Problem 69 - College Algebra with Modeling and Visualization

Question

The matrix \(A\) translates a point to the right 4 units and downward 2 units, and the matrix \(B\) translates a point to the left 3 units and upward 3 units, where $$ A=\left[\begin{array}{rrr} 1 & 0 & 4 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{array}\right] \text { and } B=\left[\begin{array}{rrr} 1 & 0 & -3 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array}\right] $$ (A) Let \(X\) represent the point \((1,1)\). Predict the result of \(\boldsymbol{Y}=\boldsymbol{A B X} .\) Check your prediction. (B) Find \(A B\) mentally, and then compute \(A B\) (C) Would you expect \(A B=B A ?\) Verify your answer. (D) Find \((A B)^{-1}\) mentally. Explain your reasoning.

Step-by-Step Solution

Answer

(A) Y = (2, 2); (B) AB = identity matrix with translation [1,1]. (C) No, AB ≠ BA. (D) Inverse is translation by [-1,-1].

Step 1: Represent the Point Using Homogeneous Coordinates

To perform the transformation using matrices, we first represent the point \(X\) using homogeneous coordinates. The point \((1,1)\) in homogeneous coordinates is written as \([1, 1, 1]^T\).

Step 2: Compute the Transformation ABX

First compute \(BX\) by applying matrix \(B\) to \([1, 1, 1]^T\): \[B = \begin{bmatrix} 1 & 0 & -3 \ 0 & 1 & 3 \ 0 & 0 & 1 \end{bmatrix}\] Applying this gives: \(BX = \begin{bmatrix} 1 & 0 & -3 \ 0 & 1 & 3 \ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} -2 \ 4 \ 1 \end{bmatrix}\). Next, apply \(A\): \(Y = A \cdot BX = \begin{bmatrix} 1 & 0 & 4 \ 0 & 1 & -2 \ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} -2 \ 4 \ 1 \end{bmatrix} = \begin{bmatrix} 2 \ 2 \ 1 \end{bmatrix}\). This shows that \(Y = (2, 2)\).

Step 3: Mental Verification and Matrix Multiplication

Mentally, when multiplying \(A\) and \(B\), observe that the translation represented by \(A\) followed by \(B\) involves only the translation vectors, since the upper left \(2 \times 2\) block of both matrices is the identity matrix. Adding the translation vectors [4, -2] and [-3, 3], \(AB\) should translate by \([4 - 3, -2 + 3] = [1, 1]\). Now actually compute \(AB\) to verify: \[AB = \begin{bmatrix} 1 & 0 & 4 \ 0 & 1 & -2 \ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & -3 \ 0 & 1 & 3 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix}\].

Step 4: Check Commutativity AB = BA

Now consider \(B\cdot A\). When these translations are reversed, \(B\) moves left/causeing successive translations different from \(A\cdot B\) (because matrix multiplication of non-diagonal matrices is generally non-commutative): \[BA = \begin{bmatrix} 1 & 0 & -3 \ 0 & 1 & 3 \ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 4 \ 0 & 1 & -2 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix}\]. Therefore, \(AB eq BA\).

Step 5: Find the Inverse of AB Mentally

The transformation \(AB\) is a single transformation that translates by \( (1, 1) \). Its inverse would translate by \(-1\) in both components. The inverse matrix \((AB)^{-1}\) is given by: \[(AB)^{-1} = \begin{bmatrix} 1 & 0 & -1 \ 0 & 1 & -1 \ 0 & 0 & 1 \end{bmatrix}\]. This matrix moves points back to their original position, reversing the translation.

Key Concepts

Homogeneous Coordinates

Homogeneous coordinates are an extension of traditional Cartesian coordinates. They are widely used in computer graphics and 3D geometry. In homogeneous coordinates, a point \(x, y\) is represented as \([x, y, 1]^T\). This additional dimension, often called the "w" component or scale factor, simplifies calculations involving translations, rotations, and perspectives.
By using homogeneous coordinates, transformations such as translation become linear operations, handled neatly by matrix multiplication. This representation also allows for uniform manipulation in 3D rendering systems without complicating the linear algebra involved.

Matrix Multiplication

Matrix multiplication is a crucial operation that combines transformations. When you multiply two matrices, you're effectively applying the transformation of one matrix and then the other. The order in which you perform these operations is important due to the non-commutative nature of matrix multiplication.
In our example, to find \(ABX\), we start by multiplying matrix \(B\) with the point \(X\) in homogeneous coordinates. The result is then multiplied by matrix \(A\) to produce the final output. Such stepwise multiplication ensures each transformation is correctly applied and is necessary when dealing with multiple transformations in sequence.

Matrix Inverses

A matrix inverse is like a reverse transformation. If applying a matrix changes a point, multiplying by its inverse will return the point to its original position. Not all matrices have inverses, but those that do can be extremely useful in reversing transformations.
For our translation matrix \(AB\), it shifts points by \[ (1, 1) \]. The inverse of this matrix, \((AB)^{-1}\), is calculated to translate by \(-1\) along both x and y axes. This matrix effectively undoes the movement of \(AB\), ensuring that any point transformed by \(AB\) can be reverted.

Non-Commutative Property

Matrix multiplication suffers from a non-commutative property. This means changing the order of matrices can lead to different results. In simple terms, \(AB eq BA\).
This property plays an important role in transformations, as seen in our example. While \(AB\) and \(BA\) might involve the same matrices, the resulting transformations are markedly different, affecting translation paths and outcomes. Always remember the sequence of operations is paramount in matrix algebra.

Translation Matrix

Translation matrices are specifically designed to move points in space by a fixed amount. They are simple to identify, as they extend the identity matrix with a translation vector.
In our exercise, matrix \(A\) moves a point right by 4 units and down by 2 units, while matrix \(B\) moves 3 units to the left and 3 units up. These movements are represented in the third column of each 3x3 matrix. By manipulating this column, translation matrices can shift points efficiently, making them vital for computer graphics and motion transformations.