Matrix and Transformation Connection

Mapping Points with a Matrix

A $2 \times 2$ matrix defines a linear transformation of the Cartesian plane. It always maps the origin to itself, so a nonzero translation needs vector addition or homogeneous coordinates instead.

A point in the Cartesian plane can be written as an ordered pair $(x,y)$ or as a position vector:

We use this vector notation to express transformations through matrix multiplication.

Consider a transformation with matrix $M$:

The image $P'(x',y')$ of a point $P(x,y)$ is obtained by matrix multiplication:

$x' = ax + by$ and $y' = cx + dy$.

Multiplying a Matrix by a Position Vector

Let the following vector represent any point in the Cartesian plane:

Calculate this matrix product and identify the transformation:

Solution:

The matrix product is:

The matrix $R_{90^\circ}$ maps $(x,y)$ to $(-y,x)$, so it rotates the point $90^\circ$ counter-clockwise about the origin:

Multiplying a Matrix by Three Points Simultaneously

Find the image of $\triangle ABC$, with vertices $A(1,1)$, $B(4,1)$, and $C(4,2)$ transformed by the matrix:

Solution:

First, write the point coordinates as columns of a matrix:

The three columns correspond to $A$, $B$, and $C$, in that order.

Next, multiply this matrix from the left by:

Compute each entry by multiplying a row by a column:

The result of the transformation is a new triangle $\triangle A'B'C'$ with vertices $A'(-1,-1)$, $B'(-4,-1)$, and $C'(-4,-2)$.

The matrix $R_{180^\circ}$ represents a $180^\circ$ rotation about the origin:

Exercises

1. Calculate the following matrix product. What transformation does the matrix represent?

   $$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$

2. A transformation has the following matrix:

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

   Find the image of the triangle with vertices $A(2,0)$, $B(2,1)$, and $C(0,1)$ under this transformation.

Solutions

1. $$\begin{aligned} &\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ &= \begin{pmatrix} (0)(x) + (1)(y) \\ (-1)(x) + (0)(y) \end{pmatrix} \\ &= \begin{pmatrix} y \\ -x \end{pmatrix} \end{aligned}$$

   The point $(x,y)$ is transformed into $(y,-x)$.

   This is a $90^\circ$ clockwise rotation, equivalently a $270^\circ$ counter-clockwise rotation, about the origin.

2. Transformation matrix:

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

   Vertices: $A(2,0)$, $B(2,1)$, $C(0,1)$.

   Point matrix:

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

   Compute each entry by multiplying a row by a column:

   $$\begin{aligned} &m_{11}=(1)(2)+(2)(0) \\ &m_{12}=(1)(2)+(2)(1) \\ &m_{13}=(1)(0)+(2)(1) \\ &m_{21}=(0)(2)+(1)(0) \\ &m_{22}=(0)(2)+(1)(1) \\ &m_{23}=(0)(0)+(1)(1) \end{aligned}$$

   $$\begin{aligned} &\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 2 & 0 \\ 0 & 1 & 1 \end{pmatrix} \\ &=\begin{pmatrix}m_{11}&m_{12}&m_{13}\\m_{21}&m_{22}&m_{23}\end{pmatrix} \end{aligned}$$

   $$\begin{aligned} &= \begin{pmatrix} 2+0 & 2+2 & 0+2 \\ 0+0 & 0+1 & 0+1 \end{pmatrix} \\ &= \begin{pmatrix} 2 & 4 & 2 \\ 0 & 1 & 1 \end{pmatrix} \end{aligned}$$

   Image vertices: $A'(2,0)$, $B'(4,1)$, $C'(2,1)$.

   This is a horizontal shear: the $x$-coordinate changes by twice the $y$-coordinate, while $y$ remains unchanged.