the correct interpretation.

What is the definition of $f$ then?
Is it perhaps related to your transformation algorithm: $(x, y) \to(\frac{x}{k} + p, ay + d)$?
It seems to fit.

 

[Evgeny.Makarov]

What is the definition of $f$ then?
Is it perhaps related to your transformation algorithm: $(x, y) \to(\frac{x}{k} + p, ay + d)$?
It seems to fit.

Yes, this transformation maps the graph of $f(x)$ into that of $g(x)$. It cannot be represented in the form
\[
(x',y')=(x,y)A\qquad(*)
\]
for a 2x2 matrix A. Indeed, (*) maps $(0,0)$ to $(0,0)$, while the transformation in the quote maps $(0,0)$ to $(p,d)$. But note that
\[
(x,y,1)
\begin{pmatrix}
a_{11}&a_{12}&0\\
a_{21}&a_{22}&0\\
a_{1}&a_{b2}&1
\end{pmatrix}
=(a_{11}x+a_{21}y+b_1, a_{12}x+a_{22}y+b_2,1)
\]
You need to come up with a matrix that produces
\[
((1/k)x +0y + p, 0x+ay + d,1)
\]

 

[eleventhxhour]

Yes, this transformation maps the graph of $f(x)$ into that of $g(x)$. It cannot be represented in the form
\[
(x',y')=(x,y)A\qquad(*)
\]
for a 2x2 matrix A. Indeed, (*) maps $(0,0)$ to $(0,0)$, while the transformation in the quote maps $(0,0)$ to $(p,d)$. But note that
\[
(x,y,1)
\begin{pmatrix}
a_{11}&a_{12}&0\\
a_{21}&a_{22}&0\\
a_{1}&a_{b2}&1
\end{pmatrix}
=a_{11}x+a_{21}y+b_1, a_{12}x+a_{22}y+b_2,1)
\]
You need to come up with a matrix that produces
\[
((1/k)x +0y + p, 0x+ay + d,1)
\]

Hmm..okay, I think I understand that a bit more. How would you create that matrix?

 

[Evgeny.Makarov]

How would you create that matrix?

You compare the two expressions
\[
(a_{11}x+a_{21}y+b_1, a_{12}x+a_{22}y+b_2,1)
\]
and
\[
((1/k)x +0y + p, 0x+ay + d,1)
\]
and come up with appropriate $a_{ij}$ and $b_i$ so that the expressions coincide.

 

[johng1]

Hi,
I hope the following discussion helps: