Question - Transformation Matrix Involving Vertical Stretch and Clockwise Rotation

We first calculate the transformation matrix for a vertical stretch with scale factor \( k \):

\[ A_{\text{stretch}} = \left( \begin{array}{cc} 1 & 0 \\ 0 & k \end{array} \right) \]

Next, we calculate the transformation matrix for a clockwise rotation through \( \frac{\pi}{4} \) radians about the origin \( O \):

\[ A_{\text{rotation}} = \left( \begin{array}{cc} \cos\left(\frac{\pi}{4}\right) & \sin\left(\frac{\pi}{4}\right) \\ -\sin\left(\frac{\pi}{4}\right) & \cos\left(\frac{\pi}{4}\right) \end{array} \right) = \left( \begin{array}{cc} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{array} \right) \]

The combined transformation matrix \( A \) for both transformations is given by the product of the individual matrices:

\[ A = A_{\text{stretch}} \cdot A_{\text{rotation}} \]

\[ A = \left( \begin{array}{cc} 1 & 0 \\ 0 & k \end{array} \right) \cdot \left( \begin{array}{cc} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{array} \right) \]

\[ A = \left( \begin{array}{cc} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ -\frac{k\sqrt{2}}{2} & \frac{k\sqrt{2}}{2} \end{array} \right) \]

Using the given transformation, we can now map the point \( (1, k) \) to \( (2\sqrt{2}, \sqrt{2}) \) using matrix multiplication with \( A \):

\[ \left( \begin{array}{cc} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ -\frac{k\sqrt{2}}{2} & \frac{k\sqrt{2}}{2} \end{array} \right) \left( \begin{array}{c} 1 \\ k \end{array} \right) = \left( \begin{array}{c} 2\sqrt{2} \\ \sqrt{2} \end{array} \right) \]

\[ \left( \begin{array}{c} \frac{\sqrt{2}}{2} \cdot 1 + \frac{\sqrt{2}}{2} \cdot k \\ -\frac{k\sqrt{2}}{2} \cdot 1 + \frac{k\sqrt{2}}{2} \cdot k \end{array} \right) = \left( \begin{array}{c} 2\sqrt{2} \\ \sqrt{2} \end{array} \right) \]

Solving the system of equations:

\[ \frac{\sqrt{2}}{2} + \frac{k\sqrt{2}}{2} = 2\sqrt{2} \]

\[ -\frac{k\sqrt{2}}{2} + \frac{k^2\sqrt{2}}{2} = \sqrt{2} \]

From the first equation, we get:

\[ 1 + k = 4 \]

\[ k = 3 \]

We only need the first equation to find the value of \( k \) because the problem gives information that the transformation is vertical, implying that \( k \) is a scalar affecting only the second coordinate's magnitude and not its sign. The second equation provided in the operation is redundant since we have the value for \( k \). Hence, \( k = 3 \).