Problem 63 - College Algebra with Modeling and Visualization

Question

Translations (Refer to the discussion in this section about translating a point.) The matrix product AX performs a translation on the point \((x, y),\) where $$ A=\left[\begin{array}{lll} 1 & 0 & h \\ 0 & 1 & k \\ 0 & 0 & 1 \end{array}\right] \text { and } \quad X=\left[\begin{array}{l} x \\ y \\ 1 \end{array}\right] $$ (A) Predict the new location of the point (x, y) when it is translated by \(A\). Compute \(Y=A X\) to verify your prediction. (B) Make a conjecture as to what \(A^{-1}\) Y represents. Find \(A^{-1}\) and calculate \(A^{-1} Y\) to test your conjecture. (C) What will \(A A^{-1}\) and \(A^{-1} A\) equal? $$ A=\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array}\right],(x, y)=(0,1), \text { and } X=\left[\begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right] $$

Step-by-Step Solution

Answer

The translated point is (2, 4). The inverse returns the point to (0, 1), and both \( AA^{-1} \) and \( A^{-1}A \) equal the identity matrix.

Step 1: Predict the new point location

The matrix \( A \) translates the point \((x, y)\) by adding \(h\) to \(x\) and \(k\) to \(y\). For the given matrix \( A \) with \( h = 2 \) and \( k = 3 \), and point \((x, y) = (0, 1)\), the new point after translation is \((x', y') = (0+2, 1+3) = (2, 4)\).

Step 2: Matrix multiplication to verify prediction

Compute \( Y = AX \) where \( A = \begin{bmatrix} 1 & 0 & 2 \ 0 & 1 & 3 \ 0 & 0 & 1 \end{bmatrix} \) and \( X = \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} \). Multiply the two matrices:\[ Y = \begin{bmatrix} 1 & 0 & 2 \ 0 & 1 & 3 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} 0*1 + 1*0 + 1*2 \ 0*0 + 1*1 + 1*3 \ 0*0 + 0*1 + 1*1 \end{bmatrix} = \begin{bmatrix} 2 \ 4 \ 1 \end{bmatrix} \].Thus, the new location \( (2, 4) \) verifies our prediction.

Step 3: Conjecture on inverse representation

The matrix \( A^{-1} \) when applied to \( Y \) should ideally reverse the transformation applied by \( A \). Thus, \( A^{-1} Y \) should return the original point \((0, 1)\).

Step 4: Compute the inverse of A

The inverse \( A^{-1} \) of a translation matrix \( A = \begin{bmatrix} 1 & 0 & h \ 0 & 1 & k \ 0 & 0 & 1 \end{bmatrix} \) is \( A^{-1} = \begin{bmatrix} 1 & 0 & -h \ 0 & 1 & -k \ 0 & 0 & 1 \end{bmatrix} \). For \( h = 2 \) and \( k = 3 \), we have \( A^{-1} = \begin{bmatrix} 1 & 0 & -2 \ 0 & 1 & -3 \ 0 & 0 & 1 \end{bmatrix} \).

Step 5: Calculate A^{-1} Y

Apply the inverse matrix to \( Y \):\[ A^{-1}Y = \begin{bmatrix} 1 & 0 & -2 \ 0 & 1 & -3 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 2 \ 4 \ 1 \end{bmatrix} = \begin{bmatrix} 1*2 + 0*4 + (-2)*1 \ 0*2 + 1*4 + (-3)*1 \ 0*2 + 0*4 + 1*1 \end{bmatrix} = \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} \].The result \((0, 1)\) confirms that \( A^{-1} Y \) returns to the original point.

Step 6: Compute A A^{-1} and A^{-1} A

Using the property of invertible matrices, \( AA^{-1} = I \) and \( A^{-1}A = I \), where \( I \) is the identity matrix. Calculate:- \( AA^{-1} = \begin{bmatrix} 1 & 0 & 2 \ 0 & 1 & 3 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & -2 \ 0 & 1 & -3 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I \).- \( A^{-1}A = \begin{bmatrix} 1 & 0 & -2 \ 0 & 1 & -3 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \ 0 & 1 & 3 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I \).Both products result in the identity matrix.

Key Concepts

Translation Matrices

A translation matrix is a special type of transformation matrix. It is mainly used to move points or objects from one location to another in a given space. In a two-dimensional space, translation matrices are 3x3 matrices that incorporate a translation vector. This vector is used to shift an object by a specified distance in the x and y directions.

For example, consider the matrix:

• \[ A = \begin{bmatrix} 1 & 0 & h \ 0 & 1 & k \ 0 & 0 & 1 \end{bmatrix} \]

Here, the values \( h \) and \( k \) represent the horizontal and vertical shifts, respectively. When this matrix multiplies a point \( (x, y) \) represented as the vector \[ X = \begin{bmatrix} x \ y \ 1 \end{bmatrix} \], it shifts the point, resulting in a new location \( (x + h, y + k) \). This operation effectively "translates" or moves the point in the xy-plane without altering its original shape or orientation.

Inverse Matrices

An inverse matrix is the key to reversing transformations. For a given matrix \( A \), the inverse is denoted as \( A^{-1} \). Multiplying \( A \) and \( A^{-1} \) results in the identity matrix, indicating a perfect reversal of transformations.

To find the inverse of a translation matrix, say \( A = \begin{bmatrix} 1 & 0 & h \ 0 & 1 & k \ 0 & 0 & 1 \end{bmatrix} \), we use the inverse formula. This results in:

• \[ A^{-1} = \begin{bmatrix} 1 & 0 & -h \ 0 & 1 & -k \ 0 & 0 & 1 \end{bmatrix} \]

By applying the inverse matrix \( A^{-1} \) to a transformed point \( Y \), you essentially "undo" the translation. It returns the point to its original position before the translation was applied, as seen in the exercise where \( A^{-1}Y \) mapped back to the initial point \( (0, 1) \).

Identity Matrix

The identity matrix is a special type of matrix that, when multiplied with another matrix, does not change the original matrix. It functions as the multiplicative identity in matrix algebra, similar to how the number 1 is the multiplicative identity in standard arithmetic.

In two-dimensional transformations, the identity matrix is a 3x3 matrix:

• \[ I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]

When you multiply any matrix by the identity matrix, the result will always be the original matrix, i.e., \( AI = A \). Similarly, the operations \( AA^{-1} \) or \( A^{-1}A \) resulting in the identity matrix demonstrates the fundamental properties of inverse matrices as shown in the exercise.

Matrix Multiplication

Matrix multiplication is the principal arithmetic operation in matrix algebra. It involves taking two matrices and producing a third matrix as a result. To multiply two matrices, the number of columns in the first must equal the number of rows in the second. This operation combines the rows of the first matrix with the columns of the second.

For translation matrices, matrix multiplication can be visualized when a transformation matrix \( A \) is multiplied with a point in homogeneous coordinates \( X = \begin{bmatrix} x \ y \ 1 \end{bmatrix} \). The resulting product, \( Y = AX \), gives a new matrix that represents the transformed position. For instance,

• \[ Y = \begin{bmatrix} 1 & 0 & 2 \ 0 & 1 & 3 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} 2 \ 4 \ 1 \end{bmatrix} \]

This example from the exercise demonstrates how matrix multiplication accurately shifts the original point \((0,1)\) by \(2\) units in the x-direction and \(3\) units in the y-direction.