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.
I assume that an improper rotation means an element of the orthogonal group with determinant $=-1$. In other words, an $n$-dimensional improper rotation is represented by a matrix $R$ such that $RR^T=I_n$ and $\det R=-1$.
All such matrices have $\lambda=-1$ as an eigenvalue. This is because
$$\det(R+I)=\det(R+RR^T)=\det R \det (I+R^T)=-\det(I+R^T)=-\det(R+I),$$
which implies that $\det(R+I)=0$.
This has the following corollary
A 2-dimensional improper rotation is just the orthogonal reflection w.r.t. a line through the origin.
Proof. We saw above that $\lambda_1=-1$ is an eigenvalue of $R$. If $\lambda_2$ is the other eigenvalue, then $\lambda_1\lambda_2=\det R=-1$, so we can conclude that $\lambda_2=1$. If $\vec{u}$ is an eigenvector belonging to $\lambda_2$, and $\vec{v}\perp\vec{u}$ is another unit vector, then (because $R$ preserves lengths and angles) we can conclude that $R\vec{v}\perp R\vec{u}$.
But here $R\vec{u}=\vec{u}$, and in 2D the only unit vectors $\perp\vec{u}$ are $\pm\vec{v}$. So we can conclude that $R\vec{v}=\pm\vec{v}$. The plus sign cannot occur, for then we would have $R=I_2$. Therefore $R\vec{v}=-\vec{v}$. This implies that $R$ is the orthogonal reflection w.r.t. the line spanned by $\vec{u}$.
Best Answer
The complex number $a+bi$ can be represented by the matrix $\displaystyle \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$.
Note that $(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i$ and
$$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \pm\begin{pmatrix} c & -d \\ d & c \end{pmatrix}=\begin{pmatrix} a\pm c & -(b\pm d) \\ b\pm d & a\pm c \end{pmatrix}$$
Also we have $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ and
$$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix}\begin{pmatrix} c & -d \\ d & c \end{pmatrix}=\begin{pmatrix} ac-bd & -(ad+bc) \\ ad+bc & ac-bd \end{pmatrix}$$