Question

Find the inverse of the transformation \(x^{\prime}=2 x-3 y, y^{\prime}=x+y,\) that is, find \(x, y\) in terms of \(x^{\prime}, y^{\prime}\). (Hint: Use matrices.) Is the transformation orthogonal?

Short answer

The inverse transformation is \( x = \frac{1}{5}x^\prime + \frac{3}{5}y^\prime \) and \( y = -\frac{1}{5}x^\prime + \frac{2}{5}y^\prime \). The transformation is not orthogonal.

Step-by-step solution

Write the transformation equations in matrix form

Express the given transformation

Set up the matrix equation

Rewrite the system of equations as a matrix equation: \[ \begin{pmatrix} x^\prime \ y^\prime \end{pmatrix} = \begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} \]

Invert the transformation matrix

Find the inverse of the transformation matrix \( \begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix} \). Calculate the determinant \(\Delta = 2(1) - (-3)(1) = 2 + 3 = 5\). The inverse matrix is given by \[ \frac{1}{\Delta} \begin{pmatrix} 1 & 3 \ -1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{5} & \frac{3}{5} \ -\frac{1}{5} & \frac{2}{5} \end{pmatrix} \]

Find the inverse transformation

Multiply the inverse matrix by the column matrix \( \begin{pmatrix} x^\prime \ y^\prime \end{pmatrix} \): \[ \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} \frac{1}{5} & \frac{3}{5} \ -\frac{1}{5} & \frac{2}{5} \end{pmatrix} \begin{pmatrix} x^\prime \ y^\prime \end{pmatrix} \] This results in the equations: \( x = \frac{1}{5}x^\prime + \frac{3}{5}y^\prime \) and \( y = -\frac{1}{5}x^\prime + \frac{2}{5}y^\prime \).

Check orthogonality

A transformation is orthogonal if its transformation matrix \( A \) satisfies \( A^T A = I \), where \( A^T \) is the transpose of \( A \). For \( A = \begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix} \), multiply \( A^T \) and \( A \): \[ \begin{pmatrix} 2 & 1 \ -3 & 1 \end{pmatrix} \begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix} = \begin{pmatrix} 5 & -1 \ -1 & 10 \end{pmatrix} eq I \] Therefore, the transformation is not orthogonal.

Key concepts

Inverse Matrix

An inverse matrix is essential in finding the solution to linear transformations and systems of equations. To determine the inverse of a matrix, follow these steps:

Calculate the determinant, \(\text{det}(A)\). For a 2x2 matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is \( \text{det}(A) = ad - bc \).
Switch the positions of \(a\) and \(d\) and change the signs of \(b\) and \(c\).
Divide each element by the determinant. The inverse matrix is \( A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \).

In our transformation, the matrix \(A = \begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix}\) has a determinant of \(5\). Following our steps, the inverse matrix is \( \frac{1}{5} \begin{pmatrix} 1 & 3 \ -1 & 2 \end{pmatrix} \). Multiplying this inverse by the matrix of the transformed coordinates, we find the original coordinates.

Orthogonal Transformation

An orthogonal transformation retains length and angle measurements, essentially preserving geometric properties. For a matrix \(A\) to represent an orthogonal transformation, it must satisfy \( A^T A = I \), where \(A^T\) is the transpose of \(A\) and \(I\) is the identity matrix.

To find \(A^T\), swap the rows and columns of \(A\). If \(A = \begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix}\), then \( A^T = \begin{pmatrix} 2 & 1 \ -3 & 1 \end{pmatrix} \).
Multiplying \(A^T \)with \(A\), we get \( \begin{pmatrix} 2 & 1 \ -3 & 1 \end{pmatrix} \begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix} = \begin{pmatrix} 5 & -1 \ -1 & 10 \end{pmatrix} \). This result is not equal to the identity matrix \(I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\), so the transformation is not orthogonal.

The important distinction is that non-orthogonal transformations can change geometrical properties such as the length of vectors and the angles between them.

Determinant Calculation

The determinant of a matrix is a special number that provides valuable information about the matrix. It is especially crucial in the context of calculating the inverse matrix. For a 2x2 matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is computed as \( ad - bc \).

Determining whether a matrix has an inverse is simple:

• If \(\text{det}(A)\) is zero, the matrix does not have an inverse.
• If \(\text{det}(A)\) is non-zero, the matrix does have an inverse.

In our example, the transformation matrix is \( \begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix} \), and its determinant is calculated as:
\( \text{det}(A) = 2 \times 1 - (-3) \times 1 = 2 + 3 = 5 \) Since \( 5 \) is not zero, the matrix is invertible.

Matrix Multiplication

Matrix multiplication is a fundamental operation when dealing with linear transformations. This operation combines rows of the first matrix with columns of the second one.
Given two matrices, \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\) and \(B = \begin{pmatrix} e & f \ g & h \end{pmatrix}\), their product is calculated by taking the dot product of rows of \(A\) with columns of \(B\). This results in:
\(AB = \begin{pmatrix} ae+bg & af+bh \ ce+dg & cf+dh \end{pmatrix} \).

In our transformed system, we use the inverse of the transformation matrix and multiply it with the column matrix of the new coordinates to derive the original coordinates:
\( \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} \frac{1}{5} & \frac{3}{5} \ -\frac{1}{5} & \frac{2}{5} \end{pmatrix} \begin{pmatrix} x^\text{'} \ y^\text{'} \end{pmatrix} \).
This multiplication yields the original coordinates \( (x, y) \) in terms of \( (x^\text{'}, y^\text{'}) \). Understanding this process helps greatly in solving systems of linear equations and transformations.