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Chapter 10: Problem 11

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

Short Answer

Expert verified
The inverse transformation is \( x = \frac{1}{5} x' + \frac{3}{5} y' \) \( y = -\frac{1}{5} x' + \frac{2}{5} y' \). The transformation is not orthogonal.

Step by step solution

01

Write the system of equations in matrix form

Express the given transformation using a matrix representation. The system of equations can be written as: \[ \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} \]
02

Find the inverse of the transformation matrix

To find the inverse transformation, first determine the inverse of the matrix: For a matrix \[ A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \ \text{, the inverse is given by} \ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \] Using this, find the inverse of \[ \begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix} \]
03

Calculate the inverse matrix

First calculate the determinant: \[ ad - bc = (2)(1) - (-3)(1) = 2 + 3 = 5 \ \therefore A^{-1} = \frac{1}{5} \begin{pmatrix} 1 & 3 \ -1 & 2 \end{pmatrix} \] The inverse matrix is: \[ A^{-1} = \begin{pmatrix} \frac{1}{5} & \frac{3}{5} \ -\frac{1}{5} & \frac{2}{5} \end{pmatrix} \]
04

Express \((x, y)\) in terms of \((x', y')\)

Use the inverse matrix to transform \((x', y')\) back to \((x, y)\): \[ \begin{pmatrix} x \ y \end{pmatrix} = A^{-1} \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} \frac{1}{5} & \frac{3}{5} \ -\frac{1}{5} & \frac{2}{5} \end{pmatrix} \begin{pmatrix} x' \ y' \end{pmatrix} \] This gives: \[ x = \frac{1}{5} x' + \frac{3}{5} y' \] \[ y = -\frac{1}{5} x' + \frac{2}{5} y' \]
05

Check if the transformation is orthogonal

To check orthogonality, determine if the original transformation matrix is orthogonal; that is if \[ A^{-1} = A^T \]. Since \[ \begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix} A^{-1} ≠ A^T \]},
06

Conclusion

Thus, the transformation is not orthogonal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

inverse matrix
An inverse matrix is like the 'reverse' of a matrix. When you multiply a matrix by its inverse, you get the identity matrix (which is like the number 1 for matrices). To find the inverse of a 2x2 matrix, you first need to calculate its determinant. For a matrix \ \ A = \begin{pmatrix} a & b \ c & d \end{pmatrix}, the determinant is calculated as \( ad - bc \). The formula for the inverse is \ \ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}. So, by finding the determinant and using this formula, we can get the inverse matrix.
linear transformation
A linear transformation is a function that takes a vector and transforms it into another vector. It has to satisfy two main properties: additivity and homogeneity. This means if you have vectors u and v, and a scalar k, then \( T(u + v) = T(u) + T(v) \) and \( T(ku) = kT(u) \). Imagine moving and scaling shapes on a plane; that's what linear transformations do. They are often represented by matrices. In our example, the equations \( x' = 2x - 3y \) and \( y' = x + y \) form a linear transformation. Using matrices simplifies these operations, allowing us to find solutions and inverses.
orthogonality test
Orthogonality is like checking if transformations keep the properties of a shape (like distances and angles). For a matrix to be orthogonal, its transpose has to be equal to its inverse \( A^{-1} = A^T \). This means the rows and columns of the matrix are orthogonal unit vectors. To test this, you can multiply the matrix by its transpose and see if you get the identity matrix. If not, the transformation is not orthogonal. In the given exercise, after finding the transpose and inverse, they do not match, confirming the transformation is not orthogonal.
determinant calculation
The determinant is a special number that can tell us a lot about a matrix. For a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is calculated as \( ad - bc \). The determinant helps in finding the inverse of a matrix and also tells if a matrix is invertible. If the determinant is zero, the matrix does not have an inverse. In the exercise, the determinant is found to be 5 (since \( 2 \cdot 1 - (-3) \cdot 1 = 5 \)), which is non-zero, so the inverse exists. This determinant is crucial in finding the inverse matrix.

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Most popular questions from this chapter

Let \(\mathbf{T}\) be a second-order tensor (dyadic) and let \(\mathrm{V}\) be a vector. Then we have seen that \(\mathbf{T} \cdot \mathrm{V}\) is another vector, say U. Write the equations giving each of the three components of \(\mathrm{U}\) as linear combinations of the three components of \(\mathrm{V} ;\) write these equations in matrix form, dyadic form, and summation form. Observe that the nine coefficients in these linear combinations are just the nine components of \(T\).

Let each of the following matrices \(M\) describe a deformation of the \((x, y)\) plane. For each given \(M\) find: the eigenvalues and eigenvectors of the transformation, the matrix \(C\) which diagonalizes \(M\) and specifies the rotation to new axes \(\left(x^{\prime}, y^{\prime}\right)\) along the eigenvectors, and the matrix \(D\) which gives the deformation relative to the new axes. Describe the deformation relative to the new axes.\(\left(\begin{array}{ll}3 & 4 \\ 4 & 9\end{array}\right)\)

The characteristic equation for a second-order matrix \(M\) is a quadratic equation. We have, considered in detail the case in which \(M\) is a real symmetric matrix and the roots of thecharacteristic equation (eigenvalues) are real, positive, and unequal. Discuss some other possibilities as follows: (a) \(M\) real and symmetric, eigenvalues real, one positive and one negative. Show that the plane is reflected in one of the cigenvector lines (as well as stretched or shrunk). Consider as a simple special case $$ M=\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right) $$ (b) \(M\) real and symmetric, eigenvalues equal (and therefore real). Show that \(M\) must he a multiple of the unit matrix. Thus show that the deformation consists of dilation or shrinkage in the radial direction (the same in all directions) with no rotation (and reflection in the origin if the root is negative). (c) \(M\) real, not symmetric, eigenvalues real and not equal. Show that in this case the eigenvectors are not orthogonal. Hint: Find their dot product. (d) \(M\) real, not symmetric, eigenvalues complex, Show that all vectors are rotated, that is, there are no (real) eigenvectors which are unchanged in direction by the transformation. Consider the characteristic equation of a rotation matrix as a special case.

Find the eigenvalues and eigenvectors of the following matrices.\(\left(\begin{array}{lll}1 & 2 & 2 \\ 2 & 3 & 0 \\ 2 & 0 & 3\end{array}\right)\)

Any rotation of axes in three dimensions can be described by giving the nine direction cosines of the angles between the \((x, y, z)\) and \(\left(x^{\prime}, y^{\prime}, z^{\prime}\right)\) axes. Show that the 3 by 3 matrix of these direction cosines [arranged as in the table in (11.1)] is an orthogonal matrix. Himt: Find \(A A^{T}\).
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