Dilation Matrix

Matching Matrix Entries to Scaled Coordinates

How to find the matrix associated with a dilation operation? Recall that a point $(x,y)$ is mapped by a dilation with a factor $k \neq 0$ and center $O$ to $(kx,ky)$ .

Suppose the matrix we are looking for is:

\begin{pmatrix} r & s \\ t & u \end{pmatrix}

Find $r, s, t, u$ such that it satisfies

\begin{pmatrix} r & s \\ t & u \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix}

From the matrix multiplication on the left side, we get:

\begin{pmatrix} rx + sy \\ tx + uy \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix}

By equating the corresponding components:

First row: $rx + sy = kx$ . For this equation to hold for all $x$ and $y$ , the coefficients of $x$ must be equal, and the coefficients of $y$ must be equal. Thus, $r = k$ and $s = 0$ .
Second row: $tx + uy = ky$ . Similarly, $t = 0$ and $u = k$ .

Dilation Matrix with Respect to the Origin

The matrix associated with a dilation by a factor $k \neq 0$ with respect to the origin $O(0,0)$ is

\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}

Matrix Operation for Dilation with Respect to an Arbitrary Point

A point $(x,y)$ dilated by a factor $k \neq 0$ and center $(a,b)$ will be mapped to $(k(x-a)+a,\, k(y-b)+b)$ .

After subtracting the center coordinates, the position vector is:

\begin{pmatrix} x-a \\ y-b \end{pmatrix}

Find the matrix operations that transform this vector into the required image:

\begin{pmatrix} k(x-a)+a \\ k(y-b)+b \end{pmatrix}

The matrix operation associated with a dilation by a factor $k \neq 0$ with respect to the point $(a,b)$ is

\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} \begin{pmatrix} x-a \\ y-b \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix}

An equivalent form is:

\begin{aligned} &\begin{pmatrix} x' \\ y' \end{pmatrix} - \begin{pmatrix} a \\ b \end{pmatrix} \\ &= \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} \left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} a \\ b \end{pmatrix} \right) \end{aligned}

Finding the Image of a Dilation Using Matrices

Determine the image of point $A(2,4)$ transformed by a dilation with a factor of $2$ with respect to the center point $P(1,1)$ !

Solution:

Given $x=2, y=4, k=2, a=1, b=1$ .

\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2-1 \\ 4-1 \end{pmatrix} + \begin{pmatrix} 1 \\ 1 \end{pmatrix}

The image of point $A(2,4)$ is $A'(3,7)$ .

Visualization of Dilation of Point

A(2,4)

with Center

P(1,1)

and Scale Factor

k=2

Point

A(2,4)

is dilated with respect to the center

P(1,1)

with a scale factor

k=2

to produce the image

A'(3,7)

. The line from the center to the original point and from the center to the image lie on the same line, and the distance

PA'

is twice the distance

PA

Exercises

Each problem either builds the dilation matrix or applies it. The third problem moves the center away from the origin, so the translation step is needed as well.

Find the coordinates of the image of the point $(a,b)$ under the dilation $[O,3]$ !
Determine the matrix corresponding to a dilation with a scale factor of $-2$ and centered at $O(0,0)$ .
A point $B(-1, 5)$ is dilated with center $P(2, -3)$ and scale factor $k = \frac{1}{2}$ . Determine the coordinates of the image of point $B$ !
A triangle $KLM$ with vertices $K(1,1)$ , $L(5,1)$ , and $M(3,4)$ is dilated with center $O(0,0)$ and scale factor $2$ . Draw the original triangle and its image, then determine the coordinates of the image vertices!

Solutions

The dilation $[O,3]$ means the center of dilation is $O(0,0)$ and the scale factor is $k=3$ .

Let the point be $Q(a,b)$ .

Then $x=a, y=b, k=3$ .

$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 3a \\ 3b \end{pmatrix}$

Thus, the coordinates of the image of point $(a,b)$ are $(3a, 3b)$ .
Scale factor $k=-2$ , center $O(0,0)$ .

The dilation matrix is:

$\begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix}$
Point $B(-1, 5)$ , center $P(2, -3)$ , scale factor $k = \frac{1}{2}$ .

$x=-1, y=5, a=2, b=-3, k=\frac{1}{2}$ .
$\begin{aligned} \begin{pmatrix} x' \\ y' \end{pmatrix} &= \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \begin{pmatrix} -1-2 \\ 5-(-3) \end{pmatrix} \\ &\quad{}+ \begin{pmatrix} 2 \\ -3 \end{pmatrix} \end{aligned}$
Calculate the two coordinates, including the translation vector:
$\begin{aligned} &x'=(\frac12)(-3)+(0)(8)+2 \\ &y'=(0)(-3)+(\frac12)(8)-3 \end{aligned}$
The coordinates of the image of point $B$ are $B'(\frac{1}{2}, 1)$ .
Triangle $KLM$ with $K(1,1)$ , $L(5,1)$ , $M(3,4)$ .

Center $O(0,0)$ , $k=2$ .

Image of point $K(1,1)$ :

$\begin{aligned} \begin{pmatrix} x_K' \\ y_K' \end{pmatrix} &= \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} 2 \\ 2 \end{pmatrix} \\ &\implies K'(2,2) \end{aligned}$

Image of point $L(5,1)$ :

$\begin{aligned} \begin{pmatrix} x_L' \\ y_L' \end{pmatrix} &= \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 5 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} 10 \\ 2 \end{pmatrix} \\ &\implies L'(10,2) \end{aligned}$

Image of point $M(3,4)$ :

$\begin{aligned} \begin{pmatrix} x_M' \\ y_M' \end{pmatrix} &= \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 3 \\ 4 \end{pmatrix} \\ &= \begin{pmatrix} 6 \\ 8 \end{pmatrix} \\ &\implies M'(6,8) \end{aligned}$
Visualization of Dilation of Triangle $KLM$ with Center $O(0,0)$ and Scale Factor $k=2$
Triangle $KLM$ is dilated to become triangle $K'L'M'$ . The center of dilation is $O(0,0)$ .