Multiplying a matrix by a column $\begin{bmatrix} x \\ y \end{bmatrix}$ gives a new point. Do that to every point of the plane and the matrix becomes a geometric move: a rotation, a reflection, a stretch or a shear. The determinant measures what happens to area.
Key idea
$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ sends $(1, 0)$ to its first column $(a, c)$ and $(0, 1)$ to its second column $(b, d)$, and that decides everything. Rotation by $\theta$: $\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$. Areas are multiplied by $|\det|$; a negative determinant also flips the plane over. Doing $F$ and then $S$ is the single matrix $SF$.
Why it works
$$\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = x\begin{bmatrix} a \\ c \end{bmatrix} + y\begin{bmatrix} b \\ d \end{bmatrix}.$$ So the image of any point is built from the images of the two unit vectors, the columns. Rotating $(1, 0)$ by $\theta$ gives $(\cos\theta, \sin\theta)$ and rotating $(0, 1)$ gives $(-\sin\theta, \cos\theta)$; those are the columns of the rotation matrix. The unit square goes to the parallelogram on the two columns, whose area is $|\det|$.
Worked examples
Worked example: An image
Where does $\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ send the point $(2, 3)$?
- $$\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 + 6 \\ 0 + 3 \end{bmatrix}.$$
- This matrix is a shear: heights stay the same and points slide sideways in proportion to their height.
Answer: $(8, 3)$
Worked example: A rotation
Find the matrix for a $30^\circ$ counterclockwise rotation.
- $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$.
Answer: $\begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$
Worked example: Area
A triangle of area $4$ is transformed by $\begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}$. What is the new area?
- $\det = 6 - 1 = 5$, so every area is multiplied by $5$.
Answer: $20$
Worked example: Composing
Reflect across the $x$-axis, then rotate $90^\circ$ counterclockwise. What single matrix does both?
- First $F = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$, then $S = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
- The first move goes on the right: $SF = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, the reflection across $y = x$.
Answer: $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
Common mistake
Multiplying in the order you read the moves, $FS$ instead of $SF$. A point $\mathbf{v}$ becomes $F\mathbf{v}$ first, then $S(F\mathbf{v}) = (SF)\mathbf{v}$, so the matrix applied first is on the right. Check with one point, such as $(1, 0)$, by applying the moves one at a time.