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In Exercises \(3-8,\) find the \(3 \times 3\) matrices that produce the described composite 2 \(\mathrm{D}\) transformations, using homogeneous coordinates. Rotate points \(30^{\circ},\) and then reflect through the \(x\) -axis.

Short Answer

Expert verified
The composite transformation matrix is \(\begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\ -\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ 0 & 0 & 1 \end{bmatrix}\).

Step by step solution

01

Represent Rotation Matrix

To rotate a point by \(30^\circ\), we use the rotation matrix for 2D transformations.\[R = \begin{bmatrix} \cos 30^\circ & -\sin 30^\circ & 0 \ \sin 30^\circ & \cos 30^\circ & 0 \ 0 & 0 & 1 \end{bmatrix}\]We know that \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\). So the matrix becomes:\[R = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \ \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \ 0 & 0 & 1 \end{bmatrix}\]
02

Represent Reflection Matrix Through x-axis

To reflect a point through the \(x\)-axis, the reflection matrix is:\[M = \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]
03

Combine the Transformations

To apply a sequence of transformations, we multiply the matrices representing each transformation. Since matrix multiplication is not commutative, the order of multiplication matters. We first rotate and then reflect, which translates to multiplying the reflection matrix by the rotation matrix.\[T = M \cdot R = \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \ \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \ 0 & 0 & 1 \end{bmatrix}\]
04

Matrix Multiplication

Let’s multiply the matrices to get the composite transformation matrix:\[T = \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \ \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \ -\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \ 0 & 0 & 1 \end{bmatrix}\]This is the final transformation matrix that first applies a rotation of \(30^\circ\) followed by a reflection through the \(x\)-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Homogeneous Coordinates
In the world of 2D transformations, homogeneous coordinates play a vital role by simplifying the math involved. They extend the usual 2D coordinates \(x, y\) to \(x, y, 1\), adding an extra dimension.
This extra element allows transformations like translation, rotation, and scaling to be represented using simple matrix multiplications.
Since translation requires shifting the object, the use of a third component facilitates this with ease.
  • Standard 2D point: \(x, y\)
  • Homogeneous coordinates: \(x, y, 1\)
The last coordinate is typically a '1', which helps keep the math uniform.
In geometric transformations, this third dimension gets around the limitations of 2D matrices, delivering seamless transformations through matrix multiplication.
Rotation Matrix
A rotation matrix allows you to turn points around the origin in a 2D plane. It operates by changing the position of a point while maintaining the same distance from the origin.
The rotation matrix for 2D transformations by an angle \theta\ is given by: \[R = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\sin \theta & \cos \theta & 0 \0 & 0 & 1 \end{bmatrix} \] For example, when rotating a point \(30\) degrees:
  • \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)
  • \(\sin(30^\circ) = \frac{1}{2}\)
After substituting these values, the rotation matrix becomes: \[R = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \0 & 0 & 1 \end{bmatrix}\] This matrix's magic lies in re-orienting the shape while preserving other properties.
Reflection Matrix
Reflection matrices flip objects over specified axes, like a mirror image across a line.
When reflecting through the x-axis, each point below the axis is mirrored to a position above the axis, and vice versa.
The matrix for reflecting a point across the x-axis is: \[M = \begin{bmatrix} 1 & 0 & 0 \0 & -1 & 0 \0 & 0 & 1 \end{bmatrix}\]
  • The x-coordinate remains unchanged (because we're flipping vertically).
  • The y-coordinate is inverted to reflect.
This matrix transforms any shape by reversing its vertical component, effectively flipping it across a horizontal plane.
Matrix Multiplication
Matrix multiplication is essential for combining multiple transformations into a single operation.
When performing operations like rotation followed by reflection, the sequence of matrix multiplication is crucial.
Essentially, multiplying two matrices involves combining rows and columns, resulting in a new matrix. However, the order isn't interchangeable; multiplying one matrix by another in reverse gives a different result!
  • First, perform the first transformation's matrix multiplication.
  • Next, apply the second transformation by multiplying the result from the first step with the second matrix.
For instance, to rotate and then reflect an object, we multiply the reflection matrix by the rotation matrix: \[T = M \cdot R = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \-\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \0 & 0 & 1 \end{bmatrix}\] This final matrix can be applied to a point to achieve both transformations in one seamless step.

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Most popular questions from this chapter

Consider the production model \(\mathbf{x}=C \mathbf{x}+\mathbf{d}\) for an economy with two sectors, where $$C=\left[\begin{array}{cc}{.0} & {.5} \\ {.6} & {.2}\end{array}\right], \quad \mathbf{d}=\left[\begin{array}{c}{50} \\ {30}\end{array}\right]$$ Use an inverse matrix to determine the production level necessary to satisfy the final demand.

Suppose columns \(1,3,5,\) and 6 of a matrix \(A\) are linearly independent (but are not necessarily pivot columns) and the rank of \(A\) is \(4 .\) Explain why the four columns mentioned must be a basis for the column space of \(A .\)

a. Verify that \(A^{2}=I\) when \(A=\left[\begin{array}{rr}{1} & {0} \\ {3} & {-1}\end{array}\right]\) b. Use partitioned matrices to show that \(M^{2}=I\) when \(M=\left[\begin{array}{rrrr}{1} & {0} & {0} & {0} \\ {3} & {-1} & {0} & {0} \\\ {1} & {0} & {-1} & {0} \\ {0} & {1} & {-3} & {1}\end{array}\right]\)

In Exercises 19–24, justify each answer or construction. If possible, construct a \(3 \times 4\) matrix \(A\) such that dim Nul \(A=2\) and \(\operatorname{dim} \operatorname{Col} A=2\)

Use partitioned matrices to prove by induction that the product of two lower triangular matrices is also lower triangular. [Hint: A \((k+1) \times(k+1)\) matrix \(A_{1}\) can be written in the form below, where \(a\) is a scalar, \(\mathbf{v}\) is in \(\mathbb{R}^{k},\) and \(A\) is a \(k \times k\) lower triangular matrix. See the Study Guide for help with induction. \(A_{1}=\left[\begin{array}{cc}{a} & {\mathbf{0}^{T}} \\ {\mathbf{v}} & {A}\end{array}\right]\)

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