Linear Transformations and Matrices — AP® Precalculus Notes

Beyond the exam. The AP® Precalculus exam does not test this unit. Many classes teach it, and it prepares you for calculus.

A matrix can do more than store numbers: it can move the whole plane. Multiply every vector by the same $2 \times 2$ matrix and the plane is rotated, reflected, stretched or slanted. Moves of this kind are called linear transformations, and the matrix is the recipe.

Linear transformations and their matrices

A linear transformation of the plane sends each vector to a new one by a rule of the form

L(\left\langle x, y \right\rangle) = \left\langle ax + by, cx + dy \right\rangle.

Write the input as a column and this is matrix multiplication: $\left[ \begin{matrix} a & b \\ c & d \end{matrix} \right]\left[ \begin{matrix} x \\ y \end{matrix} \right]$ $= \left[ \begin{matrix} ax + by \\ cx + dy \end{matrix} \right]$ . For example, $L(\left\langle x, y \right\rangle)$ $= \left\langle x + 2y, 3x - y \right\rangle$ has the matrix $\left[ \begin{matrix} 1 & 2 \\ 3 & -1 \end{matrix} \right]$ , and $L(\left\langle 1, 1 \right\rangle) = \left\langle 3, 2 \right\rangle$ .

Reading the matrix from a picture

Put in $\left\langle 1, 0 \right\rangle$ : the output is $\left\langle a, c \right\rangle$ , the first column. Put in $\left\langle 0, 1 \right\rangle$ : the output is $\left\langle b, d \right\rangle$ , the second column. So the matrix is "where the two unit vectors go, written as columns".

Figure 1. The unit square and its image under

\left[ \begin{matrix} 1 & 1 \\ 0 & 2 \end{matrix} \right]

. LevlPrep original diagram.

Worked example 1. A transformation sends $\left\langle 1, 0 \right\rangle$ to $\left\langle 2, 1 \right\rangle$ and $\left\langle 0, 1 \right\rangle$ to $\left\langle -1, 1 \right\rangle$ . Find its matrix and the image of $\left\langle 3, 2 \right\rangle$ .

Columns: $\left[ \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right]$ .

$\left[ \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right]\left[ \begin{matrix} 3 \\ 2 \end{matrix} \right]$ $= \left[ \begin{matrix} 2(3) + (-1)(2) \\ 1(3) + 1(2) \end{matrix} \right]$ $= \left[ \begin{matrix} 4 \\ 5 \end{matrix} \right]$ . Or think of it as $3$ copies of $\left\langle 2, 1 \right\rangle$ plus $2$ copies of $\left\langle -1, 1 \right\rangle$ .

What stays the same

Because every vector is moved by the same matrix, the origin always goes to the origin ( $\left[ \begin{matrix} a & b \\ c & d \end{matrix} \right]\left[ \begin{matrix} 0 \\ 0 \end{matrix} \right]$ $= \left[ \begin{matrix} 0 \\ 0 \end{matrix} \right]$ ), lines go to lines, and parallel lines stay parallel. A rule that moves the origin, such as $\left\langle x, y \right\rangle \to \left\langle x + 1, y \right\rangle$ , is not linear.

A catalog of moves

Move	Matrix	Determinant
rotation by $\theta$ counterclockwise	$\left[ \begin{matrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{matrix} \right]$	$1$
reflection across the $x$ -axis	$\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right]$	$-1$
reflection across the line $y = x$	$\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]$	$-1$
dilation by $k$	$\left[ \begin{matrix} k & 0 \\ 0 & k \end{matrix} \right]$	$k^2$
horizontal shear	$\left[ \begin{matrix} 1 & k \\ 0 & 1 \end{matrix} \right]$	$1$

The rotation matrix comes from the unit circle: $\left\langle 1, 0 \right\rangle$ turns to $\left\langle \cos\theta, \sin\theta \right\rangle$ and $\left\langle 0, 1 \right\rangle$ turns to $\left\langle -\sin\theta, \cos\theta \right\rangle$ .

Worked example 2. Rotate $\left\langle 2, 0 \right\rangle$ by $\frac{\pi}{6}$ counterclockwise.

$\left[ \begin{matrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix} \right]\left[ \begin{matrix} 2 \\ 0 \end{matrix} \right]$ $= \left[ \begin{matrix} \sqrt{3} \\ 1 \end{matrix} \right]$ , a vector of length $2$ at angle $\frac{\pi}{6}$ .

Areas and orientation

The unit square goes to the parallelogram made by the two columns, whose area is $\left| \det \right|$ . Every other region is scaled by the same factor, so a triangle of area $5$ under a matrix with determinant $3$ becomes a triangle of area $15$ . A negative determinant also flips the plane over, like a reflection: a counterclockwise trip around a shape becomes a clockwise one.