Composition of isometries: intuition I want to prove the following statement.

Let $x$ be a point in $\mathbb{R}^3$. First rotate $x$ $\theta$ radians about the $z$-axis and then reflect it through the $yz$ plane. Write the $3 \times 3$ matrix representation this composition of isometries.

Though I do not quite understand the right-hand rule or the derivation of this matrix, I believe the rotation matrix $\theta$ radians about the $z$-axis is
\begin{align*}
\begin{bmatrix} \cos \theta & - \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}.
\end{align*}
I was also able to find that a reflection through the $yz$-plane takes the form:
\begin{align*}
\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.
\end{align*}
The matrix that rotates and then reflects is:
\begin{align*}
\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos \theta & - \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos \theta & - \sin \theta & 0 \\ - \sin \theta & - \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}.
\end{align*}
My questions are: first, is this correct? Second, how does one make sense of what is going on with these matrices and where they come from? I feel that I lack any intuition.
 A: Let's consider this matrix first:
\begin{align*}
\begin{bmatrix} \cos \theta & - \sin \theta & \color{Red}0 \\ \sin \theta & \cos \theta & \color{Red}0 \\ \color{Blue}0 & \color{Blue}0 & 1 \end{bmatrix}.
\end{align*}
Notice this matrix is block-diagonal. The upper left block is a $2\times2$ rotation matrix which describes what this 3D rotation does within the $xy$-plane. (I assume you understand where this 2D matrix comes from, correct me if I'm wrong.) The lower right block is $1$ which signifies this rotation fixes the $z$-axis (and doesn't stretch it or anything). The red column of $0$s indicates the $z$-axis doesn't get tilted into the $xy$-plane, and similarly the blue row of $0$s indicates the $xy$-plane doesn't get tilted into the $z$-axis any.
In general, if a linear operator $T$ preserves two complementary subspaces $A$ and $B$, and a basis is chosen which is a union of bases for $A$ and $B$, then the matrix for $T$ is block-diagonal, with one block describing how $T$ acts on $A$ and the other describing how it acts on $B$ (with respect to their respective bases). (Note a subspaces $W$ is preserved by $T$ if $TW\subseteq W$, i.e. $Tw\in W$ for all $w\in W$.) In this case the two subspaces are the $xy$-plane and the $z$-axis, with the standard basis.
Similarly, for a reflection across the $yz$-plane, you know the $yz$-plane is fixed (so you need a $2\times2$ identity matrix for the $yz$-coordinates) and the $x$-axis is inverted (so you need a $-1$ in the upper left corner), the rest being $0$s because again the axis and the plane do not get tilted. So a reflection across the $yz$-plane will be the diagonal matrix $\mathrm{diag}(-1,1,1)$.
