# 2006_Question5

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.