Rotation along z gives the transpose of the elementary rotation matrix along z; not true along y

Question

I considered the two cases in figure:

In the first case I rotated the reference frame along z, I expressed x2,y2,z2 in function of x1,y1,z1, and I got a rotation matrix A:

       | cos(g)      sin(g)      0 |
A =    | -sin(g)     cos(g)      0 |
       |   0           0         1 |

which is the transpose of the elementary rotation matrix along z:

In the second case I rotated the reference frame along y, I expressed x2,y2,z2 in function of x1,y1,z1, and I got a rotation matrix B:

       | cos(b)      0      sin(b) |
B =    |   0        1          0   |
       | -sin(b)      0     cos(b) |

which is the elementary rotation matrix along y:

Why? I expect to get either A=transpose(R_z) and B=traspose(R_y) or A=R_z and B=R_y.

Thank you so much for your time.

Answer 1

Based on right-hand rule, in the first case the $z$ axis is coming out of the paper (if you swipe your fingers from $x_1$ to $y_1$ then your thumb points upward, indicating the direction of $z_1$). So while swiping from $x_1$ to $y_1$, your fingers pass $x_2$ indicating the positive rotation along $z$.

However, in the second case, based on the right-hand rule, the $y_1$ axis points inward so the rotation shown in the figure is actually along -$y_1$ axis not the $y_1$ itself.