The generalized quaternion algebra $D$ is defined relative to a field $F$ of characteristic $2$ with parameters $a, b \in F^\times$ by $$D = F\langle i, j \rangle/(i^2 = a, j^2 = b, ij = - ji).$$ This algebra has dimension 4, with basis $1, i, j, k = ij$ and has an involution $\overline{x+yi+zj+wk} = x-yi-zj-wk$ satisfying $\overline{p\cdot q} = \overline{q} \cdot \overline{p}$. The reduced norm and trace are defined by $n(q) = q \cdot \overline{q}$ and $t(q) = q + \overline{q}$ and the reduced characteristic polynomial is defined by $P_q(\lambda) = \lambda^2 - t(q)\lambda + n(q)$.
I'm taking an algebraic number theory class and the first homework builds up to the result that $D$ is either a division algebra or is isomorphic to $M_2(F)$, the latter case happening iff there is some nonzero $q \in D$ with $n(q) = 0$. Let $m_q : D \to D$ be the endomorphism $m_q(p) = pq$. We're asked to prove that $m_q$ has characteristic polynomial $P_q^2$ and then later that the invariant factors of $m_q$ are either $P_q, P_q$ if $q\notin F$ or $\lambda-q, \lambda-q, \lambda-q, \lambda-q$ if $q\in F$.
To show the characteristic polynomial of $m_q$ is $P_q^2$ it suffices to show that the determinant of $m_q$ is $n(q)^2$, by calculating the characteristic polynomial as a determinant of an element of $D \otimes F(\lambda)$. If we write out $q = x+yi+zj+wk$ then $m_q$ has matrix representation $$m_q = \begin{bmatrix} x & a y & b z & - abw \\ y & x & -bw & bz \\ z & aw & x & - ay \\ w & z & -y & x \end{bmatrix}$$ with respect to the standard basis $1,i,j,k$. The identity $\det(m_q) = n(q)^2$ is then $$\begin{vmatrix} x & a y & b z & - abw \\ y & x & -bw & bz \\ z & aw & x & - ay \\ w & z & -y & x \end{vmatrix} = (x^2 - ay^2 - bz^2 + abw^2)^2.$$
I'm wondering if there's a clever way to derive this identity, without directly calculating the 4 by 4 determinant. I tried to understand the defintion of the reduced norm and trace for general central simple algebras, but it seems like they involve base changing to some extension where $D \cong M_2(F)$ and establishing that such an extension exists in the specific case of the quarternion algebra is sort of the point of this exercise. It would be nice if there was an proof that used only matrix identities, or cleverly used the top exterior power, or exploited some property of conjugation (eg it's easy to show $\det(m_q) \det(m_{\overline{q}}) = n(x)^4$).
Since we're trying to show an identity of polynomials in $a, b, x, y, z, w$ with integral coefficients we could even wlog $F = \mathbb{R}$ or $F = \mathbb{C}$ and assume some polynomial constraints like $n(q) \neq 0, a, b \neq 0$. I'd be happy with a proof that uses this reduction as long as the argument doesn't involve the isomorphism $D \cong M_2(\mathbb{C})$.
It would be really really nice if one could establish that the invariant factors are what they should be by directly proving that there's some basis on which $m_q$ has representation the direct sum of two copies of the companion matrix of $P_q$ (if $q\notin F$), but so far this seems hopeless.