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# Flipping Twisty Matrices

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Age 14 to 18

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One of the ways to work with transformations is to use a matrix. If you have not met matrices before you might like to start by looking at the problem The Matrix.

Here is a reminder of how matrix multiplication works for a $2 \times 2$ matrix and a $2$D vector:

$$ \begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}ax+by\\cx+dy\end{pmatrix} $$

If the matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ represents a transformation in the $x,y$ plane, then we can find the image of a point with coordinates $(p, q)$ by multiplying the transformation matrix and the position vector of the original point:

$$\begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}p\\q\end{pmatrix}$$

**Example: Consider the transformation represented by the matrix $\begin{pmatrix}1&2\\0&1\end{pmatrix}$. By considering the image of the points $(0,0)$, $(1, 0)$, $(0, 1)$ and $(1, 1)$ describe what this transformation does to the unit square.**

The four points have images:

$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0 \end{pmatrix}$

$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}$

$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}$

$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}3\\1\end{pmatrix}$

So the unit square is transformed into a parallelogram (this type of transformation is called a*shear*).

You can find the image of the square in a single matrix multiplication as shown below:

$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}0 & 0 & 1 & 1\\0 & 1 & 1 & 0\end{pmatrix}=\begin{pmatrix}0& 2& 3&1\\0 & 1 &1&0 \end{pmatrix}$

In this case I have listed the four vertices of the unit square in clockwise order, starting at $(0,0)$.

$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0 \end{pmatrix}$

$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}$

$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}$

$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}3\\1\end{pmatrix}$

So the unit square is transformed into a parallelogram (this type of transformation is called a

You can find the image of the square in a single matrix multiplication as shown below:

$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}0 & 0 & 1 & 1\\0 & 1 & 1 & 0\end{pmatrix}=\begin{pmatrix}0& 2& 3&1\\0 & 1 &1&0 \end{pmatrix}$

In this case I have listed the four vertices of the unit square in clockwise order, starting at $(0,0)$.

Now consider matrices of the form

$$\begin{pmatrix}a&0\\0&d\end{pmatrix} $$

where $a$ and $d$ each take either the value $1$ or the value $-1$ (so there are four different matrices).

Explore what these four different transformations do. It would be a good idea to try different shapes, such as a square and a trapezium, for example the one with coordinates $(1,0)$, $(3, 0)$, $(2, 1)$ and $(0,1)$, so that you can tell whether a shape is reflected or rotated.

What happens if we consider instead matrices of the form $\begin{pmatrix}0&b\\c&0\end{pmatrix} $ where $b$ and $c$ can take the values $1$ and $-1$?

What if any of $a$, $b$, $c$ and $d$ can be equal to $1$, $-1$ or $0$?

What happens to the areas of the shapes under the different transformations?

*You might like to use this Matrix Transformation tool to help test out your ideas.*

*There are more matrix problems in this feature.*