(b) By simple calculations we observe that $\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill 0\hfill \end{array}\right)$ is the identity for addition and $\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill 1\hfill \end{array}\right)$ is the identity for multiplication. The inverses for addition and multiplication are obtained from the conditions that

i) addition of the given matrix with its inverse gives the identity for addition, so the inverse of $\left(\begin{array}{cc}\hfill x\hfill & \hfill -y\hfill \\ \multicolumn{1}{c}{y}& \hfill x\hfill \end{array}\right)$ for addition is: $\left(\begin{array}{cc}\hfill -x\hfill & \hfill y\hfill \\ \multicolumn{1}{c}{-y}& \hfill -x\hfill \end{array}\right)$ and

ii) multiplying the given matrix with its inverse gives the identity for multiplication; the inverse is $\frac{1}{(x^2+y^2)}\left(\begin{array}{cc}\hfill x\hfill & \hfill y\hfill \\ \multicolumn{1}{c}{-y}& \hfill x\hfill \end{array}\right).$ We see that both identity matrices and both inverses are from the set C*.

(c) Here, we shall consider R* as the set of matrices of the form $\left(\begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill x\hfill \end{array}\right)$. These matrices could be written as $x\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill 1\hfill \end{array}\right)={\mathrm{xI}}_2$. Evidently $\left(\begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill x\hfill \end{array}\right)\pm \left(\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill y\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill x\pm y\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill x\pm y\hfill \end{array}\right)=(x\pm y)I_2$.

The identity for addition is $\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill 0\hfill \end{array}\right)$ and the inverse of $\left(\begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill x\hfill \end{array}\right)$ is $\left(\begin{array}{cc}\hfill -x\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill -x\hfill \end{array}\right)$, which are both from R*.

For multiplication $\left(\begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill x\hfill \end{array}\right)\left(\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill y\hfill \end{array}\right)=x\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill 1\hfill \end{array}\right)·y\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill 1\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill \mathrm{xy}\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill \mathrm{xy}\hfill \end{array}\right)$.

The multiplicative identity is $\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill 1\hfill \end{array}\right)$ and the multiplicative inverse of $\left(\begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill x\hfill \end{array}\right)$ is $\left(\begin{array}{cc}\hfill \frac{1}{x}\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill \frac{1}{x}\hfill \end{array}\right)$.

The distributive law of addition and multiplication is the same as that of real numbers:

${\displaystyle \left(\begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill x\hfill \end{array}\right)·(\left(\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill y\hfill \end{array}\right)+\left(\begin{array}{cc}\hfill z\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill z\hfill \end{array}\right))=\left(\begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill x\hfill \end{array}\right)·\left(\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill y\hfill \end{array}\right)+\left(\begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill x\hfill \end{array}\right)·\left(\begin{array}{cc}\hfill z\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill z\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill x(y+z)\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill x(y+z)\hfill \end{array}\right).}$

This proves that the arithmetic of R* is the same as the arithmetic of real numbers.

(d) Considering 2 complex numbers, $x+\mathrm{iy}$ and $u+\mathrm{iv}$:
$(x+\mathrm{iy})+(u+\mathrm{iv})=(x+u)+i(y+v)$
$(x+\mathrm{iy})(u+\mathrm{iv})=(\mathrm{xu}-\mathrm{yv})+i(\mathrm{xv}+\mathrm{yu}).$

By simple computation we can show that multiplication in the set of two by two matrices C* is commutative and that the distributive law holds in C*. This is so because the laws apply to every operation on the components. We observe that addition and multiplication in C* are the same as addition and multiplication of complex numbers.

(e) Note that ${\left(\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \multicolumn{1}{c}{1}& \hfill 0\hfill \end{array}\right)}^2=\left(\begin{array}{cc}\hfill -1\hfill & \hfill 0\hfill \\ \multicolumn{1}{c}{0}& \hfill -1\hfill \end{array}\right)$ so we see that C* contains a model for $\sqrt{-}1$, the complex number $i$.

The matrix $\left(\begin{array}{cc}\hfill x\hfill & \hfill -y\hfill \\ \multicolumn{1}{c}{y}& \hfill x\hfill \end{array}\right)$ corresponds to the complex number $x+\mathrm{iy}$. Multiplying this number by $i$, we obtain $-y+\mathrm{ix}$, i.e. the matrix $\left(\begin{array}{cc}\hfill -y\hfill & \hfill -x\hfill \\ \multicolumn{1}{c}{x}& \hfill -y\hfill \end{array}\right)$. We shall consider that matrix $\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \multicolumn{1}{c}{c}& \hfill d\hfill \end{array}\right)$ corresponds to $i$: So, $\left(\begin{array}{cc}\hfill x\hfill & \hfill -y\hfill \\ \multicolumn{1}{c}{y}& \hfill x\hfill \end{array}\right)\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \multicolumn{1}{c}{c}& \hfill d\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill -y\hfill & \hfill -x\hfill \\ \multicolumn{1}{c}{x}& \hfill -y\hfill \end{array}\right)$. The only solution for this equation is $\left(\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \multicolumn{1}{c}{1}& \hfill 0\hfill \end{array}\right)$.

A different approach is the following: we associate the point $(x,y)$ in the complex plane with the complex number $(x+\mathrm{iy})$ and also with the matrix $\left(\begin{array}{cc}\hfill x\hfill & \hfill -y\hfill \\ \multicolumn{1}{c}{y}& \hfill x\hfill \end{array}\right)$. The geometrical significance of the multiplication by $i$ of a complex number is the counterclockwise rotation of the point by $\pi /2$ so $(x,y)\to (-y,x)$. In matrix notation this corresponds to

${\displaystyle \left(\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \multicolumn{1}{c}{1}& \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill x\hfill & \hfill -y\hfill \\ \multicolumn{1}{c}{y}& \hfill x\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill -y\hfill & \hfill -x\hfill \\ \multicolumn{1}{c}{x}& \hfill -y\hfill \end{array}\right).}$