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Geometric Transformation

Reflection Matrix

Reflection Matrix for X-axis

Recall that reflecting a point (x,y)(x,y) across the X-axis results in the image (x,y)(x,-y).

We are looking for a matrix (rstu)\begin{pmatrix} r & s \\ t & u \end{pmatrix} such that:

(rstu)(xy)=(xy)\begin{pmatrix} r & s \\ t & u \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ -y \end{pmatrix}
(rx+sytx+uy)=(1x+0y0x1y)\begin{pmatrix} rx + sy \\ tx + uy \end{pmatrix} = \begin{pmatrix} 1x + 0y \\ 0x - 1y \end{pmatrix}

By equating the coefficients, we get:

  • rx+sy=1x+0y    r=1,s=0rx + sy = 1x + 0y \implies r=1, s=0
  • tx+uy=0x1y    t=0,u=1tx + uy = 0x - 1y \implies t=0, u=-1

Thus, the reflection matrix for the X-axis is (1001)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.

Reflection Matrix for Y-axis

Reflecting a point (x,y)(x,y) across the Y-axis results in (x,y)(-x,y).

(rstu)(xy)=(xy)\begin{pmatrix} r & s \\ t & u \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -x \\ y \end{pmatrix}
(rx+sytx+uy)=(1x+0y0x+1y)\begin{pmatrix} rx + sy \\ tx + uy \end{pmatrix} = \begin{pmatrix} -1x + 0y \\ 0x + 1y \end{pmatrix}

This gives r=1,s=0,t=0,u=1r=-1, s=0, t=0, u=1.

The matrix is (1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}.

Reflection Matrix for Line y = x

Reflecting a point (x,y)(x,y) across the line y=xy=x results in (y,x)(y,x).

(rstu)(xy)=(yx)\begin{pmatrix} r & s \\ t & u \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} y \\ x \end{pmatrix}
(rx+sytx+uy)=(0x+1y1x+0y)\begin{pmatrix} rx + sy \\ tx + uy \end{pmatrix} = \begin{pmatrix} 0x + 1y \\ 1x + 0y \end{pmatrix}

This gives r=0,s=1,t=1,u=0r=0, s=1, t=1, u=0.

The matrix is (0110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.

Reflection Matrix for Line y = -x

Reflecting a point (x,y)(x,y) across the line y=xy=-x results in (y,x)(-y,-x).

(rstu)(xy)=(yx)\begin{pmatrix} r & s \\ t & u \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -y \\ -x \end{pmatrix}
(rx+sytx+uy)=(0x1y1x+0y)\begin{pmatrix} rx + sy \\ tx + uy \end{pmatrix} = \begin{pmatrix} 0x - 1y \\ -1x + 0y \end{pmatrix}

This gives r=0,s=1,t=1,u=0r=0, s=-1, t=-1, u=0.

The matrix is (0110)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}.

Basic Reflection Matrices

  • Reflection matrix for X-axis: (1001)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
  • Reflection matrix for Y-axis: (1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
  • Reflection matrix for line y=xy=x: (0110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
  • Reflection matrix for line y=xy=-x: (0110)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}

Finding the Image of a Point using Matrix

Find the image of (3,4)(3,-4) reflected across the X-axis.

Alternative Solution:

Using the reflection matrix for the X-axis:

(xy)=(1001)(34)=((1)(3)+(0)(4)(0)(3)+(1)(4))=(34)\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ -4 \end{pmatrix} = \begin{pmatrix} (1)(3) + (0)(-4) \\ (0)(3) + (-1)(-4) \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}

Thus, the image is (3,4)(3,4).

Finding the Image of a Triangle using Matrix

Determine the image of triangle ABC with vertices A(3,1)A(3,1), B(2,3)B(-2,3), and C(2,1)C(2,-1) reflected across the Y-axis!

Alternative Solution:

The reflection matrix for the Y-axis is (1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}.

The matrix of the triangle's vertices ABC: (322131)\begin{pmatrix} 3 & -2 & 2 \\ 1 & 3 & -1 \end{pmatrix}.

(xAxBxCyAyByC)=(1001)(322131)\begin{pmatrix} x'_A & x'_B & x'_C \\ y'_A & y'_B & y'_C \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 3 & -2 & 2 \\ 1 & 3 & -1 \end{pmatrix}
=((1)(3)+0(1)(2)+0(1)(2)+00+(1)(1)0+(1)(3)0+(1)(1))=(322131)= \begin{pmatrix} (-1)(3)+0 & (-1)(-2)+0 & (-1)(2)+0 \\ 0+(1)(1) & 0+(1)(3) & 0+(1)(-1) \end{pmatrix} = \begin{pmatrix} -3 & 2 & -2 \\ 1 & 3 & -1 \end{pmatrix}

Thus, the image is triangle ABCA'B'C' with vertices A(3,1)A'(-3,1), B(2,3)B'(2,3), and C(2,1)C'(-2,-1).

Reflection of ABC\triangle ABC over Y-axis using Matrix
Triangle ABCABC is reflected to become ABCA'B'C' over the Y-axis.

Exercises

  1. Find the image of point (3,5)(3,5) reflected across the X-axis using the matrix multiplication (1001)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.
  2. Determine the image of triangle ABC with vertices A(3,1)A(3,1), B(2,3)B(-2,3), and C(2,1)C(2,-1) reflected across the line y=xy=-x!

Key Answers

  1. The image of point (3,5)(3,5) reflected across the X-axis using the matrix multiplication (1001)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} is:

    (xy)=(1001)(35)=(35)\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 \\ -5 \end{pmatrix}

    Image: (3,5)(3,-5).

  2. The reflection matrix for y=xy=-x is (0110)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}.

    Matrix of ABC vertices: (322131)\begin{pmatrix} 3 & -2 & 2 \\ 1 & 3 & -1 \end{pmatrix}.

    (0110)(322131)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 3 & -2 & 2 \\ 1 & 3 & -1 \end{pmatrix}
    =((0)(3)+(1)(1)(0)(2)+(1)(3)(0)(2)+(1)(1)(1)(3)+(0)(1)(1)(2)+(0)(3)(1)(2)+(0)(1))= \begin{pmatrix} (0)(3)+(-1)(1) & (0)(-2)+(-1)(3) & (0)(2)+(-1)(-1) \\ (-1)(3)+(0)(1) & (-1)(-2)+(0)(3) & (-1)(2)+(0)(-1) \end{pmatrix}
    =(131322)= \begin{pmatrix} -1 & -3 & 1 \\ -3 & 2 & -2 \end{pmatrix}

    Image: A(1,3)A'(-1,-3), B(3,2)B'(-3,2), C(1,2)C'(1,-2).