03.5matrixtransform2

# Transformations Using Matrices (continued)

## Translations

Translations are obtained by adding a transformation matrix to $\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$

.

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

.

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

This represents a translation of:

• +h units parallel to the y-axis (in the x direction) and
• +k units parallel to the x-axis (in the y direction).

.

For example,
… … $T = \left[ \begin{matrix} 2 \\ -1 \\ \end{matrix} \right]$
gives
.

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

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

(or)

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

(or)
… … $x' = x + 2$ … and
… … $y' = y - 1$

This represents:

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

.

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, making x the subject in $x' = x + 2$

… … $x = x' - 2$

so

… … $y' = (x' - 2)^3 - 1$

.

In function notation, this transformation could be written as:

… … $\big\{T: R^2 \rightarrow R^2,\;\; T(x,\; y) \rightarrow (x+2, \; y-1) \big\}$
.
or in shorthand as

… … $(x', \; y') \rightarrow (x+2, \; y-1)$

.

## 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.
.

For Example

Consider the following transformations:

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

In matrix notation, this would be:

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

… … $\left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} x+3 \\ -4y \\ \end{matrix} \right]$

(or)

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

(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 making x the subject):

… … $x = x' - 3$

so:
… … $y' = -4(x' - 3)^3$

.

In function notation, this transformation could be written as:

… … $\big\{T: R^2 \rightarrow R^2, \;\; T(x,\; y) \rightarrow (x+3, \; -4y) \big\}$
.

or in shorthand as

… … $(x', \; y') \rightarrow (x+3, \; -4y)$

.

## Transformations of graphs

To transform $y = f(x)$ into $y' = af\big( n(x'-h) \big) +k$

requires the following matrix equation:

… … $\left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} \frac{1}{n}&0 \\ 0&a \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] + \left[ \begin{matrix} h \\ k \\ \end{matrix} \right]$

.

#### Example 1

The equation $y = x^2$ is transformed with:

… … $\left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} 2&0 \\ 0&-4 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] + \left[ \begin{matrix} 3 \\ 1 \\ \end{matrix} \right]$

Write the new equation

Solution 1

From the matrix equation, you may be able to write the values into the image equation:

… … $y' = -4 \big( \frac{1}{2} (x'-3) \big)^2 + 1$

which simplifies to:

… … $y' = - \big( x'-3 \big)^2 + 1$

Solution 2

When multiplied out, the matrix equation gives

… … $\left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} 2x+3 \\ -4y + 1 \\ \end{matrix} \right]$

or

… … $x’ = 2x + 3$ (and)

…. … $y’ = -4y + 1$

we have

…. … $y’ = -4y + 1$ … and
… … $y = x^2$

so
… … $y' = -4x^2 + 1$

Rearrange to make x the subject gives:

… … $x = \dfrac{x'-3}{2}$

so
… … $y' = -4 \left( \dfrac{x'-3}{2} \right)^2 + 1$

or
… … $y' = - \big( x'-3 \big)^2 + 1$

.

#### Example 2

Write in matrix form, the transformation that changed the function $y = \sqrt{x}$

into $y = 2\sqrt{5-x}+3$

Solution

first rewrite the image function into a more standard form

… … $y = 2\sqrt{-x+5}+3$

… … $y = 2\sqrt{-(x-5)}+3$

now write a matrix equation, putting the different transformations into their correct places

… … $\left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} -1&0 \\ 0&2 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] + \left[ \begin{matrix} 5 \\ 3 \\ \end{matrix} \right]$

.

## 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 $y \in (2,\; 5]$,

… … given that $y’ = -4y$

… … image range becomes $y’ \in (-8, \; -20]$

which should be written in the correct order as

… … $y’ \in [-20, \; -8)$

.

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