 ## Question:

S is the set of all 2 × 2 matrices of the form
$A=\begin{pmatrix} w & x \\ y & z \end{pmatrix}$, where wz − xy = 1.
Show that S is a group under matrix multiplication. Which element(s) have order 2? Prove that an element A has order 3 if w + z + 1 = 0.

## Step-by-step

The condition wz −xy = 1 is the same as |A| = 1; it follows that the set contains an identity element (with w = z = 1 and x = y = 0). Moreover, each matrix in S has an inverse and, since $|{A}^{−1}| |A| = | I | = 1$implies that $|{A}^{−1}| = 1$, the inverses also belong to the set.
If A and B belong to S then, since |AB| = |A| |B| = 1× 1 = 1, their product also belongs to S, i.e. the set is closed.
These observations, together with the associativity of matrix multiplication establish that the set S is, in fact, a group under this operation.
If A is to have order 2 then
$\begin{pmatrix} w & x \\ y & z \end{pmatrix}\begin{pmatrix} w & x \\ y & z \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$
i.e. ${w}^{2} + xy = 1$, x(w + z) = 0, y(w + z) = 0, $xy+ {z}^{2} = 1$.
These imply that ${w}^{2} = {z}^{2}$ and that either z = −w or x = y = 0. If z = −w, then both
${w}^{2} + xy = 1$, from the above condition,
and $− {w}^{2} − xy = 1$, from wz − xy = 1.
This is not possible and so we must have x = y = 0, implying that w and z are either both +1 or both −1. The former gives the identity (of order 1), and so the matrix given by the latter, A = −I, is the only element in S of order 2.
If w + z + 1 = 0 (as well as xy = wz − 1), ${A}^{2}$ can be written as
${ A }^{ 2 }=\begin{pmatrix} { w }^{ 2 }+xy & x(w+z) \\ y(w+z) & xy+{ z }^{ 2 } \end{pmatrix}$
$=\begin{pmatrix} { w }^{ 2 }+wz-1 & -x \\ -y & wz-1+{z}^{2} \end{pmatrix}$
$=\begin{pmatrix} -w-1 & -x \\ -y & -z-1 \end{pmatrix}$.
Multiplying again by A gives
${ A }^{ 3 }=\begin{pmatrix} -w-1 & -x \\ -y & -z-1 \end{pmatrix}\begin{pmatrix} w & x \\ y & z \end{pmatrix}$
$=-\begin{pmatrix} w(w+1)+xy & (w+1)x+xz \\ wy+y(z+1) & xy+z(z+1) \end{pmatrix}$
$=-\begin{pmatrix} w(w+1)+wz-1 & x\times 0 \\ y\times 0 & wz-1+z(z+1) \end{pmatrix}$
$=-\begin{pmatrix} (w\times 0)-1 & 0 \\ 0 & (z\times 0)-1 \end{pmatrix}$
$=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
Thus A has order 3.