= Solution
With the <Einstein summation convention>, the <magnetic force> has components $F_i=q\epsilon_{ijk}v_jB_k$. Thus its matrix acting on velocity is
$$
\boxed{M_{ij}=q\epsilon_{ijk}B_k,\qquad
M=q\begin{pmatrix}0&B_z&-B_y\\-B_z&0&B_x\\B_y&-B_x&0\end{pmatrix}.}
$$
The <Levi-Civita symbol> changes sign on exchanging $i,j$, so $M^T=-M$; the diagonal vanishes and the matrix has zero <trace>. For nonzero $\mathbf B$, $M\mathbf B=q\mathbf B\times\mathbf B=0$, exhibiting a nontrivial <kernel> and proving noninvertibility. If $\mathbf B=0$ or $q=0$, $M$ is the zero matrix and is again noninvertible.
Equivalently, an odd-dimensional real <antisymmetric matrix> has $\det M=\det M^T=\det(-M)=-\det M$, hence zero <determinant>. The same antisymmetry gives $\mathbf v^TM\mathbf v=0$, expressing that the magnetic force does no work.
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