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Geometric Transformation

Reflection Matrix

Reflection Matrix for X-axis

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

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

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

By equating the coefficients, we get:

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

Thus, the reflection matrix for the X-axis is (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}(10​0−1​).

Reflection Matrix for Y-axis

Reflecting a point (x,y)(x,y)(x,y) across the Y-axis results in (−x,y)(-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}(rt​su​)(xy​)=(−xy​)
(rx+sytx+uy)=(−1x+0y0x+1y)\begin{pmatrix} rx + sy \\ tx + uy \end{pmatrix} = \begin{pmatrix} -1x + 0y \\ 0x + 1y \end{pmatrix}(rx+sytx+uy​)=(−1x+0y0x+1y​)

This gives r=−1,s=0,t=0,u=1r=-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}(−10​01​).

Reflection Matrix for Line y = x

Reflecting a point (x,y)(x,y)(x,y) across the line y=xy=xy=x results in (y,x)(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}(rt​su​)(xy​)=(yx​)
(rx+sytx+uy)=(0x+1y1x+0y)\begin{pmatrix} rx + sy \\ tx + uy \end{pmatrix} = \begin{pmatrix} 0x + 1y \\ 1x + 0y \end{pmatrix}(rx+sytx+uy​)=(0x+1y1x+0y​)

This gives r=0,s=1,t=1,u=0r=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}(01​10​).

Reflection Matrix for Line y = -x

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

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

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

The matrix is (0−1−10)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}(0−1​−10​).

Basic Reflection Matrices

  • Reflection matrix for X-axis: (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}(10​0−1​)
  • Reflection matrix for Y-axis: (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}(−10​01​)
  • Reflection matrix for line y=xy=xy=x: (0110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}(01​10​)
  • Reflection matrix for line y=−xy=-xy=−x: (0−1−10)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}(0−1​−10​)

Finding the Image of a Point using Matrix

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

Alternative Solution:

Using the reflection matrix for the X-axis:

(x′y′)=(100−1)(3−4)=((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}(x′y′​)=(10​0−1​)(3−4​)=((1)(3)+(0)(−4)(0)(3)+(−1)(−4)​)=(34​)

Thus, the image is (3,4)(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)A(3,1), B(−2,3)B(-2,3)B(−2,3), and C(2,−1)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}(−10​01​).

The matrix of the triangle's vertices ABC: (3−2213−1)\begin{pmatrix} 3 & -2 & 2 \\ 1 & 3 & -1 \end{pmatrix}(31​−23​2−1​).

(xA′xB′xC′yA′yB′yC′)=(−1001)(3−2213−1)\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}(xA′​yA′​​xB′​yB′​​xC′​yC′​​)=(−10​01​)(31​−23​2−1​)
=((−1)(3)+0(−1)(−2)+0(−1)(2)+00+(1)(1)0+(1)(3)0+(1)(−1))=(−32−213−1)= \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}=((−1)(3)+00+(1)(1)​(−1)(−2)+00+(1)(3)​(−1)(2)+00+(1)(−1)​)=(−31​23​−2−1​)

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

Reflection of △ABC\triangle ABC△ABC over Y-axis using Matrix
Triangle ABCABCABC is reflected to become A′B′C′A'B'C'A′B′C′ over the Y-axis.

Exercises

  1. Find the image of point (3,5)(3,5)(3,5) reflected across the X-axis using the matrix multiplication (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}(10​0−1​).
  2. Determine the image of triangle ABC with vertices A(3,1)A(3,1)A(3,1), B(−2,3)B(-2,3)B(−2,3), and C(2,−1)C(2,-1)C(2,−1) reflected across the line y=−xy=-xy=−x!

Key Answers

  1. The image of point (3,5)(3,5)(3,5) reflected across the X-axis using the matrix multiplication (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}(10​0−1​) is:

    (x′y′)=(100−1)(35)=(3−5)\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}(x′y′​)=(10​0−1​)(35​)=(3−5​)

    Image: (3,−5)(3,-5)(3,−5).

  2. The reflection matrix for y=−xy=-xy=−x is (0−1−10)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}(0−1​−10​).

    Matrix of ABC vertices: (3−2213−1)\begin{pmatrix} 3 & -2 & 2 \\ 1 & 3 & -1 \end{pmatrix}(31​−23​2−1​).

    (0−1−10)(3−2213−1)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 3 & -2 & 2 \\ 1 & 3 & -1 \end{pmatrix}(0−1​−10​)(31​−23​2−1​)
    =((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}=((0)(3)+(−1)(1)(−1)(3)+(0)(1)​(0)(−2)+(−1)(3)(−1)(−2)+(0)(3)​(0)(2)+(−1)(−1)(−1)(2)+(0)(−1)​)
    =(−1−31−32−2)= \begin{pmatrix} -1 & -3 & 1 \\ -3 & 2 & -2 \end{pmatrix}=(−1−3​−32​1−2​)

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

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Reflection Matrix over Center Point

  • Reflection MatrixMaster reflection matrices for X-axis, Y-axis, y=x, and y=-x lines. Learn to transform points and triangles with detailed matrix calculations.
On this page
  • Reflection Matrix for X-axis
  • Reflection Matrix for Y-axis
  • Reflection Matrix for Line y = x
  • Reflection Matrix for Line y = -x
  • Basic Reflection Matrices
  • Finding the Image of a Point using Matrix
  • Finding the Image of a Triangle using Matrix
  • Exercises
    • Key Answers
  • Comments
  • Report
  • Source code