02Gmatrixtransform

toc = Transformations with Matrices =

We can use matrices to represent and manipulate the transformations we have been performing on graphs.

Notation
If we start with a set of points (x, y) Then we transform them using a process, T, to get a new set of points (x’, y’) (sometimes called the image) We can show this in matrix notation as:

math T \left( \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] \right) = \left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] math

Reflections and Dilations
(multiply by transformation matrix)

math \left[ \begin{matrix} 1&0 \\ 0&1 \\ \end{matrix} \right] \text { causes no change at all} math

This would be written as: math \left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} 1&0 \\ 0&1 \\ \end{matrix} \right] \, \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] math

so {by multiplying matrices} math \left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] math

or x' ``=`` x and y' ``=`` y {ie no change}

Remember that the matrix shown here is called the ** Identity Matrix ** (or I) For any matrix, A, it follows from the definition of I that I × A = A

Reflections
math \left[ \begin{matrix} -1&0 \\ 0&1 \\ \end{matrix} \right] \text { causes a reflection in the x direction (across the y-axis or on the y-axis)} math math \left[ \begin{matrix} 1&0 \\ 0&-1 \\ \end{matrix} \right] \text { causes a reflection in the y direction (across the x-axis or on the x-axis)} math

A reflection in the x-direction (across the y-axis or on the y-axis) could be written as: math \left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} -1&0 \\ 0&1 \\ \end{matrix} \right] \, \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] math

so {by multiplying matrices} math \left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} -x \\ y \\ \end{matrix} \right] math

or x' ``=`` –x and y' ``=`` y

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

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

Dilations
math \left[ \begin{matrix} a&0 \\ 0&1 \\ \end{matrix} \right] \text { causes a dilation by a factor of a in the x direction (from the y-axis) } math math \left[ \begin{matrix} 1&0 \\ 0&a \\ \end{matrix} \right] \text { causes a dilation by a factor of a in the y direction (from the x-axis) } math

For example:

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

This represents a dilation by a factor of 3 in the x direction (all x-values are multiplied by 3).

This transformation could be written as:

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

or x' ``=`` 3x and y' ``=`` y

For example, if we start with the point (5. 3) under this transformation, the image is (15, 3)

Compare this to starting with the cubic graph y = x 3 under this transformation, the image is: math y' = \big( \frac{1}{3}x' \big)^3 math

{Remember that y = f(nx) causes a dilation by a factor of 1/n in the x-direction (from the y-axis)}

We can arrive at the new equation alegbraically by making x the subject and substituting: y' ``=`` y and y ``=`` x 3 so: y' = x 3 math \\ \text{but } x = \frac{1}{3}x' \text{ so:} \\ y' = \big( \frac{1}{3}x' \big)^3 math

In ** __function notation__ **, this transformation could be written as: math \big\{ T: R^2 \rightarrow R^2, \; T(x,\,y) \rightarrow (3x,\, y) \big\} math or in shorthand as: math (x',\,y') \rightarrow (3x,\, y) math

{Read the arrow as "maps onto". Don't confuse it with the arrow used in limits which means "approaches the limit". The R 2 is because we are working in 2 dimensions (x, y)}

These operations can be combined to cause more than one transformation. math \text{For example, the transformation matrix } \left[ \begin{matrix} -2&0 \\ 0&3 \\ \end{matrix} \right] \text{ gives:} math

math \left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} -2&0 \\ 0&3 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \left[ \begin{matrix} -2x \\ 3y \\ \end{matrix} \right] math

(or)

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

or x' ``=`` –2x and y' ``=`` 3y

This represents
 * a reflection in the x direction (across the y-axis) and
 * a dilation by a factor of 2 in the x-direction (away from the y-axis) and
 * a dilation by a factor of 3 in the y direction (away from the x-axis).

For example, if we start with the point (5, 3) under this transformation, the image is (–10, 9)

Compare this to starting with a standard cubic graph y = x 3 math \text{under this transformation, the image is: } \; y'=3 \big( -\frac{1}{2}x' \big)^3 math

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

but making x the subject of x' = –2x gives us: math \\ x = -\frac{1}{2}x' \qquad \text{so:} \\ y' = 3\big( -\frac{1}{2}x' \big)^3 math

This equation could be simplified by moving the ½ out of the cube, so we get: math \\ y' = 3 \big( -\frac{1}{2} \big)^3 \big( x' \big)^3 \\ y' = -\frac{3}{8} \big( x' \big)^3 math

In __**function notation**__, this transformation could be written as: {T: R 2 → R 2, T(x, y) → (–2x, 3y)} or in shorthand as (x', y') → (–2x, 3y) {Read the arrow as “maps onto”. The R 2 is because we are working in 2 dimensions (x, y)}

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