Question

In Problems 14 to 17, multiply matrices to find the resultant transformation. Caution: Be sure you are multiplying the matrices in the right order.

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

Short answer

The resultant of the transformation for the multiplication of the given transformations is$x^{\text{'}\text{'}}=y,y^{\text{'}\text{'}}=-x,z^{\text{'}\text{'}}=z$ and the rotation is${270}^{\text{o}}$ counter clockwise.

Step-by-step solution

Given information.

Given matrix in the question is,

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

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.

The matrix form of $\left\{\begin{array}{c}x\text{'}=\left(x+y\sqrt{2}+z\right)/2\\ y\text{'}=\left(x\sqrt{2}-z\sqrt{2}\right)/2\\ z\text{'}=\left(-x+y\sqrt{2}-z\right)/2\end{array}\right.$can be written as,

$\left[\begin{array}{c}x\text{'}\\ y\text{'}\\ z\text{'}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& \frac{\sqrt{2}}{2}& \frac{1}{2}\\ \frac{\sqrt{2}}{2}& 0& -\frac{\sqrt{2}}{2}\\ -\frac{1}{2}& \frac{\sqrt{2}}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

The matrix form of $\left\{\begin{array}{c}x\text{'}\text{'}=\left(x\text{'}\sqrt{2}+z\text{'}\sqrt{2}\right)/2\\ y\text{'}\text{'}=\left(-x\text{'}-y\text{'}+z\text{'}\right)/2\\ z\text{'}\text{'}=\left(x\text{'}-y\text{'}\sqrt{2}-z\text{'}\right)/2\end{array}\right.$can be written as,

$\left[\begin{array}{c}x\text{'}\text{'}\\ y\text{'}\text{'}\\ z\text{'}\text{'}\end{array}\right]=\left[\begin{array}{ccc}\frac{\sqrt{2}}{2}& 0& \frac{\sqrt{2}}{2}\\ -\frac{1}{2}& -\frac{\sqrt{2}}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{\sqrt{2}}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}x\text{'}\\ y\text{'}\\ z\text{'}\end{array}\right]$

Now, multiply the matrices to find the resultant transformation.

$$\begin{gathered} \left[\begin{array}{c}x\text{'}\text{'}\\ y\text{'}\text{'}\\ z\text{'}\text{'}\end{array}\right]=\left[\begin{array}{ccc}\frac{\sqrt{2}}{2}& 0& \frac{\sqrt{2}}{2}\\ -\frac{1}{2}& -\frac{\sqrt{2}}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{\sqrt{2}}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right] \\ \left[\begin{array}{c}x\text{'}\text{'}\\ y\text{'}\text{'}\\ z\text{'}\text{'}\end{array}\right]=\left[\begin{array}{ccc}\frac{\sqrt{2}}{2}& 0& \frac{\sqrt{2}}{2}\\ -\frac{1}{2}& -\frac{\sqrt{2}}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{\sqrt{2}}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{2}& \frac{\sqrt{2}}{2}& \frac{1}{2}\\ \frac{\sqrt{2}}{2}& 0& -\frac{\sqrt{2}}{2}\\ -\frac{1}{2}& \frac{\sqrt{2}}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right] \\ \left[\begin{array}{c}x\text{'}\text{'}\\ y\text{'}\text{'}\\ z\text{'}\text{'}\end{array}\right]=\left[\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right] \end{gathered}$$

Therefore,

$x^{\text{'}\text{'}}=y,y^{\text{'}\text{'}}=-x,z^{\text{'}\text{'}}=z$

The above matrix equation can be written as,

$\left[\begin{array}{c}x\text{'}\text{'}\\ y\text{'}\text{'}\\ z\text{'}\text{'}\end{array}\right]=\left[\begin{array}{ccc}\cos \left({270}^{\text{o}}\right)& -\sin \left({270}^{\text{o}}\right)& 0\\ \sin \left({270}^{\text{o}}\right)& \cos \left({270}^{\text{o}}\right)& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

This shows that${270}^{\text{o}}$ rotation.