Coordinate transformations
3-&gt;2 symmetric component transforms (a.k.a. &quot;Clarke&quot;) are not unique -- in fact there are infinitely many, depending on how you define the axes and how you define the transform (whether it preserves power or magnitude, or whether it is an orthonormal matrix which preserves neither but has certain useful mathematical properties). What is important is that the matrix has orthogonal rows (the way you've shown it) and one row is along the &quot;common-mode&quot; axis with identical components.
I've always used
$$
\begin{bmatrix} 2/3 &amp; -\frac{1}{3} &amp; -\frac{1}{3} \\ 
0 &amp; \frac{\sqrt{3}}{3} &amp; -\frac{\sqrt{3}}{3} \\ 
\frac{1}{3} &amp; \frac{1}{3} &amp; \frac{1}{3} \\ 
\end{bmatrix}
$$
myself, which matches the Wikipedia entry you mentioned.
The first row maps a vector ABC onto the two-phase component X, where if ABC is a 3-phase sine wave set with 0 mean, then X is equal to A in magnitude and phase. The second row maps ABC onto the two-phase component Y, where if ABC is a 3-phase sine-wave set with 0 mean, then Y is equal to A in magnitude but 90 degrees out of phase (and I can never remember whether it leads or lags so I have to derive that every time). The third row computes the common-mode voltage of the set.
(The verification test is this: Use $$ABC = \begin{bmatrix}\cos 0 &amp;  \cos(2\pi/3) &amp; \cos(4\pi/3)\end{bmatrix}^T$$ for θ = 0 and θ = π/2, multiply by the transformation matrix to get the X, Y, and common mode components, and make sure you get unit vectors in the X and Y directions.
If $$ABC = \begin{bmatrix}\cos 0 &amp;  \cos(2\pi/3) &amp; \cos(4\pi/3)\end{bmatrix}^T = \begin{bmatrix}1 &amp; -1/2 &amp; -1/2\end{bmatrix}^T$$
then multiplication by the transformation matrix should yield $$XYn = \begin{bmatrix}1 &amp; 0 &amp; 0\end{bmatrix}$$
and if $$ABC = \begin{bmatrix}\sin 0 &amp;  \sin(2\pi/3) &amp; \sin(4\pi/3)\end{bmatrix}^T = \begin{bmatrix}0 &amp; \sqrt 3/2 &amp; -\sqrt 3/2\end{bmatrix}^T$$
then multiplication by the transformation matrix should yield $$XYn = \begin{bmatrix}0 &amp; 1 &amp; 0\end{bmatrix}$$
which if you do the matrix multiplication, you should be able to verify this.
The rotation matrix that transforms from XY &lt;-&gt; DQ (a.k.a. &quot;Park transform&quot;) is also not unique and depends on your axes reference. It should only rotate and preserve vector magnitude without any anisotropy, so it must be a 2x2 orthonormal matrix. In my systems, I use a transform matrix defined by X = Q when the rotation angle is 0 and Y = Q when the rotation angle is 90. The resulting matrix is a little bit different than the one cited in most motor control literatore.
The most egregious error, in my opinion, in using Park and Clarke transforms without citing a definition that you have verified is appropriate for your use. (And even worse is someone who implements a Park and Clark transform for someone else's use without defining it.) If you just use what's on Wikipedia or what's in a motor control software library, you may end up not understanding it, and creating signals that are incompatible with someone else's system unless you swap X and Y or D and Q, or negate one of the signals, or rotate it in the opposite direction.

As to your specific question:

Which one is right?

They differ in two ways: the missing factor of 2/3, and the negation of the middle row. The 2/3 is for translating ABC -&gt; XY (the inverse transform mapping XY -&gt; ABC has no 2/3) to maintain constant amplitude. When you negate the middle row, you reverse the phase sequence. (C lags B lags A, rather than C leads B leads A as in the Wikipedia entry)
So neither is &quot;wrong&quot; unless you have a specific phase sequence definition and you have an application that requires amplitude or power invariance.