Respuesta :
Answer:
The required matrix is[tex]A = \left[\begin{array}{ccc}1.07&-0.21\\-0.86&1.35\end{array}\right][/tex]
Step-by-step explanation:
Matrix of rotation:
[tex]P = \left[\begin{array}{ccc}cos\pi/4&-sin\pi/4\\sin\pi/4&cos\pi/4\end{array}\right][/tex]
[tex]P = \left[\begin{array}{ccc}1/\sqrt{2} &-1/\sqrt{2} \\1/\sqrt{2} &1/\sqrt{2}\end{array}\right][/tex]
x' + iy' = (x + iy)(cosθ + isinθ)
x' = x cosθ - ysinθ
y' = x sinθ + ycosθ
In matrix form:
[tex]\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][/tex]
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
[tex]\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][/tex]
[tex]Q = \left[\begin{array}{ccc}1.8&0\\0&0.7\end{array}\right][/tex]
According to the question, the result is rotated by pi/3 clockwise radians
[tex]R = \left[\begin{array}{ccc}cos(-\pi/3)& -sin(-\pi/3)\\-sin(\pi/3)&cos(\pi/3)\end{array}\right][/tex]
[tex]R = \left[\begin{array}{ccc}1/2&\sqrt{3}/2 \\-\sqrt{3}/2 &1/2\end{array}\right][/tex]
To get the matrix A, we would multiply matrices R, Q and P together.
[tex]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][/tex]
[tex]A = \left[\begin{array}{ccc}1.07&-0.21\\-0.86&1.35\end{array}\right][/tex]
