Matrices as Transformations

In linear algebra, matrices are not just grids of numbers—they are actions. A $2 \times 2$ matrix can represent a linear transformation of the 2D plane, mapping any point $(x, y)$ to a new point $(x', y')$.

How It Works

To apply a transformation, we multiply the transformation matrix by a column vector representing our point:

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

Common Transformations

Scaling

A scaling matrix stretches or compresses the plane along the x and y axes.

$\begin{bmatrix} k_x & 0 \\ 0 & k_y \end{bmatrix}$

If $k_x = 2$, the x-coordinates are doubled. If $k_y = 0.5$, the y-coordinates are halved.

Reflections

Matrices can reflect points across specific lines.

• Over the x-axis: $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$
• Over the y-axis: $\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$

Rotations

To rotate a point counterclockwise around the origin by an angle $\theta$, use the rotation matrix:

$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$

Composing Transformations

If you want to apply transformation $A$ and then transformation $B$, you can simply multiply their matrices. The combined transformation matrix is $BA$. Note that matrix multiplication is not commutative, so the order matters: $BA$ means transformation $A$ happens first, followed by $B$.

Example Problems

Example 1: Find the matrix that rotates the plane by $90^\circ$ counterclockwise.

Using the rotation matrix formula with $\theta = 90^\circ$:

$\cos(90^\circ) = 0, \quad \sin(90^\circ) = 1$

Substituting these into the rotation matrix gives:

$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

Example 2: Apply the matrix $\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$ to the triangle with vertices $(1,1)$, $(2,0)$, $(0,2)$.

We multiply the matrix by each vertex (written as column vectors):

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

$\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \end{bmatrix}$

$\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 6 \end{bmatrix}$

The new vertices of the transformed triangle are $(2,3)$, $(4,0)$, and $(0,6)$.