Matrix rows or columns are traditionally listed under $(x,y,z)$ order.
Cyclically change the pairs under consideration i.e $(x,y)\to(y,z)\to(z,x)$. The pairs $(x,y)$ and $(y,z)$ show up in the same order in the matrix but the $(z,x)$ shows up in reverse in the matrix. That is the cause of apparent discrepancy but really there is no discrepancy.
For example write
$x'=x\cos \alpha - y \sin \alpha$
$y'=x\sin \alpha + y \cos \alpha$
now change $(x,y)\to(y,z)\to(z,x)$ and $\alpha\to \beta \to \gamma$ and write the three matrices to see how $(z,x)$ part gets flipped.
Edit:
If you want them to look alike then give up the matrix notation and instead write
$y'=y\cos \beta - z \sin \beta$
$z'=y\sin \beta + z \cos \beta$
And
$z'=z\cos \gamma - x \sin \gamma$
$x'=z\sin \gamma + x \cos \gamma$
In each instance if you try to write $\left[ \matrix{ x' \cr y' \cr z'}\right]$ in terms of $\left[ \matrix{ x \cr y \cr z}\right]$ you will see that the mystery goes away.
If you know the sequence of roll, pitch and yaw that you want to do in succession, you can find their matrices and multiply them together in the right order, and the three operations are performed "simultaneously."
The order is important because the group of rotations is nonabelian. If you can produce a matrix for each of these operations, then you can produce a single matrix combining all of them.
Best Answer
So a rotation matrix is always orthonormal, so the transpose of your rotation matrix is the same as your inverse. So if your input point was $\vec v$ and your output point was $\vec v_{rot}$, then you know that (depending on which order you applied the rotations):
$$ \vec v_{rot} = \underbrace{R(\text{yaw}) R(\text{pitch}) R(\text{roll})}_{\text{order matters}}\vec v$$
But when you multiply rotation matrices, you always get a new rotation matrix. So you can write:
$$ R = R(\text{yaw}) R(\text{pitch}) R(\text{roll}) $$
Now, the inverse is:
$$ R^{-1} = R^T = \left(R(\text{yaw}) R(\text{pitch}) R(\text{roll})\right)^T = R(\text{roll})^T R(\text{pitch})^T R(\text{yaw})^T$$
Since you used rodrigues' formula, you got R directly. But the point is that $R^T$ is the inverse of R, so you shouldn't have calculated $R^{-1}$ you should have just used R^T. The reason the answer is not identical is due to numerical error introduced when calculating the inverse of a matrix.
Because you are dealing with floating point numbers, it is very unlikely that you will get EXACTLY the same output after inverting.