= Solution
The <Drude model> treats mobile <electric charge> carriers as independent particles with density $n$, signed <electric charge> $q$, isotropic <mass> $m$, and <momentum relaxation time> $\tau$. In uniform steady fields,
$$
0=q(\mathbf E+\mathbf v\times\mathbf B)-\frac{m\mathbf v}{\tau},\qquad \mathbf J=nq\mathbf v.
$$
At zero <magnetic field> the <drift velocity> is $q\tau\mathbf E/m$, giving <electrical conductivity> and <electrical resistivity>
$$
\boxed{\sigma_0=\frac{nq^2\tau}{m},\qquad \rho_0=\frac{m}{nq^2\tau}.}
$$
The <electrical conductivity> is positive for either carrier sign. Scattering dissipates directed drift <energy> into the environment; a finite steady <electric current> is compatible with a short-lived drift <momentum>. The independent-carrier relaxation-time approximation omits detailed band anisotropy, quantum <interference> and microscopic collision correlations.
For <electric current> along $x$ and <magnetic field> $B\hat{\mathbf z}$, the magnetic <force> deflects carriers toward one side of the sample. <electric charge> accumulates until a transverse Hall field cancels that <force> in an open transverse circuit. With $J_y=0$, $E_y=v_xB$, so
$$
\boxed{R_H=\frac{E_y}{J_xB}=\frac1{nq},\qquad \rho_{yx}=\frac{B}{nq}.}
$$
<Electrons> have negative $R_H$ in this convention, whereas positive carriers have positive $R_H$. For a strip of thickness $d$ along $z$, the magnitude of the Hall <electric potential difference> across its width is $|V_H|=|IB/(nqd)|$. Thus Hall measurements determine carrier sign and, within this one-carrier model, density, while combining $R_H$ with $\sigma_0$ determines <charge carrier mobility> $\mu=|q|\tau/m=\sigma_0|R_H|$.
Keeping both in-plane <electric current> components, the <resistivity tensor> is
$$
\boldsymbol\rho=\begin{pmatrix}\rho_0&-B/(nq)\\B/(nq)&\rho_0\end{pmatrix},\qquad \boldsymbol\sigma=\boldsymbol\rho^{-1}.
$$
This fixes the off-diagonal sign convention and shows that inversion is not componentwise reciprocation. Under the Hall boundary condition $J_y=0$, the ideal one-carrier isotropic model gives $E_x/J_x=\rho_0$, independent of <magnetic field>: there is no ordinary longitudinal <magnetoresistance> in this minimal model. Multiple carrier types or anisotropic bands can change both this conclusion and the simple $R_H=1/(nq)$ inference. Their <electrical conductivity> must be added before inverting to obtain the total <electrical resistivity>.
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