Problem 64 - 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? $$ A=\left[\begin{array}{rrr} 1 & 0 & -4 \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{array}\right],(x, y)=(4,2), \text { and } X=\left[\begin{array}{l} 4 \\ 2 \\ 1 \end{array}\right] $$

Step-by-Step Solution

Answer

The new location is (0, 7). A^{-1} Y returns the original point (4, 2). Both AA^{-1} and A^{-1}A equal the identity matrix.

Step 1: Predict New Location of the Point

The transformation matrix \(A\) translates a point \((x, y)\) by \((h, k)\). In this case, \(h = -4\) and \(k = 5\). Therefore, the new location of the point \((4, 2)\) will be computed by adding \(h\) and \(k\) to \(x\) and \(y\) respectively: \((4 + (-4), 2 + 5) = (0, 7)\).

Step 2: Compute Y = AX

To verify the prediction, compute the matrix product \(Y = AX\). Given \(A = \begin{bmatrix} 1 & 0 & -4 \ 0 & 1 & 5 \ 0 & 0 & 1 \end{bmatrix}\) and \(X = \begin{bmatrix} 4 \ 2 \ 1 \end{bmatrix}\), the multiplication yields: \[Y = \begin{bmatrix} 1 \cdot 4 + 0 \cdot 2 - 4 \cdot 1 \ 0 \cdot 4 + 1 \cdot 2 + 5 \cdot 1 \ 0 \cdot 4 + 0 \cdot 2 + 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} 0 \ 7 \ 1 \end{bmatrix}\] Thus, the result confirms that the new location of the point is \((0, 7)\).

Step 3: Conjecture About A^(-1)Y

The inverse matrix \(A^{-1}\) would perform the reverse transformation of \(A\). Therefore, \(A^{-1}Y\) would return the transformed point back to its original location \((x, y) = (4, 2)\).

Step 4: Find A^(-1) and Calculate A^(-1)Y

To find \(A^{-1}\), we replace \(h\) and \(k\) in \(A\) with \((-h, -k)\) and negate the third column. Thus, \[A^{-1} = \begin{bmatrix} 1 & 0 & 4 \ 0 & 1 & -5 \ 0 & 0 & 1 \end{bmatrix}\]Substitute \(Y = \begin{bmatrix} 0 \ 7 \ 1 \end{bmatrix}\) in the matrix product \(A^{-1}Y\) to get: \[A^{-1}Y = \begin{bmatrix} 1 \cdot 0 + 0 \cdot 7 + 4 \cdot 1 \ 0 \cdot 0 + 1 \cdot 7 - 5 \cdot 1 \ 0 \cdot 0 + 0 \cdot 7 + 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} 4 \ 2 \ 1 \end{bmatrix}\]This confirms that \(A^{-1}Y\) returns the original point \((4, 2)\).

Step 5: Calculate AA^{-1} and A^{-1}A

The product \(AA^{-1}\) and \(A^{-1}A\) for any invertible matrix \(A\) should equal the identity matrix. Calculate: \[AA^{-1} = \begin{bmatrix} 1 & 0 & -4 \ 0 & 1 & 5 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 4 \ 0 & 1 & -5 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]\[A^{-1}A = \begin{bmatrix} 1 & 0 & 4 \ 0 & 1 & -5 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & -4 \ 0 & 1 & 5 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]Both products yield the identity matrix, confirming our calculations.

Key Concepts

Translation Matrices

Translation matrices are used to shift points or geometric figures in space without altering their orientation or shape. They are fundamental in graphics and animations, providing a simple method to move objects around a coordinate plane.
In the given problem, we use a translation matrix \(A\), which includes the translation values in its first two rows and third column. For example, the matrix \(A\) in the problem is:\[A = \begin{bmatrix}1 & 0 & h \0 & 1 & k \0 & 0 & 1\end{bmatrix}\]Here, \(h\) is the translation along the x-axis and \(k\) is along the y-axis. The point \((x, y)\) is expressed in homogeneous coordinates as \(X\):\[X = \begin{bmatrix}x \y \1\end{bmatrix}\]To find the translated point, we compute the matrix product \(Y = AX\), resulting in a new point \((x', y') = (x + h, y + k)\). This approach is both practical and efficient for translating points in linear algebra.

Inverse Matrix

An inverse matrix is used to reverse the effect of its original matrix on a transformed point or figure. Finding the inverse of a matrix essentially 'undoes' whatever transformation the original matrix applied.For the given translation matrix \(A\), the inverse matrix \(A^{-1}\) is as follows:\[A^{-1} = \begin{bmatrix}1 & 0 & -h \0 & 1 & -k \0 & 0 & 1\end{bmatrix}\]This inverse matrix changes the direction of the translation. For instance, if \(A\) shifts a point to the right by 4 units and up by 5 units, \(A^{-1}\) would shift it left by 4 units and down by 5 units. By applying \(A^{-1}\) to a point that has already been transformed by \(A\), you essentially return the point to its original position.Thus, in the exercise, calculating \(A^{-1}Y\) allows us to verify that the transformation has indeed been reversed, taking the point back from its new location to its original one.

Matrix Multiplication

Matrix multiplication is a cornerstone of linear transformations, allowing us to combine transformations and efficiently handle numerous points.It involves the systematic multiplication of corresponding rows and columns between two matrices. Specifically, each element in the resulting matrix is produced by multiplying elements of the rows of the first matrix by the corresponding elements of the columns of the second matrix and summing these products.For example, performing matrix multiplication with the translation matrix \(A\) and the vector \(X\) involves:\[Y = \begin{bmatrix}1 & 0 & -4 \0 & 1 & 5 \0 & 0 & 1\end{bmatrix} \begin{bmatrix}4 \2 \1\end{bmatrix} = \begin{bmatrix}0 \7 \1\end{bmatrix}\]This results in the new coordinates \((0, 7, 1)\), effectively translating the point. It beautifully illustrates how matrix multiplication can apply complex transformations quickly and dependably in linear algebra.

Identity Matrix

The identity matrix is a special matrix that acts like the number 1 in regular multiplication. It leaves any matrix unchanged when multiplied.In matrix form, the identity matrix looks like:\[I = \begin{bmatrix}1 & 0 & 0 \0 & 1 & 0 \0 & 0 & 1\end{bmatrix}\]This matrix is especially important in the context of invertible matrices. When a matrix \(A\) is multiplied by its inverse \(A^{-1}\), the result is always the identity matrix. This property is used to check the accuracy of matrix inversions and transformations.Calculating \(AA^{-1}\) or \(A^{-1}A\) for the exercise yields:\[AA^{-1} = \begin{bmatrix}1 & 0 & -4 \0 & 1 & 5 \0 & 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & 4 \0 & 1 & -5 \0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \0 & 1 & 0 \0 & 0 & 1\end{bmatrix}\]Identifying that the result is the identity matrix confirms the correctness and consistency of our calculations, affirming the roles of the inverse matrix and matrix multiplication in linear transformations.