In Problems14 to 17, multiply matrices to find the resultant transformation. Caution: Be sure you are multiplying the matrices in the right order.

$\left[\begin{array}{c}x\text{'}=\left(x+y\sqrt{3}\right)/2\\ y\text{'}=\left(-x\sqrt{3}+y\right)/2\end{array}\right]\left[\begin{array}{c}x\text{'}\text{'}=\left(-x\text{'}+y\text{'}\sqrt{3}\right)/2\\ y\text{'}\text{'}=-\left(x\text{'}\sqrt{3}+y\text{'}\right)/2\end{array}\right]$

Given matrix in the question is,

$\left[\begin{array}{c}\mathrm{x}\text{'}=(\mathrm{x}+\mathrm{y}\sqrt{3})/2\\ \mathrm{y}\text{'}=(-\mathrm{x}\sqrt{3}+\mathrm{y})/2\end{array}\right]\left[\begin{array}{c}\mathrm{x}\text{'}\text{'}=(-\mathrm{x}\text{'}+\mathrm{y}\text{'}\sqrt{3})/2\\ \mathrm{y}\text{'}\text{'}=-(\mathrm{x}\text{'}\sqrt{3}+\mathrm{y}\text{'})/2\end{array}\right]$

A transformation matrix is a matrix that, through the process of matrix multiplication, turns one vector into another vector.

Express the transformation in the matrix form for the first transformation:

$\left(\begin{array}{c}x\text{'}\\ y\text{'}\end{array}\right)=\left[\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$ ......(1)

Similarly, the matrix is represented below for the transformation of second case:

$\left[\begin{array}{c}x\text{'}\text{'}\\ y\text{'}\text{'}\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}x\text{'}\\ y\text{'}\end{array}\right]$ ......(2)

Now substitute the $\left[\begin{array}{c}x\text{'}\\ y\text{'}\end{array}\right]$ in the Equation 1 and 2:

$\begin{array}{rcl}\left[\begin{array}{c}x\text{'}\text{'}\\ y\text{'}\text{'}\end{array}\right]& =& \left[\begin{array}{cc}-\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\\ & =& \left[\begin{array}{cc}-\frac{1}{4}-\frac{3}{4}& -\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\\ -\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}& -\frac{3}{4}-\frac{1}{4}\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\\ & =& \left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\end{array}$

Therefore, the resultant of the transformation for the multiplication of the given transformations of the equation 1 and 2 is evaluated as:

$$\begin{gathered} x\text{'}\text{'}=-x \\ y\text{'}\text{'}=-y \end{gathered}$$