Question

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

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

Short answer

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

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

Step-by-step solution

Given information.

Given matrix in the question is,

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

Transformation matrix.

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

Find the resultant transformation for the multiplication of the matrices.

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:

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