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Remembering that $\cos \left(\theta +\frac{\pi }{2}\right)=-\sin \left(\theta \right)$ and $\sin \left(\theta +\frac{\pi }{2}\right)=\cos \left(\theta \right)$, the following matrix is the three-dimensional Givens Rotation from the $y$ -axis to the $z$ -axis (rotation counter-clockwise around the $x$ -axis):

$\begin{array}{c}\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}+{90}^{\circ }\right)\\ 0& \sin \left({\theta }_{y\to z}\right)& \sin \left({\theta }_{y\to z}+{90}^{\circ }\right)\end{array}\right)\\ \left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}+\frac{\pi }{2}\right)\\ 0& \sin \left({\theta }_{y\to z}\right)& \sin \left({\theta }_{y\to z}+\frac{\pi }{2}\right)\end{array}\right)\\ \left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{y\to z}\right)\\ 0& \sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)\end{array}$

This matrix rotates three-dimensional column vectors about the origin of the $y\to z$ plane:

 

Similarly, the following is the three-dimensional Givens Rotation from the $x$ -axis to the $z$ -axis (rotation clockwise around the $y$ -axis):

$\begin{array}{c}\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}+{90}^{\circ }\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \sin \left({\theta }_{x\to z}+{90}^{\circ }\right)\end{array}\right)\\ \left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}+\frac{\pi }{2}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \sin \left({\theta }_{x\to z}+\frac{\pi }{2}\right)\end{array}\right)\\ \left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)\end{array}$

This matrix rotates three-dimensional column vectors about the origin of the $x\to z$ plane:

 

Notice that this is not the same as the traditional rotation matrix where we rotate around the $y$ axis using the right-hand rule (i.e. where we rotate in the opposite direction about the origin of the $x\to z$ plane).  In other words, given $\theta =-\varphi$ and remembering that $\cos \left(-\varphi \right)=\cos \left(\varphi \right)$ and $\sin \left(-\varphi \right)=-\sin \left(\varphi \right)$, we have the following:

$\begin{array}{c}\left(\begin{array}{ccc}\cos \left(\theta \right)& 0& \cos \left(\theta +{90}^{\circ }\right)\\ 0& 1& 0\\ \sin \left(\theta \right)& 0& \sin \left(\theta +{90}^{\circ }\right)\end{array}\right)\\ \left(\begin{array}{ccc}\cos \left(\left(-\varphi \right)\right)& 0& \cos \left(\left(-\varphi \right)+{90}^{\circ }\right)\\ 0& 1& 0\\ \sin \left(\left(-\varphi \right)\right)& 0& \sin \left(\left(-\varphi \right)+{90}^{\circ }\right)\end{array}\right)\\ \left(\begin{array}{ccc}\cos \left(-\varphi \right)& 0& \sin \left(-\varphi +\frac{\pi }{2}\right)\\ 0& 1& 0\\ \sin \left(-\varphi \right)& 0& \sin \left(-\varphi +\frac{\pi }{2}\right)\end{array}\right)\\ \left(\begin{array}{ccc}\cos \left(-\varphi \right)& 0& -\sin \left(-\varphi \right)\\ 0& 1& 0\\ \sin \left(-\varphi \right)& 0& \cos \left(-\varphi \right)\end{array}\right)\\ \left(\begin{array}{ccc}\cos \left(\varphi \right)& 0& \sin \left(\varphi \right)\\ 0& 1& 0\\ -\sin \left(\varphi \right)& 0& \cos \left(\varphi \right)\end{array}\right)\end{array}$

Lastly, the following is the three-dimensional Givens Rotation from the $x$ -axis to the $y$ -axis (rotation counter-clockwise around the $z$ -axis):

$\begin{array}{c}\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}+{90}^{\circ }\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \sin \left({\theta }_{x\to y}+{90}^{\circ }\right)& 0\\ 0& 0& 1\end{array}\right)\\ \left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}+\frac{\pi }{2}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \sin \left({\theta }_{x\to y}+\frac{\pi }{2}\right)& 0\\ 0& 0& 1\end{array}\right)\\ \left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)\end{array}$

This matrix rotates three-dimensional column vectors about the origin of the $x\to y$ plane:

 

Observe that given $n$ dimensions, we have the following number of distinct Givens Rotation matrices:

$\begin{array}{c}\left(n-1\right)+\left(n-2\right)+\cdots +1\\ {\displaystyle \sum _{k=1}^{n-1}k}\\ \frac{\left(n-1\right)\left(\left(n-1\right)+1\right)}{2}\\ \frac{\left(n-1\right)\left(n-1+1\right)}{2}\\ \frac{\left(n-1\right)\left(n\right)}{2}\\ \frac{n\left(n-1\right)}{2}\end{array}$

For instance, when we’re in three dimensions, we have $n=3$ and $\frac{n\left(n-1\right)}{2}=\frac{\left(3\right)\left(\left(3\right)-1\right)}{2}=\frac{3\left(\cancel{2}\right)}{\cancel{2}}=3$ distinct Givens Rotation matrices:

 

Moreover, notice the order in which we’re considering these distinct Givens Rotation matrices:

 

 

We’ll leverage this order in a moment.  Before that, let’s see what happens when we rotate the unit column vector $\left(1,0,0\right)$ using each of these Givens Rotations:

As expected, this simply extracts the first column vector from each Givens Rotation matrix.  Because we’re extracting the first column vector, we’ll only see $\theta$ in those vectors … we’ll never see $\theta +{90}^{\circ }$; so for the Givens Rotation matrices that modify $\left(1,0,0\right)$, $\cos \left(\theta \right)$ will always be in the $x$ position and $\sin \left(\theta \right)$ will walk its way up the remaining dimensions based on the order established above.  For the Givens Rotation matrices that do not modify $\left(1,0,0\right)$ (it’s just that first matrix in three dimensions), those rotations do not impact the $x$ position.  Looking back at our established order:

 

we see that the non- $x$ rows cannot impact the $x$ position.  In this three-dimensional case, we see that the Givens Rotation matrix associated with the $y$ row cannot modify the $x$ position.  To make this a little clearer, let’s rotate the arbitrarily positioned column vector $\left(x,y,z\right)$:

We immediately see that rotating $\left(x,y,z\right)$ about the $y\to z$ plane (multiplying by the first Givens Rotation matrix) does not impact the $x$ position … and only impacts the $y$ and $z$ positions if $y\ne 0$ or $z\ne 0$ … which only happens when we have a column vector off of the $x$ -axis.  For the column vector $\left(r,0,0\right)$ on the $x$ -axis, rotating about the $y\to z$ plane (multiplying by that first Givens Rotation matrix) has no effect … we’ll still have $\left(r,0,0\right)$ after that multiplication.

This brings up the question: given the column vector $\left(r,0,0\right)$ where $r=\sqrt{x^2+y^2+z^2}$ and $r\ge 0$, can all column vectors $\left(x,y,z\right)$ be recovered using just these Givens Rotation matrices?

First, we’ll start by using all the Givens Rotation matrices in our established order … and notice that the first Givens Rotation matrix doesn’t really help out that much:

$\begin{array}{rrr}\hfill \stackrel{3}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\stackrel{2}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}}\stackrel{1}{\overbrace{\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{y\to z}\right)\\ 0& \sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)}}\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}x\\ y\\ z\end{array}\right)\\ \hfill \stackrel{3}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\stackrel{2}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}}\left(\stackrel{1}{\overbrace{\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{y\to z}\right)\\ 0& \sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)}}\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\right)& \hfill =& \hfill \left(\begin{array}{c}x\\ y\\ z\end{array}\right)\\ \hfill \stackrel{3}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\stackrel{2}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}}\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}x\\ y\\ z\end{array}\right)\end{array}$

Then, we have:

$\begin{array}{rrr}\hfill \stackrel{3}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\stackrel{2}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}}\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}x\\ y\\ z\end{array}\right)\\ \hfill \left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& -\sin \left({\theta }_{x\to y}\right)& -\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\\ \sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}x\\ y\\ z\end{array}\right)\\ \hfill \left(\begin{array}{c}r\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)\\ r\sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)\\ r\sin \left({\theta }_{x\to z}\right)\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}x\\ y\\ z\end{array}\right)\end{array}$

For $r=\sqrt{x^2+y^2+z^2}$, $r=0$ implies $x=0$, $y=0$, and $z=0$, so given $r=0$, we have:

$\begin{array}{rrr}\hfill \left(\begin{array}{c}r\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)\\ r\sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)\\ r\sin \left({\theta }_{x\to z}\right)\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}x\\ y\\ z\end{array}\right)\\ \hfill \left(\begin{array}{c}\left(0\right)\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)\\ \left(0\right)\sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)\\ \left(0\right)\sin \left({\theta }_{x\to z}\right)\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\\ \hfill \left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\end{array}$

So, we can recover the column vector $\left(0,0,0\right)$.

Given $r=\sqrt{x^2+y^2+z^2}$ with $r>0$ and $-{90}^{\circ }\le {\theta }_{x\to z}\le {90}^{\circ }$, we have:

$\begin{array}{ccc}r\sin \left({\theta }_{x\to z}\right)& =& z\\ \left(\sqrt{x^2+y^2+z^2}\right)\sin \left({\theta }_{x\to z}\right)& =& z\\ \sin \left({\theta }_{x\to z}\right)& =& \frac{z}{\sqrt{x^2+y^2+z^2}}\\ {\theta }_{x\to z}& =& \arcsin \left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\end{array}$

Moreover, since $\arcsin \left(w\right)=\arctan \left(\frac{w}{\sqrt{1-w^2}}\right)$, we also have:

$\begin{array}{lll}{\theta }_{x\to z}\hfill & =\hfill & \arcsin \left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\hfill \\ \hfill & =\hfill & \arctan \left(\frac{\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)}{\sqrt{1-{\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)}^2}}\right)\hfill \\ \hfill & =\hfill & \arctan \left(\frac{\frac{z}{\sqrt{x^2+y^2+z^2}}}{\sqrt{\frac{x^2+y^2+z^2}{x^2+y^2+z^2}-\frac{z^2}{x^2+y^2+z^2}}}\right)\hfill \\ \hfill & =\hfill & \arctan \left(\frac{z}{\sqrt{x^2+y^2+z^2}\sqrt{\frac{x^2+y^2+\cancel{z^2}-\cancel{z^2}}{x^2+y^2+z^2}}}\right)\hfill \\ \hfill & =\hfill & \arctan \left(\frac{z}{\cancel{\sqrt{x^2+y^2+z^2}}\frac{\sqrt{x^2+y^2}}{\cancel{\sqrt{x^2+y^2+z^2}}}}\right)\hfill \\ \hfill & =\hfill & \arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\hfill \end{array}$

Remember: $\arctan \left(\frac{y}{x}\right)$ is atan2( y, x ) in C/C++ and ArcTan[ x, y ] in Mathematica.

Given $r=\sqrt{x^2+y^2+z^2}$ with $r>0$, $-{90}^{\circ }\le {\theta }_{x\to z}\le {90}^{\circ }$, and remembering that $\cos \left(\arctan \left(w\right)\right)=\frac{1}{\sqrt{1+w^2}}$ and $\sin \left(\arctan \left(w\right)\right)=\frac{w}{\sqrt{1+w^2}}$, if ${\theta }_{x\to z}=\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)$, we have:

Given $r=\sqrt{x^2+y^2+z^2}$ with $r>0$ and $-{180}^{\circ }\le {\theta }_{x\to y}\le {180}^{\circ }$, since $\cos \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}$, if ${\theta }_{x\to z}=\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)$, we have:

$\begin{array}{ccc}r\sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& =& y\\ r\sin \left({\theta }_{x\to y}\right)\cos \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)& =& y\\ \left(\sqrt{x^2+y^2+z^2}\right)\sin \left({\theta }_{x\to y}\right)\left(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)& =& y\\ \cancel{\sqrt{x^2+y^2+z^2}}\sin \left({\theta }_{x\to y}\right)\frac{\sqrt{x^2+y^2}}{\cancel{\sqrt{x^2+y^2+z^2}}}& =& y\\ \sin \left({\theta }_{x\to y}\right)& =& \frac{y}{\sqrt{x^2+y^2}}\\ {\theta }_{x\to y}& =& \arcsin \left(\frac{y}{\sqrt{x^2+y^2}}\right)\end{array}$

Moreover, since $\arcsin \left(w\right)=\arctan \left(\frac{w}{\sqrt{1-w^2}}\right)$, we also have:

$\begin{array}{lll}{\theta }_{x\to y}\hfill & =\hfill & \arcsin \left(\frac{y}{\sqrt{x^2+y^2}}\right)\hfill \\ \hfill & =\hfill & \arctan \left(\frac{\left(\frac{y}{\sqrt{x^2+y^2}}\right)}{\sqrt{1-{\left(\frac{y}{\sqrt{x^2+y^2}}\right)}^2}}\right)\hfill \\ \hfill & =\hfill & \arctan \left(\frac{y}{\sqrt{x^2+y^2}\sqrt{\frac{x^2+y^2}{x^2+y^2}-\frac{y^2}{x^2+y^2}}}\right)\hfill \\ \hfill & =\hfill & \arctan \left(\frac{y}{\sqrt{x^2+y^2}\sqrt{\frac{x^2+\cancel{y^2}-\cancel{y^2}}{x^2+y^2}}}\right)\hfill \\ \hfill & =\hfill & \arctan \left(\frac{y}{\cancel{\sqrt{x^2+y^2}}\frac{\sqrt{x^2}}{\cancel{\sqrt{x^2+y^2}}}}\right)\hfill \\ \hfill & =\hfill & \arctan \left(\frac{y}{\sqrt{x^2}}\right)\hfill \\ \hfill & =\hfill & \arctan \left(\frac{y}{x}\right)\hfill \end{array}$

Given $r=\sqrt{x^2+y^2+z^2}$ with $r>0$, $-{180}^{\circ }\le {\theta }_{x\to y}\le {180}^{\circ }$, and remembering that $\cos \left(\arctan \left(w\right)\right)=\frac{1}{\sqrt{1+w^2}}$ and $\sin \left(\arctan \left(w\right)\right)=\frac{w}{\sqrt{1+w^2}}$ , we have:

Given $r=\sqrt{x^2+y^2+z^2}$ with $r>0$, $-{180}^{\circ }\le {\theta }_{x\to y}\le {180}^{\circ }$, and $-{90}^{\circ }\le {\theta }_{x\to z}\le {90}^{\circ }$, since $\cos \left(\arctan \left(\frac{y}{x}\right)\right)=\frac{x}{\sqrt{x^2+y^2}}$ and $\cos \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}$, if  ${\theta }_{x\to y}=\arctan \left(\frac{y}{x}\right)$ and ${\theta }_{x\to z}=\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)$, we have:

$\begin{array}{c}r\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)\\ r\cos \left(\arctan \left(\frac{y}{x}\right)\right)\cos \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)\\ \left(\sqrt{x^2+y^2+z^2}\right)\left(\frac{x}{\sqrt{x^2+y^2}}\right)\left(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)\\ \cancel{\sqrt{x^2+y^2+z^2}}\frac{x}{\cancel{\sqrt{x^2+y^2}}}\frac{\cancel{\sqrt{x^2+y^2}}}{\cancel{\sqrt{x^2+y^2+z^2}}}\\ x\end{array}$

So, we have $r\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)=x$.

Obviously, since $\sin \left(\arctan \left(\frac{y}{x}\right)\right)=\frac{y}{\sqrt{x^2+y^2}}$, $\cos \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}$, and $\sin \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)=\frac{z}{\sqrt{x^2+y^2+z^2}}$, if ${\theta }_{x\to y}=\arctan \left(\frac{y}{x}\right)$ and ${\theta }_{x\to z}=\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)$, we also have:

So, we also have $r\sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)=y$ and $r\sin \left({\theta }_{x\to z}\right)=z$ such that:

$\begin{array}{rrr}\hfill \left(\begin{array}{c}r\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)\\ r\sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)\\ r\sin \left({\theta }_{x\to z}\right)\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}x\\ y\\ z\end{array}\right)\\ \hfill \left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}x\\ y\\ z\end{array}\right)\end{array}$

Therefore, all column vectors $\left(x,y,z\right)$ can be recovered by applying the Givens Rotation matrices associated with the $x$ row to the column vector $\left(r,0,0\right)$ where $r=\sqrt{x^2+y^2+z^2}$ and $r\ge 0$:

 

Notice that the $\left(r,{\theta }_{x\to y},{\theta }_{x\to z}\right)$ coordinate system implied here is not the same as the traditional spherical coordinate system $\left(\rho ,\theta ,\varphi \right)$.  In the traditional spherical coordinate system, we have:

$\begin{array}{lll}x\hfill & =\hfill & \rho \cos \left(\theta \right)\sin \left(\varphi \right)\hfill \\ y\hfill & =\hfill & \rho \sin \left(\theta \right)\sin \left(\varphi \right)\hfill \\ z\hfill & =\hfill & \rho \cos \left(\varphi \right)\hfill \end{array}$

where $\rho =\sqrt{x^2+y^2+z^2}$, $0^{\circ }\le \theta \le {360}^{\circ }$ where $\theta$ is the angle measured from the positive $x$ -axis (toward the positive $y$ -axis), and $0^{\circ }\le \varphi \le {180}^{\circ }$ where $\varphi$ is the angle measured from the positive $z$ -axis (toward the negative $z$ -axis); however, what we have here is the following:

$\begin{array}{lll}x\hfill & =\hfill & r\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)\hfill \\ y\hfill & =\hfill & r\sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)\hfill \\ z\hfill & =\hfill & r\sin \left({\theta }_{x\to z}\right)\hfill \end{array}$

where $r=\sqrt{x^2+y^2+z^2}$, $-{180}^{\circ }\le {\theta }_{x\to y}\le {180}^{\circ }$ where ${\theta }_{x\to y}$ is the angle measured from the positive $x$ -axis (toward the positive $y$ -axis), and $-{90}^{\circ }\le {\theta }_{x\to z}\le {90}^{\circ }$ where ${\theta }_{x\to z}$ is the angle measured from the positive $x$ -axis (toward the positive $z$ -axis).

Now, let’s rotate a column vector $\left(x,y,z\right)$ back to the $x$ -axis so that we get the column vector $\left(r,0,0\right)$.  In order to do that, let’s first take a look at each inverse of these Givens Rotation matrices using augmentation and finding the reduced row echelon forms … which as you’ll see is very similar for each matrix.  Remembering that ${\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)=1$ which implies that $1-{\sin }^2\left(\theta \right)={\cos }^2\left(\theta \right)$, we have:

$\begin{array}{c}\boxed{\left(\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{y\to z}\right)& 0& 1& 0\\ 0& \sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)& 0& 0& 1\end{array}\right)}\\ \left(\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& 1& -\frac{\sin \left({\theta }_{y\to z}\right)}{\cos \left({\theta }_{y\to z}\right)}& 0& \frac{1}{\cos \left({\theta }_{y\to z}\right)}& 0\\ 0& \sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)& 0& 0& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& 1& -\frac{\sin \left({\theta }_{y\to z}\right)}{\cos \left({\theta }_{y\to z}\right)}& 0& \frac{1}{\cos \left({\theta }_{y\to z}\right)}& 0\\ 0& 0& \cos \left({\theta }_{y\to z}\right)+\frac{{\sin }^2\left({\theta }_{y\to z}\right)}{\cos \left({\theta }_{y\to z}\right)}& 0& -\frac{\sin \left({\theta }_{y\to z}\right)}{\cos \left({\theta }_{y\to z}\right)}& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& 1& -\frac{\sin \left({\theta }_{y\to z}\right)}{\cos \left({\theta }_{y\to z}\right)}& 0& \frac{1}{\cos \left({\theta }_{y\to z}\right)}& 0\\ 0& 0& \frac{{\cos }^2\left({\theta }_{y\to z}\right)+{\sin }^2\left({\theta }_{y\to z}\right)}{\cos \left({\theta }_{y\to z}\right)}& 0& -\frac{\sin \left({\theta }_{y\to z}\right)}{\cos \left({\theta }_{y\to z}\right)}& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& 1& -\frac{\sin \left({\theta }_{y\to z}\right)}{\cos \left({\theta }_{y\to z}\right)}& 0& \frac{1}{\cos \left({\theta }_{y\to z}\right)}& 0\\ 0& 0& \frac{1}{\cos \left({\theta }_{y\to z}\right)}& 0& -\frac{\sin \left({\theta }_{y\to z}\right)}{\cos \left({\theta }_{y\to z}\right)}& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& 1& -\frac{\sin \left({\theta }_{y\to z}\right)}{\cos \left({\theta }_{y\to z}\right)}& 0& \frac{1}{\cos \left({\theta }_{y\to z}\right)}& 0\\ 0& 0& 1& 0& -\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& 1& 0& 0& \frac{1}{\cos \left({\theta }_{y\to z}\right)}-\frac{{\sin }^2\left({\theta }_{y\to z}\right)}{\cos \left({\theta }_{y\to z}\right)}& \sin \left({\theta }_{y\to z}\right)\\ 0& 0& 1& 0& -\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& 1& 0& 0& \frac{1-{\sin }^2\left({\theta }_{y\to z}\right)}{\cos \left({\theta }_{y\to z}\right)}& \sin \left({\theta }_{y\to z}\right)\\ 0& 0& 1& 0& -\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& 1& 0& 0& \frac{{\cos }^{\cancel{2}}\left({\theta }_{y\to z}\right)}{\cancel{\cos \left({\theta }_{y\to z}\right)}}& \sin \left({\theta }_{y\to z}\right)\\ 0& 0& 1& 0& -\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)\\ \boxed{\left(\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& 1& 0& 0& \cos \left({\theta }_{y\to z}\right)& \sin \left({\theta }_{y\to z}\right)\\ 0& 0& 1& 0& -\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)}\end{array}$

$\begin{array}{c}\boxed{\left(\begin{array}{cccccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)& 1& 0& 0\\ 0& 1& 0& 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)& 0& 0& 1\end{array}\right)}\\ \left(\begin{array}{cccccc}1& 0& -\frac{\sin \left({\theta }_{x\to z}\right)}{\cos \left({\theta }_{x\to z}\right)}& \frac{1}{\cos \left({\theta }_{x\to z}\right)}& 0& 0\\ 0& 1& 0& 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)& 0& 0& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& -\frac{\sin \left({\theta }_{x\to z}\right)}{\cos \left({\theta }_{x\to z}\right)}& \frac{1}{\cos \left({\theta }_{x\to z}\right)}& 0& 0\\ 0& 1& 0& 0& 1& 0\\ 0& 0& \cos \left({\theta }_{x\to z}\right)+\frac{{\sin }^2\left({\theta }_{x\to z}\right)}{\cos \left({\theta }_{x\to z}\right)}& -\frac{\sin \left({\theta }_{x\to z}\right)}{\cos \left({\theta }_{x\to z}\right)}& 0& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& -\frac{\sin \left({\theta }_{x\to z}\right)}{\cos \left({\theta }_{x\to z}\right)}& \frac{1}{\cos \left({\theta }_{x\to z}\right)}& 0& 0\\ 0& 1& 0& 0& 1& 0\\ 0& 0& \frac{{\cos }^2\left({\theta }_{x\to z}\right)+{\sin }^2\left({\theta }_{x\to z}\right)}{\cos \left({\theta }_{x\to z}\right)}& -\frac{\sin \left({\theta }_{x\to z}\right)}{\cos \left({\theta }_{x\to z}\right)}& 0& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& -\frac{\sin \left({\theta }_{x\to z}\right)}{\cos \left({\theta }_{x\to z}\right)}& \frac{1}{\cos \left({\theta }_{x\to z}\right)}& 0& 0\\ 0& 1& 0& 0& 1& 0\\ 0& 0& \frac{1}{\cos \left({\theta }_{x\to z}\right)}& -\frac{\sin \left({\theta }_{x\to z}\right)}{\cos \left({\theta }_{x\to z}\right)}& 0& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& -\frac{\sin \left({\theta }_{x\to z}\right)}{\cos \left({\theta }_{x\to z}\right)}& \frac{1}{\cos \left({\theta }_{x\to z}\right)}& 0& 0\\ 0& 1& 0& 0& 1& 0\\ 0& 0& 1& -\sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& \frac{1}{\cos \left({\theta }_{x\to z}\right)}-\frac{{\sin }^2\left({\theta }_{x\to z}\right)}{\cos \left({\theta }_{x\to z}\right)}& 0& \sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0& 0& 1& 0\\ 0& 0& 1& -\sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& \frac{1-{\sin }^2\left({\theta }_{x\to z}\right)}{\cos \left({\theta }_{x\to z}\right)}& 0& \sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0& 0& 1& 0\\ 0& 0& 1& -\sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& \frac{{\cos }^{\cancel{2}}\left({\theta }_{x\to z}\right)}{\cancel{\cos \left({\theta }_{x\to z}\right)}}& 0& \sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0& 0& 1& 0\\ 0& 0& 1& -\sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)\\ \boxed{\left(\begin{array}{cccccc}1& 0& 0& \cos \left({\theta }_{x\to z}\right)& 0& \sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0& 0& 1& 0\\ 0& 0& 1& -\sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}\end{array}$

$\begin{array}{c}\boxed{\left(\begin{array}{cccccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0& 1& 0& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0& 0& 1& 0\\ 0& 0& 1& 0& 0& 1\end{array}\right)}\\ \left(\begin{array}{cccccc}1& -\frac{\sin \left({\theta }_{x\to y}\right)}{\cos \left({\theta }_{x\to y}\right)}& 0& \frac{1}{\cos \left({\theta }_{x\to y}\right)}& 0& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0& 0& 1& 0\\ 0& 0& 1& 0& 0& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& -\frac{\sin \left({\theta }_{x\to y}\right)}{\cos \left({\theta }_{x\to y}\right)}& 0& \frac{1}{\cos \left({\theta }_{x\to y}\right)}& 0& 0\\ 0& \cos \left({\theta }_{x\to y}\right)+\frac{{\sin }^2\left({\theta }_{x\to y}\right)}{\cos \left({\theta }_{x\to y}\right)}& 0& -\frac{\sin \left({\theta }_{x\to y}\right)}{\cos \left({\theta }_{x\to y}\right)}& 1& 0\\ 0& 0& 1& 0& 0& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& -\frac{\sin \left({\theta }_{x\to y}\right)}{\cos \left({\theta }_{x\to y}\right)}& 0& \frac{1}{\cos \left({\theta }_{x\to y}\right)}& 0& 0\\ 0& \frac{{\cos }^2\left({\theta }_{x\to y}\right)+{\sin }^2\left({\theta }_{x\to y}\right)}{\cos \left({\theta }_{x\to y}\right)}& 0& -\frac{\sin \left({\theta }_{x\to y}\right)}{\cos \left({\theta }_{x\to y}\right)}& 1& 0\\ 0& 0& 1& 0& 0& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& -\frac{\sin \left({\theta }_{x\to y}\right)}{\cos \left({\theta }_{x\to y}\right)}& 0& \frac{1}{\cos \left({\theta }_{x\to y}\right)}& 0& 0\\ 0& \frac{1}{\cos \left({\theta }_{x\to y}\right)}& 0& -\frac{\sin \left({\theta }_{x\to y}\right)}{\cos \left({\theta }_{x\to y}\right)}& 1& 0\\ 0& 0& 1& 0& 0& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& -\frac{\sin \left({\theta }_{x\to y}\right)}{\cos \left({\theta }_{x\to y}\right)}& 0& \frac{1}{\cos \left({\theta }_{x\to y}\right)}& 0& 0\\ 0& 1& 0& -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1& 0& 0& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& \frac{1}{\cos \left({\theta }_{x\to y}\right)}-\frac{{\sin }^2\left({\theta }_{x\to y}\right)}{\cos \left({\theta }_{x\to y}\right)}& \sin \left({\theta }_{x\to y}\right)& 0\\ 0& 1& 0& -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1& 0& 0& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& \frac{1-{\sin }^2\left({\theta }_{x\to y}\right)}{\cos \left({\theta }_{x\to y}\right)}& \sin \left({\theta }_{x\to y}\right)& 0\\ 0& 1& 0& -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1& 0& 0& 1\end{array}\right)\\ \left(\begin{array}{cccccc}1& 0& 0& \frac{{\cos }^{\cancel{2}}\left({\theta }_{x\to y}\right)}{\cancel{\cos \left({\theta }_{x\to y}\right)}}& \sin \left({\theta }_{x\to y}\right)& 0\\ 0& 1& 0& -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1& 0& 0& 1\end{array}\right)\\ \boxed{\left(\begin{array}{cccccc}1& 0& 0& \cos \left({\theta }_{x\to y}\right)& \sin \left({\theta }_{x\to y}\right)& 0\\ 0& 1& 0& -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1& 0& 0& 1\end{array}\right)}\end{array}$

Therefore, the inverse of each three-dimensional Givens Rotation matrix is simply the transpose of that matrix:

$\begin{array}{ccccc}{\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{y\to z}\right)\\ 0& \sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)}^{-1}& =& {\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{y\to z}\right)\\ 0& \sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)}^T& =& \left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& \sin \left({\theta }_{y\to z}\right)\\ 0& -\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)\\ {\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}^{-1}& =& {\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}^T& =& \left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& \sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ -\sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)\\ {\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}^{-1}& =& {\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}^T& =& \left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& \sin \left({\theta }_{x\to y}\right)& 0\\ -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)\end{array}$

and it shouldn’t be too hard to see that the inverse of each $n$ -dimensional Givens Rotation matrix is also the transpose of that matrix … like it is for any other $n$ -dimensional rotation matrix.

Also, if we consider rotations in the opposite direction where $\theta =-\varphi$ while remembering that $\cos \left(-\varphi \right)=\cos \left(\varphi \right)$ and $\sin \left(-\varphi \right)=-\sin \left(\varphi \right)$, then we have:

$\begin{array}{ccccc}\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& \sin \left({\theta }_{y\to z}\right)\\ 0& -\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)& =& \left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left(-{\varphi }_{y\to -z}\right)& \sin \left(-{\varphi }_{y\to -z}\right)\\ 0& -\sin \left(-{\varphi }_{y\to -z}\right)& \cos \left(-{\varphi }_{y\to -z}\right)\end{array}\right)& =& \left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\varphi }_{y\to -z}\right)& -\sin \left({\varphi }_{y\to -z}\right)\\ 0& \sin \left({\varphi }_{y\to -z}\right)& \cos \left({\varphi }_{y\to -z}\right)\end{array}\right)\\ \left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& \sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ -\sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)& =& \left(\begin{array}{ccc}\cos \left(-{\varphi }_{x\to -z}\right)& 0& \sin \left(-{\varphi }_{x\to -z}\right)\\ 0& 1& 0\\ -\sin \left(-{\varphi }_{x\to -z}\right)& 0& \cos \left(-{\varphi }_{x\to -z}\right)\end{array}\right)& =& \left(\begin{array}{ccc}\cos \left({\varphi }_{x\to -z}\right)& 0& -\sin \left({\varphi }_{x\to -z}\right)\\ 0& 1& 0\\ \sin \left({\varphi }_{x\to -z}\right)& 0& \cos \left({\varphi }_{x\to -z}\right)\end{array}\right)\\ \left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& \sin \left({\theta }_{x\to y}\right)& 0\\ -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)& =& \left(\begin{array}{ccc}\cos \left(-{\varphi }_{x\to -y}\right)& \sin \left(-{\varphi }_{x\to -y}\right)& 0\\ -\sin \left(-{\varphi }_{x\to -y}\right)& \cos \left(-{\varphi }_{x\to -y}\right)& 0\\ 0& 0& 1\end{array}\right)& =& \left(\begin{array}{ccc}\cos \left({\varphi }_{x\to -y}\right)& -\sin \left({\varphi }_{x\to -y}\right)& 0\\ \sin \left({\varphi }_{x\to -y}\right)& \cos \left({\varphi }_{x\to -y}\right)& 0\\ 0& 0& 1\end{array}\right)\end{array}$

We see that each inverse amounts to just rotating in the opposite direction. In other words, since $\cos \left(-\theta \right)=\cos \left(\theta \right)$ and $\sin \left(-\theta \right)=-\sin \left(\theta \right)$ implies that $\cos \left(\theta \right)=\cos \left(-\theta \right)$ and $\sin \left(\theta \right)=-\sin \left(-\theta \right)$ , we have:

$\begin{array}{ccccc}{\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{y\to z}\right)\\ 0& \sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)}^{-1}& =& \left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& \sin \left({\theta }_{y\to z}\right)\\ 0& -\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)& =& \left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left(-{\theta }_{y\to z}\right)& -\sin \left(-{\theta }_{y\to z}\right)\\ 0& \sin \left(-{\theta }_{y\to z}\right)& \cos \left(-{\theta }_{y\to z}\right)\end{array}\right)\\ {\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}^{-1}& =& \left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& \sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ -\sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)& =& \left(\begin{array}{ccc}\cos \left(-{\theta }_{x\to z}\right)& 0& -\sin \left(-{\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left(-{\theta }_{x\to z}\right)& 0& \cos \left(-{\theta }_{x\to z}\right)\end{array}\right)\\ {\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}^{-1}& =& \left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& \sin \left({\theta }_{x\to y}\right)& 0\\ -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)& =& \left(\begin{array}{ccc}\cos \left(-{\theta }_{x\to y}\right)& -\sin \left(-{\theta }_{x\to y}\right)& 0\\ \sin \left(-{\theta }_{x\to y}\right)& \cos \left(-{\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)\end{array}$

Now, let’s use those inverse Givens Rotation matrices to rotate a column vector $\left(x,y,z\right)$ in those opposite directions back to the $x$ -axis so that we get the column vector $\left(r,0,0\right)$.

First, we’ll start by using all the Givens Rotation matrices in our established order … and get rid of the ones that don’t matter here:

$\begin{array}{lll}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)\hfill & =\hfill & \stackrel{3}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\stackrel{2}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}}\stackrel{1}{\overbrace{\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{y\to z}\right)\\ 0& \sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)}}\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\hfill \\ \left(\begin{array}{c}x\\ y\\ z\end{array}\right)\hfill & =\hfill & \stackrel{3}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\stackrel{2}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}}\left(\stackrel{1}{\overbrace{\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{y\to z}\right)\\ 0& \sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)}}\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\right)\hfill \\ \left(\begin{array}{c}x\\ y\\ z\end{array}\right)\hfill & =\hfill & \stackrel{3}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\stackrel{2}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}}\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\hfill \end{array}$

Next, remembering that the inverse of a rotation matrix is its transpose, then we have:

$\begin{array}{c}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\stackrel{3}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\stackrel{2}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}}\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \stackrel{\text{Inverse of 3}}{\overbrace{{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}^{-1}}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\stackrel{2}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}}\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \stackrel{\text{Inverse of 2}}{\overbrace{{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& -\sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}^{-1}}}\stackrel{\text{Inverse of 3}}{\overbrace{{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& -\sin \left({\theta }_{x\to y}\right)& 0\\ \sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}^{-1}}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \stackrel{\text{Inverse of 2}}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& \sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ -\sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}}\stackrel{\text{Inverse of 3}}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& \sin \left({\theta }_{x\to y}\right)& 0\\ -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\end{array}$

Keeping in mind that $\cos \left(\theta \right)=\cos \left(-\theta \right)$ and $\sin \left(\theta \right)=-\sin \left(-\theta \right)$, we have:

$\begin{array}{rrr}\hfill \stackrel{\text{Inverse of 2}}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& \sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ -\sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}}\stackrel{\text{Inverse of 3}}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& \sin \left({\theta }_{x\to y}\right)& 0\\ -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \stackrel{\text{Inverse of 2}}{\overbrace{\left(\begin{array}{ccc}\cos \left(-{\theta }_{x\to z}\right)& 0& -\sin \left(-{\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left(-{\theta }_{x\to z}\right)& 0& \cos \left(-{\theta }_{x\to z}\right)\end{array}\right)}}\stackrel{\text{Inverse of 3}}{\overbrace{\left(\begin{array}{ccc}\cos \left(-{\theta }_{x\to y}\right)& -\sin \left(-{\theta }_{x\to y}\right)& 0\\ \sin \left(-{\theta }_{x\to y}\right)& \cos \left(-{\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \stackrel{\text{2 for }-{\theta }_{x\to z}}{\overbrace{\left(\begin{array}{ccc}\cos \left(\left(-{\theta }_{x\to z}\right)\right)& 0& -\sin \left(\left(-{\theta }_{x\to z}\right)\right)\\ 0& 1& 0\\ \sin \left(\left(-{\theta }_{x\to z}\right)\right)& 0& \cos \left(\left(-{\theta }_{x\to z}\right)\right)\end{array}\right)}}\stackrel{\text{3 for }-{\theta }_{x\to y}}{\overbrace{\left(\begin{array}{ccc}\cos \left(\left(-{\theta }_{x\to y}\right)\right)& -\sin \left(\left(-{\theta }_{x\to y}\right)\right)& 0\\ \sin \left(\left(-{\theta }_{x\to y}\right)\right)& \cos \left(\left(-{\theta }_{x\to y}\right)\right)& 0\\ 0& 0& 1\end{array}\right)}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\end{array}$

Since $\theta =-\varphi$ implies $-\theta =\varphi$, we also have:

$\begin{array}{rrr}\hfill \stackrel{\text{2 for }-{\theta }_{x\to z}}{\overbrace{\left(\begin{array}{ccc}\cos \left(\left(-{\theta }_{x\to z}\right)\right)& 0& -\sin \left(\left(-{\theta }_{x\to z}\right)\right)\\ 0& 1& 0\\ \sin \left(\left(-{\theta }_{x\to z}\right)\right)& 0& \cos \left(\left(-{\theta }_{x\to z}\right)\right)\end{array}\right)}}\stackrel{\text{3 for }-{\theta }_{x\to y}}{\overbrace{\left(\begin{array}{ccc}\cos \left(\left(-{\theta }_{x\to y}\right)\right)& -\sin \left(\left(-{\theta }_{x\to y}\right)\right)& 0\\ \sin \left(\left(-{\theta }_{x\to y}\right)\right)& \cos \left(\left(-{\theta }_{x\to y}\right)\right)& 0\\ 0& 0& 1\end{array}\right)}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \stackrel{\text{2 for }{\varphi }_{x\to -z}}{\overbrace{\left(\begin{array}{ccc}\cos \left({\varphi }_{x\to -z}\right)& 0& -\sin \left({\varphi }_{x\to -z}\right)\\ 0& 1& 0\\ \sin \left({\varphi }_{x\to -z}\right)& 0& \cos \left({\varphi }_{x\to -z}\right)\end{array}\right)}}\stackrel{\text{3 for }{\varphi }_{x\to -y}}{\overbrace{\left(\begin{array}{ccc}\cos \left({\varphi }_{x\to -y}\right)& -\sin \left({\varphi }_{x\to -y}\right)& 0\\ \sin \left({\varphi }_{x\to -y}\right)& \cos \left({\varphi }_{x\to -y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\end{array}$

In other words, given a column vector $\left(x,y,z\right)$, we can rotate that vector back to the $x$ -axis by applying Givens Rotation matrices in the opposite order using the opposite angles. Instead of the following:

 

 

repeated again with subscripts to keep things clear:

 

we now have the following:

 

 

In terms of this new backward ordering and keeping in mind that $\varphi =-\theta$, we now have the following:

$\begin{array}{rrr}\hfill \stackrel{{\text{3}}_{\text{backward}}\text{ for }{\varphi }_{y\to -z}}{\overbrace{\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\varphi }_{y\to -z}\right)& -\sin \left({\varphi }_{y\to -z}\right)\\ 0& \sin \left({\varphi }_{y\to -z}\right)& \cos \left({\varphi }_{y\to -z}\right)\end{array}\right)}}\stackrel{{\text{2}}_{\text{backward}}\text{ for }{\varphi }_{x\to -z}}{\overbrace{\left(\begin{array}{ccc}\cos \left({\varphi }_{x\to -z}\right)& 0& -\sin \left({\varphi }_{x\to -z}\right)\\ 0& 1& 0\\ \sin \left({\varphi }_{x\to -z}\right)& 0& \cos \left({\varphi }_{x\to -z}\right)\end{array}\right)}}\stackrel{{\text{1}}_{\text{backward}}\text{ for }{\varphi }_{x\to -y}}{\overbrace{\left(\begin{array}{ccc}\cos \left({\varphi }_{x\to -y}\right)& -\sin \left({\varphi }_{x\to -y}\right)& 0\\ \sin \left({\varphi }_{x\to -y}\right)& \cos \left({\varphi }_{x\to -y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \stackrel{{\text{3}}_{\text{backward}}\text{ for }-{\theta }_{y\to z}}{\overbrace{\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left(-{\theta }_{y\to z}\right)& -\sin \left(-{\theta }_{y\to z}\right)\\ 0& \sin \left(-{\theta }_{y\to z}\right)& \cos \left(-{\theta }_{y\to z}\right)\end{array}\right)}}\stackrel{{\text{2}}_{\text{backward}}\text{ for }-{\theta }_{x\to z}}{\overbrace{\left(\begin{array}{ccc}\cos \left(-{\theta }_{x\to z}\right)& 0& -\sin \left(-{\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left(-{\theta }_{x\to z}\right)& 0& \cos \left(-{\theta }_{x\to z}\right)\end{array}\right)}}\stackrel{{\text{1}}_{\text{backward}}\text{ for }-{\theta }_{x\to y}}{\overbrace{\left(\begin{array}{ccc}\cos \left(-{\theta }_{x\to y}\right)& -\sin \left(-{\theta }_{x\to y}\right)& 0\\ \sin \left(-{\theta }_{x\to y}\right)& \cos \left(-{\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \stackrel{{\text{1}}_{\text{forward}}\text{ for }-{\theta }_{y\to z}}{\overbrace{\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left(-{\theta }_{y\to z}\right)& -\sin \left(-{\theta }_{y\to z}\right)\\ 0& \sin \left(-{\theta }_{y\to z}\right)& \cos \left(-{\theta }_{y\to z}\right)\end{array}\right)}}\stackrel{{\text{2}}_{\text{forward}}\text{ for }-{\theta }_{x\to z}}{\overbrace{\left(\begin{array}{ccc}\cos \left(-{\theta }_{x\to z}\right)& 0& -\sin \left(-{\theta }_{x\to z}\right)\\ 0& 1& 0\\ \sin \left(-{\theta }_{x\to z}\right)& 0& \cos \left(-{\theta }_{x\to z}\right)\end{array}\right)}}\stackrel{{\text{3}}_{\text{forward}}\text{ for }-{\theta }_{x\to y}}{\overbrace{\left(\begin{array}{ccc}\cos \left(-{\theta }_{x\to y}\right)& -\sin \left(-{\theta }_{x\to y}\right)& 0\\ \sin \left(-{\theta }_{x\to y}\right)& \cos \left(-{\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \stackrel{{\text{Inverse of 1}}_{\text{forward}}\text{ for }{\theta }_{y\to z}}{\overbrace{\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \left({\theta }_{y\to z}\right)& \sin \left({\theta }_{y\to z}\right)\\ 0& -\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)\end{array}\right)}}\stackrel{{\text{Inverse of 2}}_{\text{forward}}\text{ for }{\theta }_{x\to z}}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& \sin \left({\theta }_{x\to z}\right)\\ 0& 1& 0\\ -\sin \left({\theta }_{x\to z}\right)& 0& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}}\stackrel{{\text{Inverse of 3}}_{\text{forward}}\text{ for }{\theta }_{x\to y}}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& \sin \left({\theta }_{x\to y}\right)& 0\\ -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& \sin \left({\theta }_{x\to z}\right)\\ -\sin \left({\theta }_{x\to z}\right)\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{x\to z}\right)\sin \left({\theta }_{y\to z}\right)\\ -\sin \left({\theta }_{x\to z}\right)\cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{x\to z}\right)\cos \left({\theta }_{y\to z}\right)\end{array}\right)\stackrel{{\text{Inverse of 3}}_{\text{forward}}\text{ for }{\theta }_{x\to y}}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& \sin \left({\theta }_{x\to y}\right)& 0\\ -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \left(\begin{array}{ccc}\cos \left({\theta }_{x\to z}\right)& 0& \sin \left({\theta }_{x\to z}\right)\\ -\sin \left({\theta }_{x\to z}\right)\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{x\to z}\right)\sin \left({\theta }_{y\to z}\right)\\ -\sin \left({\theta }_{x\to z}\right)\cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{x\to z}\right)\cos \left({\theta }_{y\to z}\right)\end{array}\right)\stackrel{{\text{Inverse of 3}}_{\text{forward}}\text{ for }{\theta }_{x\to y}}{\overbrace{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)& \sin \left({\theta }_{x\to y}\right)& 0\\ -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ 0& 0& 1\end{array}\right)}}\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\end{array}$

So, we have:

$\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \sin \left({\theta }_{x\to z}\right)\\ -\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\sin \left({\theta }_{y\to z}\right)-\sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\sin \left({\theta }_{y\to z}\right)+\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{x\to z}\right)\sin \left({\theta }_{y\to z}\right)\\ -\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\cos \left({\theta }_{y\to z}\right)+\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\cos \left({\theta }_{y\to z}\right)-\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{x\to z}\right)\cos \left({\theta }_{y\to z}\right)\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)$

Given $r=\sqrt{x^2+y^2+z^2}$, we choose ${\theta }_{x\to y}=\arctan \left(\frac{y}{x}\right)$, ${\theta }_{x\to z}=\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)$, and ${\theta }_{y\to z}=0$.

After applying ${\theta }_{y\to z}=0$, we have:

$\begin{array}{c}\boxed{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \sin \left({\theta }_{x\to z}\right)\\ -\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\sin \left({\theta }_{y\to z}\right)-\sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\sin \left({\theta }_{y\to z}\right)+\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{x\to z}\right)\sin \left({\theta }_{y\to z}\right)\\ -\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\cos \left({\theta }_{y\to z}\right)+\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{y\to z}\right)& -\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\cos \left({\theta }_{y\to z}\right)-\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{y\to z}\right)& \cos \left({\theta }_{x\to z}\right)\cos \left({\theta }_{y\to z}\right)\end{array}\right)}\\ \left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \sin \left({\theta }_{x\to z}\right)\\ -\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\sin \left(0\right)-\sin \left({\theta }_{x\to y}\right)\cos \left(0\right)& -\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\sin \left(0\right)+\cos \left({\theta }_{x\to y}\right)\cos \left(0\right)& \cos \left({\theta }_{x\to z}\right)\sin \left(0\right)\\ -\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\cos \left(0\right)+\sin \left({\theta }_{x\to y}\right)\sin \left(0\right)& -\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\cos \left(0\right)-\cos \left({\theta }_{x\to y}\right)\sin \left(0\right)& \cos \left({\theta }_{x\to z}\right)\cos \left(0\right)\end{array}\right)\\ \left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \sin \left({\theta }_{x\to z}\right)\\ -\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\left(0\right)-\sin \left({\theta }_{x\to y}\right)\left(1\right)& -\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\left(0\right)+\cos \left({\theta }_{x\to y}\right)\left(1\right)& \cos \left({\theta }_{x\to z}\right)\left(0\right)\\ -\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\left(1\right)+\sin \left({\theta }_{x\to y}\right)\left(0\right)& -\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)\left(1\right)-\cos \left({\theta }_{x\to y}\right)\left(0\right)& \cos \left({\theta }_{x\to z}\right)\left(1\right)\end{array}\right)\\ \boxed{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \sin \left({\theta }_{x\to z}\right)\\ -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ -\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)& -\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}\end{array}$

Then after applying ${\theta }_{x\to y}=\arctan \left(\frac{y}{x}\right)$ and ${\theta }_{x\to z}=\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)$, since $\cos \left(\arctan \left(\frac{y}{x}\right)\right)=\frac{x}{\sqrt{x^2+y^2}}$, $\sin \left(\arctan \left(\frac{y}{x}\right)\right)=\frac{y}{\sqrt{x^2+y^2}}$, $\cos \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}$, and $\sin \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)=\frac{z}{\sqrt{x^2+y^2+z^2}}$, we have:

$\begin{array}{c}\boxed{\left(\begin{array}{ccc}\cos \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \sin \left({\theta }_{x\to y}\right)\cos \left({\theta }_{x\to z}\right)& \sin \left({\theta }_{x\to z}\right)\\ -\sin \left({\theta }_{x\to y}\right)& \cos \left({\theta }_{x\to y}\right)& 0\\ -\cos \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)& -\sin \left({\theta }_{x\to y}\right)\sin \left({\theta }_{x\to z}\right)& \cos \left({\theta }_{x\to z}\right)\end{array}\right)}\\ \left(\begin{array}{ccc}\cos \left(\arctan \left(\frac{y}{x}\right)\right)\cos \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)& \sin \left(\arctan \left(\frac{y}{x}\right)\right)\cos \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)& \sin \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)\\ -\sin \left(\arctan \left(\frac{y}{x}\right)\right)& \cos \left(\arctan \left(\frac{y}{x}\right)\right)& 0\\ -\cos \left(\arctan \left(\frac{y}{x}\right)\right)\sin \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)& -\sin \left(\arctan \left(\frac{y}{x}\right)\right)\sin \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)& \cos \left(\arctan \left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)\end{array}\right)\\ \left(\begin{array}{ccc}\left(\frac{x}{\cancel{\sqrt{x^2+y^2}}}\right)\left(\frac{\cancel{\sqrt{x^2+y^2}}}{\sqrt{x^2+y^2+z^2}}\right)& \left(\frac{y}{\cancel{\sqrt{x^2+y^2}}}\right)\left(\frac{\cancel{\sqrt{x^2+y^2}}}{\sqrt{x^2+y^2+z^2}}\right)& \left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\\ -\left(\frac{y}{\sqrt{x^2+y^2}}\right)& \left(\frac{x}{\sqrt{x^2+y^2}}\right)& 0\\ -\left(\frac{x}{\sqrt{x^2+y^2}}\right)\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)& -\left(\frac{y}{\sqrt{x^2+y^2}}\right)\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)& \left(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)\end{array}\right)\\ \boxed{\left(\begin{array}{ccc}\frac{x}{\sqrt{x^2+y^2+z^2}}& \frac{y}{\sqrt{x^2+y^2+z^2}}& \frac{z}{\sqrt{x^2+y^2+z^2}}\\ -\frac{y}{\sqrt{x^2+y^2}}& \frac{x}{\sqrt{x^2+y^2}}& 0\\ -\frac{xz}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}& -\frac{yz}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}& \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\end{array}\right)}\end{array}$

Therefore, since $r=\sqrt{x^2+y^2+z^2}$, we have:

$\begin{array}{rrr}\hfill \left(\begin{array}{ccc}\frac{x}{\sqrt{x^2+y^2+z^2}}& \frac{y}{\sqrt{x^2+y^2+z^2}}& \frac{z}{\sqrt{x^2+y^2+z^2}}\\ -\frac{y}{\sqrt{x^2+y^2}}& \frac{x}{\sqrt{x^2+y^2}}& 0\\ -\frac{xz}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}& -\frac{yz}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}& \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \left(\begin{array}{c}\frac{x^2}{\sqrt{x^2+y^2+z^2}}+\frac{y^2}{\sqrt{x^2+y^2+z^2}}+\frac{z^2}{\sqrt{x^2+y^2+z^2}}\\ -\cancel{\frac{xy}{\sqrt{x^2+y^2}}}+\cancel{\frac{xy}{\sqrt{x^2+y^2}}}+0\\ -\frac{x^{2}z}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}-\frac{y^{2}z}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}+\frac{\left(\sqrt{x^2+y^2}\right)z}{\sqrt{x^2+y^2+z^2}}\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \left(\begin{array}{c}\frac{x^2+y^2+z^2}{\sqrt{x^2+y^2+z^2}}\\ 0\\ \frac{-x^{2}z-y^{2}z+\left(x^2+y^2\right)z}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \left(\begin{array}{c}\sqrt{x^2+y^2+z^2}\\ 0\\ \frac{-\cancel{x^{2}z}-\cancel{y^{2}z}+\cancel{x^{2}z}+\cancel{y^{2}z}}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \left(\begin{array}{c}\sqrt{x^2+y^2+z^2}\\ 0\\ 0\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\\ \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)& \hfill =& \hfill \left(\begin{array}{c}r\\ 0\\ 0\end{array}\right)\end{array}$