Dilation Matrix

Finding the Matrix Associated with Dilation

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

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

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

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 $(k(x-a)+a,\, k(y-b)+b)$.

Find the combination of matrix operations on the position vector $\begin{pmatrix} x-a \\ y-b \end{pmatrix}$ such that the result is $\begin{pmatrix} k(x-a)+a \\ k(y-b)+b \end{pmatrix}$.

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

or it can also be written as:

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 $P(1,1)$!

Alternative Solution:

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

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

Exercises

1. Find the coordinates of the image of the point $(a,b)$ under the dilation $[O,3]$!
2. Determine the matrix corresponding to a dilation with a scale factor of $-2$ and centered at $O(0,0)$.
3. 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 $B$!
4. 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!

Key Answers

1. 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 $(3a, 3b)$.

2. Scale factor $k=-2$, center $O(0,0)$.

   The dilation matrix is:

   $$\begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix}$$
   .

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

   $x=-1, y=5, a=2, b=-3, k=\frac{1}{2}$.

   $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \begin{pmatrix} -1-2 \\ 5-(-3) \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix}$$

   $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \begin{pmatrix} -3 \\ 8 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix}$$

   $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} (\frac{1}{2})(-3) + (0)(8) \\ (0)(-3) + (\frac{1}{2})(8) \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix}$$

   $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -\frac{3}{2} \\ 4 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} -\frac{3}{2} + \frac{4}{2} \\ 4 - 3 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \\ 1 \end{pmatrix}$$

   The coordinates of the image of point $B$ are $B'(\frac{1}{2}, 1)$.

4. Triangle $KLM$ with $K(1,1)$, $L(5,1)$, $M(3,4)$.

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

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

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

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

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

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

   $$\begin{pmatrix} x_M' \\ y_M' \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 \\ 8 \end{pmatrix} \implies M'(6,8)$$

   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 $O(0,0)$.

   Toggle GridToggle Play