 Solution

With the Einstein summation convention, the magnetic force has components $F_i=q{ϵ}_{ijk}{v}_j{B}_k$. Thus its matrix acting on velocity isThe 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 ${\mathbf{v}}^{T}M\mathbf{v}=0$, expressing that the magnetic force does no work.