Chapter 7.2, Problem 71E

Solution

a.

To prove: The inverse of the matrix A is $\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$ .

The statement has proven.

Given Information:

The matrix is defined as,

  $\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$

Calculation:

Consider the given function,

  $\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$

Now, find the inverse of the given matrix.

  $\begin{array}{c}{\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]}^{-1}=\frac{1}{\cos \alpha \cdot \cos \alpha +\sin \alpha \sin \alpha }\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\\ =\frac{1}{{\cos }^2\alpha +{\sin }^2\alpha }\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\\ =\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\end{array}$

Therefore, the inverse of the matrix $A$ is $\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$ .

b.

To prove: The coordinates $\left(x,y\right)$ of $P$ are related to the $\left(x\text{'},y\text{'}\right)$ coordinates of $P$ .

The statement has proven.

Given Information:

The equations are defined as,

  $\begin{array}{c}x=x\text{'}\cos \alpha -y\text{'}\sin \alpha \\ y=x\text{'}\sin \alpha +y\text{'}\cos \alpha \end{array}$

Explanation:

Consider the given equations,

  $\begin{array}{c}x=x\text{'}\cos \alpha -y\text{'}\sin \alpha \\ y=x\text{'}\sin \alpha +y\text{'}\cos \alpha \end{array}$

So, the inverse of the matrix is defined as,

  $\left[\begin{array}{cc}x\text{'}& y\text{'}\end{array}\right]A^{-1}=\left[\begin{array}{cc}x& y\end{array}\right]$

First simplify the left side of the above equation.

  $\begin{array}{c}\left[\begin{array}{cc}x\text{'}& y\text{'}\end{array}\right]A^{-1}=\left[\begin{array}{cc}x\text{'}& y\text{'}\end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\\ =\left[\begin{array}{cc}x\text{'}\cos \alpha -y\text{'}\sin \alpha & x\text{'}\sin \alpha +y\text{'}\cos \alpha \end{array}\right]\end{array}$

This implies that,

  $\begin{array}{l}x=x\text{'}\cos \alpha -y\text{'}\sin \alpha \\ y=x\text{'}\sin \alpha +y\text{'}\cos \alpha \end{array}$

Hence, the given has proved.

c.

To prove: The coordinates $\left(x,y\right)$ can be obtained from the $\left(x\text{'},y\text{'}\right)$ coordinates by matrix multiplication and also explain the relation with A of this matrix.

The coordinate can be obtained by matrix multiplication and the transition matrix is $A^{-1}$ .

Given Information:

The equations are defined as,

  $\begin{array}{c}x=x\text{'}\cos \alpha -y\text{'}\sin \alpha \\ y=x\text{'}\sin \alpha +y\text{'}\cos \alpha \end{array}$

Calculation:

Consider the given information,

  $\begin{array}{c}x=x\text{'}\cos \alpha -y\text{'}\sin \alpha \\ y=x\text{'}\sin \alpha +y\text{'}\cos \alpha \end{array}$

It can be seen that the condition $\left[\begin{array}{cc}x\text{'}& y\text{'}\end{array}\right]A^{-1}=\left[\begin{array}{cc}x& y\end{array}\right]$ has proved already in the previous part.

This implies that the coordinate $\left(x,y\right)$ can be obtained from the coordinates $\left(x\text{'},y\text{'}\right)$ by the matrix multiplication and the transition matrix is $A^{-1}$ .