# Matrices – Geometric Transformations

Hello and welcome to this video about using matrices to transform figures on the **coordinate plane**! In this video, we will cover translations dilations, reflections, and rotations

Using addition, subtraction, scalar multiplication, and matrix multiplication, we can transform figures on the coordinate plane. All we need are the coordinates of the figure, which can be any shape and does need to be closed. This triangle will be used to demonstrate each transformation:

First, we need to create the **coordinate matrix** for the figure. The general form of a coordinate matrix is x1, x2, x3 , y1, y2, y3, and so on.

The key to remembering this **matrix** is remembering that row 1 represents the x-coordinates and row 2 represents the y-coordinates.

The coordinate matrix for our triangle, we’ll call it T, is T equals 1, 2, 3, 0, 3, 1.

Now we can perform operations on our coordinate matrix in order to discover the coordinates of our triangle transformed in various ways.

**Translations**, or slides, can be performed simply by adding the amount and direction of the slide to the x- and y- coordinates separately.

Suppose we wanted to slide our triangle 2 units to the right. Since this is a change involving the x-coordinates moving in the positive direction, we would simply perform a matrix addition that looks like this, which causes 2 to be added to each x-coordinate.

The coordinates of our translated triangle are (3, 0), (4, 3) and (5, 1), as shown on the graph.

Pretty simple, right? Suppose we wanted to slide our triangle 3 units down from its original position. The matrix addition would look like this, which causes 3 to be subtracted from each y-coordinate.

Of course, if we wanted to perform both transformations at once, it would look like this:

We put our positive 2s in the top row to move the triangle along the x-axis, and put negative 3s in the bottom row to also move it along the y-axis.

Suppose we wanted to make our triangle twice as large. We would simply multiply our coordinate matrix by a factor of 2, like this.

This transformation is called a **dilation**, which, in the case of this example, looks like this:

Dilations can be either expansions or reductions. Instead of expanding our shape like we just did, we might have reduced its size by a factor of one-half like this:

We can also use matrices to **reflect** figures in various ways. This is done with matrix multiplication. Let’s briefly examine the big picture first:

If we multiply T by the 2×2 identity matrix, we don’t change our triangle’s size or position. Let’s take a look:

Essentially, the x-coordinates are multiplied by 1 and the y-coordinates are multiplied by 1, so they don’t change. Remember, the identity matrix is the multiplicative identity, so this is what’s supposed to happen.

Suppose we wanted to reflect our triangle over the y-axis. None of the y-coordinates would change and all the x-coordinates would become the inverse. Here’s what the multiplication looks like:

All the x-coordinates are now opposite the originals, and our triangle has been reflected:

If you wanted to reflect the triangle over the x-axis, you would go through the same process, but use this matrix to multiply:

The new x-coordinates are the inverse of the original. (x,y) becomes (x,-y).

If you wanted to reflect the triangle over the origin, meaning reflect it simultaneously over both axes, you would use this matrix to multiply:

The new x- and y-coordinates are both inverse of the originals. (x,y) becomes (-x,-y).

Lastly, let’s look at rotation. We can **rotate** our triangle by any angle measure in the counterclockwise direction. First, let’s look at some common rotations, then we’ll see how to rotate any angle amount.

Suppose we want to rotate our triangle by 90 degrees. We would simply multiply our coordinate matrix by this identity matrix, which gives us this matrix.

which gives us:

This is how the triangle is now positioned:

A 180-degree rotation is the same as reflecting about the origin, so we use the same matrix:

For a 270-degree counterclockwise rotation, we would use this matrix:

There are some clear parallels to reflections here, but there is also a trigonometry connection that allows us to rotate any angle amount we want. The general matrix to use to multiply looks like this:

We can quickly see where our three common matrices come from:

A 90-degree counterclockwise rotation would look like this,

whereas a 180-degree counterclockwise rotation might look like this

And finally, a 270-degree counterclockwise rotation looks like this

But now we can rotate by other angles, like 30 or 45 degrees if we want to.

Thanks for watching, and happy studying!

## Practice Questions

**Question #1:**

The coordinates of the vertices of quadrilateral Q are shown in the matrix.

\(Q=\begin{bmatrix}

3 & 1 & 2 & 4\\

-3 & 0 & 0 & -2

\end{bmatrix}

\)

What are the coordinates of the vertices of quadrilateral Q after it has been translated \(4\) units left and \(3\) units up?

-1 & -3 & -2 & 0\\

0 & 3 & 3 & 1

\end{bmatrix}

\)

7 & 5 & 6 & 8\\

0 & 3 & 3 & 1

\end{bmatrix}

\)

-1 & -3 & -2 & 0\\

-6 & -3 & -3 & -5

\end{bmatrix}

\)

7 & 5 & 6 & 8\\

-6 & -3 & -3 & -5

\end{bmatrix}

\)

**Answer:**

To translate the triangle \(4\) units left and \(3\) units up we will add the original matrix to the matrix reflecting the translation:

\(\begin{bmatrix}

-4 & -4 & -4 &-4 \\

3& 3 & 3 & 3

\end{bmatrix}+\begin{bmatrix}

3 & 1 & 2 & 4\\

-3 & 0 & 0 & -2

\end{bmatrix}=\begin{bmatrix}

-1 & -3 & -2 & 0\\

0 & 3 & 3 & 1

\end{bmatrix}

\)

Therefore, the coordinates of the vertices of quadrilateral Q after it has been translated \(4\) units up and \(3\) units left are \(\begin{bmatrix}

-1 & -3 & -2 & 0\\

0 & 3 & 3 & 1

\end{bmatrix}

\).

**Question #2:**

Matrix K shows the coordinates of the vertices of a triangle.

\(K=\begin{bmatrix}

-1 & 2 &-1 \\

3& 4 & 1

\end{bmatrix}

\)

What are the coordinates of the vertices of triangle K after a dilation by a factor of \(3\)?

2 & 5 & 2\\

6& 7 & 4

\end{bmatrix}

\)

-3 & 6 & -3\\

9& 12 & 3

\end{bmatrix}

\)

4 & 1 & 4\\

0& -1 & 2

\end{bmatrix}

\)

3 & -6 & 3\\

-9& -12 & -3

\end{bmatrix}

\)

**Answer:**

To dilate by a factor of \(3\) we must multiply all the entries in the matrix by \(3\), which will result in the matrix \(\begin{bmatrix}

-3 & 6 & -3\\

9& 12 & 3

\end{bmatrix}

\), which are the coordinates of triangle K after it has been dilated by a factor of \(3\).

**Question #3:**

The vertices of the coordinates of triangle W are shown in the matrix.

\(W=\begin{bmatrix}

-4 & -1 & 0\\

-1 & 1 & -2

\end{bmatrix}

\)

What are the coordinates of the vertices of triangle W after the size is reduced to \(\frac{1}{3}\) of the original size?

-7 & -4 & -3\\

-4 & -2 & -5

\end{bmatrix}

\)

-1 & 2 & 3\\

2 & 4 & 1

\end{bmatrix}

\)

\frac{4}{3} & \frac{1}{3} & 0\\

\frac{1}{3} & -\frac{1}{3} & \frac{2}{3}

\end{bmatrix}

\)

-\frac{4}{3} & -\frac{1}{3} & 0\\

-\frac{1}{3} & \frac{1}{3} & -\frac{2}{3}

\end{bmatrix}

\)

**Answer:**

To reduce the size of a triangle we multiply the original coordinates by the factor it is being reduced by, in this case \(\frac{1}{3}\), which results in the following coordinates for the vertices \(\begin{bmatrix}

-\frac{4}{3} & -\frac{1}{3} & 0\\

-\frac{1}{3} & \frac{1}{3} & -\frac{2}{3}

\end{bmatrix}

\).

**Question #4:**

The coordinates of the vertices of quadrilateral L are shown in the matrix.

\(L=\begin{bmatrix}

0 & 3 & 2 & -3\\

1 & 3 & 4 & 5

\end{bmatrix}

\)

What are the coordinates of the vertices of quadrilateral L after it has been reflected across the \(y\)-axis?

0 & 3 & 2 & -3\\

-1 & -3 & -4 & -5

\end{bmatrix}

\)

0 & -3 & -2 & 3\\

-1 & -3 & -4 & -5

\end{bmatrix}

\)

0 & -3 & -2 & 3\\

1 & 3 & 4 & 5

\end{bmatrix}

\)

1 & 3 & 4 & 5\\

0 & 3 & 2 & -3

\end{bmatrix}

\)

**Answer:**

We can find the coordinates of the vertices of quadrilateral L after a reflection across the \(y\)-axis by multiplying the matrix by \(\begin{bmatrix}

-1 & 0\\

0 & 1

\end{bmatrix}\), which results in the value of the y-coordinates staying the same and the x-values change sign, \(\begin{bmatrix}

0 & -3 & -2 & 3\\

1 & 3 & 4 & 5

\end{bmatrix}

\).

**Question #5:**

The coordinates of the vertices of triangle T are shown in the matrix.

\(T=\begin{bmatrix}

2 & 5 & 3\\

-3 & 1 & 0

\end{bmatrix}

\)

What are the coordinates of the vertices of triangle T after it has been rotated \(90°\) counterclockwise?

3 & -1 & 0\\

2 & 5 & 3

\end{bmatrix}

\)

-3 & 1 & 0\\

2 & 5 & 3

\end{bmatrix}

\)

-3 & 1 & 0\\

-2 & -5 & -3

\end{bmatrix}

\)

3 & -1 & 0\\

-2 & -5 & -3

\end{bmatrix}

\)

**Answer:**

To use matrices to rotate the coordinates of the vertices of triangle T by \(90°\) we will multiply the coordinates of the vertices of triangle T by \(\begin{bmatrix}

0 & -1\\

1 & 0

\end{bmatrix}

\), which results in the following coordinates, \(\begin{bmatrix}

3 & -1 & 0\\

2 & 5 & 3

\end{bmatrix}

\).