Question

Make a matrix A whose action is described as follows: The hit by A rotates everything Pi/4 counterclockwise radians, then stretches by a factor of 1.8 along the x-axis and a factor of 0.7 along the y-axis and then rotates the result by Pi/3 clockwise radians.

Answer 1

Answer:

The required matrix is$A = \left[\begin{array}{ccc}1.07&-0.21\\-0.86&1.35\end{array}\right]$

Step-by-step explanation:

Matrix of rotation:

$P = \left[\begin{array}{ccc}cos\pi/4&-sin\pi/4\\sin\pi/4&cos\pi/4\end{array}\right]$

$P = \left[\begin{array}{ccc}1/\sqrt{2} &-1/\sqrt{2} \\1/\sqrt{2} &1/\sqrt{2}\end{array}\right]$

x' + iy' = (x + iy)(cosθ + isinθ)

x' = x cosθ - ysinθ

y' = x sinθ + ycosθ

In matrix form:

$\left[\begin{array}{ccc}x'\\y'\end{array}\right] = \left[\begin{array}{ccc}cos\theta&-sin\theta\\sin \theta&cos\theta\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right]$

The matrix stretches by 1.8 on the x axis and 0.7 on the y axis

i.e. x' = 1.8x

y' = 0.7y

$\left[\begin{array}{ccc}x'\\y'\end{array}\right] = \left[\begin{array}{ccc}1.8&0\\0&0.7\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right]$

$Q = \left[\begin{array}{ccc}1.8&0\\0&0.7\end{array}\right]$

According to the question, the result is rotated by pi/3 clockwise radians

$R = \left[\begin{array}{ccc}cos(-\pi/3)& -sin(-\pi/3)\\-sin(\pi/3)&cos(\pi/3)\end{array}\right]$

$R = \left[\begin{array}{ccc}1/2&\sqrt{3}/2 \\-\sqrt{3}/2 &1/2\end{array}\right]$

To get the matrix A, we would multiply matrices R, Q and P together.

$A = RQP = \left[\begin{array}{ccc}1/2&\sqrt{3}/2 \\-\sqrt{3}/2 &1/2\end{array}\right] \left[\begin{array}{ccc}1.8&0\\0&0.7\end{array}\right] \left[\begin{array}{ccc}1/\sqrt{2} &-1/\sqrt{2} \\1/\sqrt{2} &1/\sqrt{2}\end{array}\right]$

$A = \left[\begin{array}{ccc}1.07&-0.21\\-0.86&1.35\end{array}\right]$