Problem 67 - College Algebra with Modeling and Visualization

Question

Translations (Refer to the discussion in this section about translating a point.) Find a \(3 \times 3\) matrix A that performs the following translation of a point \((x, y)\) represented by \(X .\) Find \(A^{-1}\) and describe what it computes. (Refer to Example 2.) The matrix \(B\) rotates the point \((x, y)\) clockwise about the origin \(45^{\circ},\) where $$ B=\left[\begin{array}{rrr} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{array}\right] \text { and } B^{-1}=\left[\begin{array}{rrr} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{array}\right] $$ (A) Let \(X\) represent the point \((-\sqrt{2},-\sqrt{2}) .\) Compute \(\boldsymbol{Y}=\boldsymbol{B} \boldsymbol{X}\) (B) Find \(B^{-1} Y\). Interpret what \(B^{-1}\) computes.

Step-by-Step Solution

Answer

(A) \(Y = (0, -2)\); (B) \(B^{-1}\) returns the point to a \(\sqrt{2}\) x-coordinate and \(-\sqrt{2}\) y-coordinate, counter-rotating by \(45^{\circ}\).

Step 1: Define Point Vector X

Given the point \((x, y) = (-\sqrt{2}, -\sqrt{2})\), we can represent it in homogeneous coordinates as \(X = \begin{bmatrix} -\sqrt{2} \ -\sqrt{2} \ 1 \end{bmatrix}\). This allows us to use a \(3 \times 3\) matrix for transformations.

Step 2: Perform the Rotation Transformation

To rotate the point \((-\sqrt{2}, -\sqrt{2})\) by \(45^{\circ}\) clockwise, multiply the vector \(X\) by the matrix \(B\):\[Y = B \cdot X = \left[ \begin{array}{rrr} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \ 0 & 0 & 1 \end{array} \right] \cdot \left[ \begin{array}{c} -\sqrt{2} \ -\sqrt{2} \ 1 \end{array} \right] = \left[ \begin{array}{c} 0 \ -2 \ 1 \end{array} \right].\]

Step 3: Compute Inverse Rotation

To reverse the rotation performed by \(B\), multiply \(Y\) by \(B^{-1}\):\[B^{-1} Y = \left[ \begin{array}{rrr} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \ 0 & 0 & 1 \end{array} \right] \cdot \left[ \begin{array}{c} 0 \ -2 \ 1 \end{array} \right] = \left[ \begin{array}{c} \sqrt{2} \ -\sqrt{2} \ 1 \end{array} \right].\]

Step 4: Interpret the Inverse Transformation

The matrix \(B^{-1}\) performs the reverse of the \(45^{\circ}\) clockwise rotation, thus applying it to \(Y\) results in the original point moved back before rotation, effectively counter-rotating by \(45^{\circ}\). The point \(\left(\sqrt{2}, -\sqrt{2}\right)\) gives the coordinates after this transformation.

Key Concepts

Matrix Inversion

In mathematics, a matrix inversion is a process important for solving systems of linear equations, among other applications. Simply put, inverting a matrix means finding another matrix, which when multiplied with the original matrix, results in the identity matrix. If you think of a matrix as a type of complex number, the inverse does a similar job to division.
In the context of transformation matrices, such as rotation matrices, the inverse matrix effectively "undoes" the transformation applied by the original matrix. For instance, if you rotate a point using a matrix, applying the inverse matrix would rotate it back to where it was.

• Finding an inverse of a matrix is similar to reversing a process, like unwinding a spiral.
• Not all matrices are invertible — just like zero cannot have a reciprocal in regular arithmetic.
• The inverse is useful in many fields such as computer graphics, physics, and engineering.

When dealing with the \(3 \times 3\) matrix in our exercise, such as matrix \(B\), its inverse \(B^{-1}\) is used to reverse a 45-degree clockwise rotation that the original matrix \(B\) performs.
Thus, if \(Y = B \cdot X\) is the point after rotation, \(B^{-1} \cdot Y\) will give back the original point coordinates before the rotation.

Homogeneous Coordinates

Homogeneous coordinates are a way of using extra dimensions to make calculations involving transformations simpler. If you’ve ever seen a point in regular 2D space as \((x, y)\), homogeneous coordinates add a third value, making it look like \((x, y, 1)\). This little trick opens up a world of possibilities in mathematical computations.
By using homogeneous coordinates:

• You can easily represent points at infinity. Great for certain geometry or 3D visualizations.
• They allow you to use matrix operations to perform transformations like translations, which aren’t possible with standard 2D vectors and matrices.
• This method simplifies the math for many complex operations in computer graphics.

In the example exercise, the point \((-\sqrt{2}, -\sqrt{2})\) is represented in homogeneous coordinates as \(([ -\sqrt{2}, -\sqrt{2}, 1])\). The extra \(1\) allows the use of the transformation matrix \(B\), which is \(3 \times 3\), to execute rotations easily and to utilize matrix inversion precisely.

Rotation Matrices

Rotation matrices are a special type of transformation matrix used to rotate points in space. In 2D, a rotation matrix can rotate points around the origin. The matrix \(B\) from the exercise rotates a point \(45^{\circ}\) clockwise.
These matrices are often defined as:\[R_{\theta} = \left[\begin{array}{cc}\cos \theta & -\sin \theta \\sin \theta & \cos \theta \end{array}\right]\]Rotation matrices are useful because:

• Their determinant is always 1, meaning they preserve the area of shapes.
• They ensure only the angle and not the point's position or size are changed during rotation.
• They can be easily inverted to undo the rotation, as shown with matrix \(B^{-1}\) in the exercise.

When using rotation matrices like \(B\) in homogeneous coordinates, as seen in the exercise, you have the added advantage of managing translations and other transformations in a unified framework. The inverse, \(B^{-1}\), then effortlessly reverses the rotation, showcasing the elegance and power of using rotation matrices in mathematical computations.