Confusion in rotation matrix - rotation about $y$ axis The rotation matrix about y axis should look like
$$\left[
\begin{array}{ccc}
\cos\frac{\pi}{2} & 0 &\sin\frac{\pi}{2}\\
0 & 1 & 0\\
-\sin\frac{\pi}{2} & 0 &\cos\frac{\pi}{2}\\
\end{array}
\right]
=
\left[
\begin{array}{ccc}
0 & 0 &1\\
0 & 1 & 0\\
-1& 0 &0\\
\end{array}
\right]
$$
Now, I pictured this rotation, and it should look like rotation about y axis, counterclockwise

I drew $(1,1,1)$ vector in original coordinate system, and in the new coordinate system, the vector should be $(-1,1,1)$

BUT! the matrix calculation gives otherwise.
$$
\left[
\begin{array}{ccc}
0 & 0 &1\\
0 & 1 & 0\\
-1& 0 &0\\
\end{array}
\right]
\left[
\begin{array}{c}
1\\
1\\
1
\end{array}
\right]
=\left[
\begin{array}{c}
1\\
1\\
-1
\end{array}
\right]
$$
Am I missing something?

Okay, it seems like I should rotate vector instead of coordinate system.
Then, How can you explain rotation about z axis?

The rotation matrix is 
$$
\left[
\begin{array}{ccc}
\cos\frac{\pi}{2} &\sin\frac{\pi}{2} & 0\\
-\sin\frac{\pi}{2} &\cos\frac{\pi}{2} & 0\\
0 & 0 & 1
\end{array}
\right]
=
\left[
\begin{array}{ccc}
0 & 1 &0\\
-1 & 0 & 0\\
0& 0 &1\\
\end{array}
\right]
$$
But... calculation is correct.
$$
\left[
\begin{array}{ccc}
0 & 1 &0\\
-1 & 0 & 0\\
0& 0 &1\\
\end{array}
\right]
\left[
\begin{array}{c}
1\\
1\\
1
\end{array}
\right]
=\left[
\begin{array}{c}
1\\
-1\\
1
\end{array}
\right]
$$
IF you rotate the vector, then in the picture, it should be $(-1,1,1)$


Oh. god.... I think I am looking at the wrong source.
My book tells

well, clearly, googling gives me 

http://what-when-how.com/the-3-d-global-spatial-data-model/rotation-matrix-derivation-the-3-d-global-spatial-data-model/
I think I solved my case. Thank you all.

I believe that you are viewing the problem backwards. The matrix
$$
A=\begin{pmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
-1 & 0 & 0
\end{pmatrix}
$$
rotates a vector into a new, rotated frame, according to what you are describing with your images. However, what you are asking is, "What matrix brings me from the rotated frame back to the original frame", which would be the inverse of $A$, i.e.
$$
A^{-1} = \begin{pmatrix}
0 & 0 & -1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{pmatrix}.
$$
