Rotations are weird.

Assuming we are rotating about the x-axis and y-axis by 90 degrees in the ccw sense.

Let's consider a vector $<a,b,c>$ . A 90 degree rotation about the x-axis means the component along x is unchanged, but the y-z part is rotated by 90 degrees ccw. That is equivalent to moving the y component to z component, and the z component to -y component. The inverse rotation just does the opposite, so z to y, and y to -z.

The matrices that represent the $R_x$ and $R_x^{-1}$ rotation are give by:

$\large R_x = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix}$ and $\large R_x^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix}$

Note that by inspection, we know $R_x R_x^{-1} = R_x^{-1} R_x=I$ where I is the identity matrix:

Along the same line of reasoning, the matrix that represents the $R_y$ rotation is given by

$\large R_y = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$

Let's compute $R_x^{-1}R_yR_x$ (note the reverse order, since the $R_x$ is applied first):

$\large R_x = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix}= \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = R_z^{-1}\neq R_y$

These rotations are done via matrix multiplication, and matrix multiplication is non-commutative. In other words, the order matters