The anisotropic conductivity of the Hall configuration. We will only explore the case of perpendicular electric and magnetic fields, throughout the course, with the convention that: $\boldsymbol{\mathcal{E}}= \left(\mathcal{E}_{x}, \mathcal{E}_{y},0 \right)$ and $\boldsymbol{\mathcal{B}} = (0,0,\mathcal{B}_{z})$. We can thus write down the steady-state transport equation for a particular carrier type, following Eqn. $10.3$. We will do this for our definition of mobility as $ = /^*$. If you remember from Lecture 5, we stated that mobility should be positive valued, while $ = \pm $ for electrons and holes. We can write this as a 2D matrix equation since we know carrier motion will be in the $$, $$ plane. $$\begin{pmatrix}v_{dx} \\v_{dy} \\ \end{pmatrix}=\frac{q}{e}\mu\left[\begin{pmatrix}_{x}\\_{y}\end{pmatrix}+\color{red}{\begin{pmatrix}0 & \mathcal{B}_{z} \\-\mathcal{B}_{z} & 0\end{pmatrix}\begin{pmatrix}v_{dx} \\ v_{dy}\end{pmatrix}}\right]$$ where $v_{dx},v_{dy}$ are components of the drift velocity.
My question is a rather simple one, I would just like to know why the matrix $\begin{pmatrix}0 & \mathcal{B}_{z} \\-\mathcal{B}_{z} & 0\end{pmatrix}$ takes this form with zeros on the leading diagonal (and why one of the elements is negative). Equation $10.3$ mentioned in the quote above (from Imperial College London Dept. of Physics) is $$\boldsymbol{\mathcal{v}}_{d}=\frac{q\tau}{m^*}\left(\boldsymbol{\mathcal{E}}+\boldsymbol{\mathcal{v}}_{d}\times\boldsymbol{\mathcal{B}}\right)\tag{1}$$
I know that if we have a cross product of two vectors $\boldsymbol{A}$ & $\boldsymbol{B}$, then, $$\boldsymbol{A}\times\boldsymbol{B}=\begin{vmatrix}\hat i & \hat j & \hat k \\ A_x & A_y & A_z \\ B_x & B_y & B_z \\ \end{vmatrix}$$
Specializing to the case of the cross-product in $(1)$ gives $$\boldsymbol{\mathcal{v}}_{d}\times\boldsymbol{\mathcal{B}}=\begin{pmatrix}v_{dx}\\v_{dy}\\0\end{pmatrix}\times\begin{pmatrix}0\\0\\B_z\end{pmatrix}=\begin{vmatrix}\hat i & \hat j & \hat k \\ v_{dx} & v_{dy} & 0 \\ 0 & 0 & B_z \\ \end{vmatrix}= \color{red}{B_z\left(v_{dy}\hat i-v_{dx}\hat j\right)}$$
But how are (the two red parts equal) $${B_z\left(v_{dy}\hat i-v_{dx}\hat j\right)}={\begin{pmatrix}0 & \mathcal{B}_{z} \\-\mathcal{B}_{z} & 0\end{pmatrix}\begin{pmatrix}v_{dx} \\ v_{dy}\end{pmatrix}}?$$
On the LHS of your last equation, $$ {B_z\left(v_{dy}\hat i-v_{dx}\hat j\right)}=B_z\,v_{dy}\begin{pmatrix}1\\0\end{pmatrix} -B_z\,v_{dx}\begin{pmatrix}0\\1\end{pmatrix} =\begin{pmatrix}B_z\,v_{dy}\\-B_z v_{dx}\end{pmatrix}; $$ on the RHS, $$ {\begin{pmatrix}0 & \mathcal{B}_{z} \\ -\mathcal{B}_{z} & 0\end{pmatrix}\begin{pmatrix}v_{dx} \\ v_{dy}\end{pmatrix}} =\begin{pmatrix}B_z\,v_{dy}\\-B_z\,v_{dx}\end{pmatrix}. $$ Therefore, $$ B_z\left(v_{dy}\hat i-v_{dx}\hat j\right)=\begin{pmatrix}0 & \mathcal{B}_{z} \\ -\mathcal{B}_{z} & 0\end{pmatrix}\begin{pmatrix}v_{dx} \\ v_{dy}\end{pmatrix}. $$