02Gmorematrices

= **Transformations Using Matrices (continued)** = toc Return for Dilations and Reflections of Matrices

Translations
(add transformation matrix)

math \left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] + \left[ \begin{matrix} h \\ k \\ \end{matrix} \right] = \left[ \begin{matrix} x+h \\ y+k \\ \end{matrix} \right] math

This represents a translation of +h units in the x direction and +k units in the y direction.

math \text{ For example, the transformation matrix } \left[ \begin{matrix} 2 \\ -1 \\ \end{matrix} \right] \text{ gives:} math

math \left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] + \left[ \begin{matrix} 2 \\ -1 \\ \end{matrix} \right] = \left[ \begin{matrix} x+2 \\ y-1 \\ \end{matrix} \right] math

(or)

math T \left( \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] \right) = \left[ \begin{matrix} x+2 \\ y-1 \\ \end{matrix} \right] math

(or) x' ``=`` x + 2 and y' ``=`` y – 1

This represents:
 * a translation of 2 units to the right (+2 in the x direction) and
 * a translation of 1 unit down (–1 unit in the y direction).

For example, if we start with the point (5, 3) under this transformation, the image is (7, 2)

Compare this to starting with a standard cubic graph y = x 3 under this transformation, the image is y' = (x' – 2) 3 – 1

We can arrive at the new equation algebraically: y' ``=`` y – 1 and y ``=`` x 3 so: y' = x 3 – 1

but (by rearranging x' = x + 2 to make x the subject) x = x' – 2 so y' = (x' – 2) 3 – 1

In __**function notation**__, this transformation could be written as: {T: R 2 → R 2, T(x, y) → (x+2, y–1)} or in shorthand as (x', y') → (x+2, y–1)

Combinations
Reflections, dilations and translations can be combined into one matrix, provided care is taken to express them in the correct order as specified by the question.

** Example **

 * a reflection across the x-axis (in the y-direction)
 * a dilation by a factor of 4 in the y-direction (from the x-axis)
 * a translation of +3 units in the x-direction

In matrix notation, this would be:

math \left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} 1&0 \\ 0&-4 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] + \left[ \begin{matrix} 3 \\ 0 \\ \end{matrix} \right] = \left[ \begin{matrix} x+3 \\ -4y \\ \end{matrix} \right] math

(or)

math T \left( \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] \right) = \left[ \begin{matrix} x+3 \\ -4y \\ \end{matrix} \right] math

(or) x' ``=`` x + 3 and y' ``=`` –4y

for example, if we start with the point (5, 3) under this transformation, the image is (8, –12)

Compare this to starting with a standard cubic graph y = x 3 under this transformation, the image is y' = –4(x' – 3) 3

We could arrive at the new equation algebraically: y' ``=`` –4y and y ``=`` x 3 so: y' = –4x 3

but (by rearranging to make x the subject): x = x' – 3 so:

y' = –4(x' – 3) 3

In __**function notation**__, this transformation could be written as: {T: R 2 → R 2, T(x, y) → (x+3, –4y)} or in shorthand as (x', y') → (x+3, –4y)

Transformations of graphs
To transform y = f(x) into math y' = af\big( n(x'-h) \big) +k math

requires the following matrix equation: math \left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} \frac{1}{n}&0 \\ 0&a \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] + \left[ \begin{matrix} h \\ k \\ \end{matrix} \right] math

** Example 1 **
The equation y = x 2 is transformed with:

math T \left( \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] \right) = \left[ \begin{matrix} 2x+3 \\ -4y + 1 \\ \end{matrix} \right] math

Using matrices, the transformation would look like this: math \left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} 2&0 \\ 0&-4 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] + \left[ \begin{matrix} 3 \\ 1 \\ \end{matrix} \right] math

From this, you should be able to write the values into the image equation: math y' = -4 \big( \frac{1}{2} (x'-3) \big)^2 + 1 math which simplifies to: math y' = - \big( x'-3 \big)^2 + 1 math

Or, you may want to do this using algebra: x’ = 2x + 3 (and) y’ = –4y + 1

Rearrange to make x the subject gives: math x = \dfrac{x'-3}{2} \qquad \textbf{(1)} math

Using y' ``=`` –4y and y ``=`` x 2 to get: y' = –4x 2

Now substitute **(1)** to get: math \\ y' = -4 \left( \dfrac{x'-3}{2} \right)^2 \\ \\ y' = - \big( x'-3 \big)^2 math

** Example 2 **
math \\ \text{Write in matrix form, the transformation that changed} \\ . \qquad \text{the function } y = \sqrt{x} \\ . \qquad \text{into } y = 2\sqrt{5-x}+3 math

{first rewrite the image function into a more standard form} math \\ y = 2\sqrt{-x+5}+3 \\. \\ y = 2\sqrt{-(x-5)}+3 math

{now write a matrix equation, putting the different transformations into their correct places} math \left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} -1&0 \\ 0&2 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] + \left[ \begin{matrix} 5 \\ 3 \\ \end{matrix} \right] math

Domain and Range
The domain of x’ and the range of y’ can be found by applying the same transformations to the domain of x and the range of y.

For example, if we had: x' = x + 3 and y' = –4y

a) if the original domain was x > 0, given that x’ = x + 3, image domain is x’ > 3

b) if the original range was y Î (2, 5], given that y’ = –4y image range is y’ Î (–8, –20] or (more correctly), y’ Î [–20, –8)

Return to Part 1 of Transformations with Matrices
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