I’ll write:
\begin{pmatrix}a & b \\ c & d\end{pmatrix}
instead because I’m not really sure how to typset it in \LaTeX.
(i)
We have:
\displaystyle \begin{pmatrix}0 & 0 \\ c & d\end{pmatrix} = \begin{pmatrix}0 \times a & 0 \times b \\ c & d\end{pmatrix} = 0 \times \begin{pmatrix}a & b \\ c & d\end{pmatrix} = 0
using property (1)
(ii)
Using property (2) on part (i) we have:
0 = \begin{pmatrix}0 & 0 \\ c & d\end{pmatrix} = \begin{pmatrix}d & 0 \\ c & 0\end{pmatrix}
Using property (3) on part (i) we have:
0 = \begin{pmatrix}0 & 0 \\ c & d\end{pmatrix} = \begin{pmatrix}0 & c \\ 0 & d\end{pmatrix}
Using property (2) on this gives:
0 = \begin{pmatrix}0 & c \\ 0 & d\end{pmatrix} = \begin{pmatrix}d & c \\ 0 & 0 \end{pmatrix}
(iii)
Applying property (2) and (3) we have:
\begin{pmatrix}a & b \\ s \times c & s \times d\end{pmatrix} = \begin{pmatrix}s \times d & s \times c \\ b & a\end{pmatrix} = s \times \begin{pmatrix}d & c \\ b & a\end{pmatrix}
Applying property (2) and (3) we have:
\begin{pmatrix}a & b \\ s \times c & s \times d\end{pmatrix} = s \times \begin{pmatrix}d & c \\ b & a\end{pmatrix} = s \times \begin{pmatrix}a & b \\ c & d\end{pmatrix}
(iv)
Applying property (2) and (3) we have:
\begin{pmatrix}a & b \\ c + x & d + y\end{pmatrix} = \begin{pmatrix}d + y & c + x \\ b & a\end{pmatrix}
Then using property (4):
\begin{pmatrix}d + y & c + x \\ b & a\end{pmatrix} = \begin{pmatrix}d & c \\ b & a\end{pmatrix} + \begin{pmatrix}y & x \\ b & a\end{pmatrix}
Applying properties (2) and (3) on both we get:
\begin{pmatrix}d & c \\ b & a\end{pmatrix} + \begin{pmatrix}y & x \\ b & a\end{pmatrix} = \begin{pmatrix}a & b \\ c & d\end{pmatrix} + \begin{pmatrix}a & b \\ x & y\end{pmatrix}
Oddly easy? I’m not sure what part is supposed to be particularly discriminating.