Matrix calculation with sinusoids If $0 < \theta < \pi$ , $0 < \phi$ , $\psi < 2\pi$ and 
$$B=\left(
\begin{array}{ccc}
 \cos (\phi ) & -\sin (\phi ) & 0 \\
 \sin (\phi ) & \cos (\phi ) & 0 \\
 0 & 0 & 1 \\
\end{array}
\right) \left(
\begin{array}{ccc}
 \cos (\theta ) & 0 & \sin (\theta ) \\
 0 & 1 & 0 \\
 -\sin (\theta ) & 0 & \cos (\theta ) \\
\end{array}
\right) \left(
\begin{array}{ccc}
 \cos (\psi ) & -\sin (\psi ) & 0 \\
 \sin (\psi ) & \cos (\psi ) & 0 \\
 0 & 0 & 1 \\
\end{array}
\right)$$
Show by calculation that $||B \mathbf{x}|| = ||\mathbf{x}||,\;\mathbf{x}\in\mathbb{R}^{3\times 1}$ and how can $B$ be interpreted graphically.
If we split $B$ into three sub-matrices and and calculate the determinant:
\begin{align*}\operatorname{Det}(B)&=\operatorname{Det}(B_1)\operatorname{Det}(B_2)\operatorname{Det}(B_3)\\
&=(\cos^2\phi+\sin^2\phi)(\cos^2\theta+\sin^2\theta)(\cos^2 \psi + \sin^2 \psi)\\
&=1\times1\times1=1\end{align*}
So I get that if the determinant is 1 $B$ doesn't "extend" any of the unit vectors in the space and only somehow rotates it. But via calculation with $\mathbf{x}=[x_1 \;x_2 \;x_3]^T$ it becomes a mess.
\begin{align*}
&\begin{bmatrix}\cos\theta\cos\phi\cos\psi-\sin\phi\sin\psi&-\cos\psi\sin\phi-\cos\theta\cos\phi\sin\psi&\cos\phi\sin\theta\\
\cos\theta\cos\psi\sin\phi+\cos\phi\sin\psi&\cos\phi\cos\psi-\cos\theta\sin\phi\sin\psi&\sin\theta\sin\phi\\
-\cos\psi\sin\theta&\sin\theta\sin\psi&\cos\theta \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}
\end{align*}
I don't see how that simplifies to $||x|| = (x_1^2+x_2^2+x_3^2)^{1/2}$ so is there any trigonometric tricks to simplify this calculation?
Also I now that the first sub-matrix $B_1$ rotates $(x_1,x_2)$ $\phi$ radians. $B_2$ rotates $(x_1,x_3)$ $\theta$ radians and $B_3$ again rotates $(x_1,x_2)$ $\psi$ radians. So it simply rotates the point in space three times?
EDIT: I'm not sure if this holds ... Surely not.. edited
$$||B x|| \neq \operatorname{Det}(B) ||x||$$
or ...
$$||B x|| \neq || \operatorname{Det}(B)x||$$
 A: I'd approach the problem in steps. 
Write $B = B_3B_2B_1$, where 
$B_3 = \begin{bmatrix}
\cos\phi & -\sin\phi & 0 \\
\sin\phi & \cos\phi & 0 \\
0 & 0 & 1
\end{bmatrix}$, for example. 
If you can show that each of $B_1, B_2,$ and $B_3$ preserve the lengths of vectors, then you know that, as a composition of length-preserving transformations, $B$ must preserve length as well (e.g. because $\lVert B_1 x \rVert = \lVert x \rVert$ and $\lVert B_2 v \rVert = \lVert v \rVert$ for all $x, v$, we see $\lVert B_2\underbrace{B_1 x}_{v} \rVert = \lVert \underbrace{B_1x}_{v} \rVert = \lVert x \rVert$.)
This should make the problem much more tractable, and the usual trig identities will work. If you're not sure how to apply them, I can edit some work in, but I bet you can prove that $B_1, B_2,$ and $B_3$ are length-preserving.
You're exactly right about the geometric description of $B$: It first rotates the vector in a plane parallel to the $xy$-plane, then in a plane parallel to the $xz$-plane, and finally in a plane parallel to the $xy$-plane again.
