# Inverse of a $2\times2$ matrix with noncommuting entries

I want to invert a $2 \times 2$ matrix with quaternionic entries. Since non-commutativity, the determinant is not well defined and I've seen in http://en.wikipedia.org/wiki/Quaternionic_matrix that, to test the invertibility, I can use the determinat of the $4 \times 4$ complex matrix that correspond to the rapresentation of each quaternion by a $2 \times 2$ complex matrix. But that page of Wikipedia does not give any rule to write the inverse matrix. There is some rule to do so?

Here is what I found.

If an inverse exists it's both left and right, since $M(2,\mathbb{H})$ is a group. So I try a matrix such that: $$\left[ \begin{array}{ccccc} a&b\\ c&d\\ \end{array} \right] \left[ \begin{array}{ccccc} x&y\\ z&t\\ \end{array} \right] = \left[ \begin{array}{ccccc} ax+bz&ay+bt\\ cx+dz&cy+dt\\ \end{array} \right] = \left[ \begin{array}{ccccc} 1&0\\ 0&1\\ \end{array} \right]$$ equivalent to $$\begin{cases} ax +bz=1\\ cx+dz=0\\ \end{cases} \qquad \land \qquad \begin{cases} ay+bt=0\\ cy+dt=1 \end{cases}$$ If one of a,b,c,d are $0$ the solution is simple. For $a,b,c,d \ne 0$, we find

$$\begin{cases} x=-c^{-1}dz\\ z=\left(b-ac^{-1}d\right)^{-1} \end {cases} \qquad \begin{cases} y=-a^{-1}bt\\ t=\left(d-ca^{-1}b\right)^{-1} \end{cases}$$ I fear that there is no simpler solution... or someone knows?

• I'm not entirely positive but I would say, convert the 2x2 quaternionic matrix into a 4x4 complex matrix, invert the 4x4 and then convert it back to a 2x2 quaternionic matrix. – Mastrel Nov 29 '14 at 22:45
• @Mastrel it's a long way.... I'm searching a shortcut. – Emilio Novati Nov 29 '14 at 22:56
• Let $M_4$ be the complex $4\times 4$ matrix and $D = \det M_4$. Then the inverse of the $2\times 2$ matrix of quaternions will be a matrix of quaternions times $D$. The entries themselves are rather complicated. – Alexander Vlasev Nov 30 '14 at 19:40
• @AleksVlasev. I added some calculations to the question. But I don't see the determinant $D$. – Emilio Novati Dec 1 '14 at 12:37
• The suggestion to transplant the problem into $\Bbb M_4( C)$ is certainly pretty good considering that computers would be able to do the translation and inverse easily. An actual formula for the inverse looks really messy, not nearly as simple as a $2\times 2$ matrix over a field. – rschwieb Dec 2 '14 at 15:59

Let $X=\begin{pmatrix}x&y\\z&w\end{pmatrix}$ be in $M_2(\mathbb H)$ then $$X^{-1}=\frac{1}{n(x)n(w)+n(y)n(z)-\operatorname{tr}(\overline{y}x\overline{z}w)}\begin{pmatrix}n(w)\overline{x}-\overline{z}w\overline{y}&n(y)\overline{z}-\overline{x}y\overline{w}\\n(z)\overline{y}-\overline{w}z\overline{x}&n(x)\overline{w}-\overline{y}x\overline{z}\end{pmatrix}.$$ Note that $n(x)n(w)+n(y)n(z)-\operatorname{tr}(\overline{y}x\overline{z}w)$ is the reduced norm of $X$.