Axis Transformation Matrix Apologies if this question has been asked, but I could not find the right answer. I have a transformation matrix between 2 frames. 
In my environment, the following is the standard.

Front : +x-axis. Left : +y-axis. Top : +z-axis

All my transformation matrices follow this convention.
However, I need to convert my transformation matrices to the following convention, for another problem. 

Right: +x-axis. Top: +y-axis. Front: +z-axis

(This, I believe is the standard, for many applications)
What is the transform for this?
(If I split into rotation and translation, the transform for translation is quite straightforward, but I could not figure out for rotation, due to the sign changes as well).
Edit: Just to be clear, I wish to obtain the transformation matrix that I have to apply to my older rotational matrix (and translation matrix) to get the new rotational matrix (which I will later apply to my points).
Any solution with quaternions is also feasible.
Thanks!
 A: If I understand your question, here is an approach to consider:

If the above figure is consistent with your definitions, we want:
$
x\rightarrow z'\\
\begin{bmatrix}
1 \\
0 \\
0 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
0 \\
0 \\
1 \\
\end{bmatrix}
$
$
y \rightarrow -x'\\  
\begin{bmatrix}
0 \\
1 \\
0 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
-1 \\
0 \\
0 \\
\end{bmatrix}
$
$
z \rightarrow y'\\
\begin{bmatrix}
0 \\
0 \\
1 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
0 \\
1 \\
0 \\
\end{bmatrix}
$
We can now determine the transformation matrix as a product of permutation matrices.  Using a test vector $\begin{bmatrix} 1 \\ 2 \\ 3\\ \end{bmatrix}$:
$
\begin{bmatrix}
0 & -1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}      
\begin{bmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{bmatrix}
\begin{bmatrix}
1 \\
2 \\
3 \\
\end{bmatrix}
=
\begin{bmatrix}
-2 \\
3 \\
1 \\
\end{bmatrix}
$
where the innermost matrix on the left hand side swaps rows 1 and 3, and the outermost matrix then swaps rows 1 and 2 with a sign change.  This leads to:
$
\begin{bmatrix}
0 & -1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0 \\
\end{bmatrix}      
\begin{bmatrix}
1 \\
2 \\
3 \\
\end{bmatrix}
=
\begin{bmatrix}
-2 \\
3 \\
1 \\
\end{bmatrix}
$
which I believe is what you are looking for.  I hope this helps.
