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

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

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/√(2) &-1/√(2) \\1/√(2) &1/√(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&√(3)/2 \\-√(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&√(3)/2 \\-√(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/√(2) &-1/√(2) \\1/√(2) &1/√(2)\end{array}\right]`

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