Solution:

(i) AB = BA

Taking Left Hand Side: AB

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times 1+3\times \left( -3 \right) & -1\times 2+3\times \left( -5 \right) \\ 2\times 1+0\times \left( -3 \right) & 2\times 2+0\times \left( -5 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-9 & -2-15 \\ 2+0 & 4+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10 & -17 \\ 2 & 4 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: BA

= $\left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]\times \left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1\times \left( -1 \right)+2\times 2 & 1\times 3+2\times 0 \\ -3\times \left( -1 \right)+\left( -5 \right)\times 2 & -3\times 3+\left( -5 \right)\times 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1+4 & 3+0 \\ 3-10 & -9+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & 3 \\ -7 & -9 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(1)$

 

From $(1)$ and $(2)$, it is proved that: AB $\neq$ BA

 

(ii) A(BC) = (AB)C

Taking Left Hand Side: A(BC)

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\left( \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]~\times ~\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\left( \left[ \begin{matrix} 1\times 2+2\times 1 & 1\times 1+2\times 3 \\ -3\times 2+\left( -5 \right)\times 1 & -3\times 1+\left( -5 \right)\times 3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]~\times \left[ \begin{matrix} 2+2 & 1+6 \\ -6-5 & -3-15 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]~\times \left[ \begin{matrix} 4 & 7 \\ -11 & -18 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times 4+3\times \left( -11 \right) & -1\times 7+3\times \left( -18 \right) \\ 2\times 4+0\times \left( -11 \right) & 2\times 7+0\times \left( -18 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -4-33 & -7-54 \\ 8+0 & 14+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -37 & -61 \\ 8 & 14 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: (AB)C

= $\left( \left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]~\times \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]~ \right)\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

= $\left( \left[ \begin{matrix} -1\times 1+3\times \left( -3 \right) & -1\times 2+3\times \left( -5 \right) \\ 2\times 1+0\times \left( -3 \right) & 2\times 2+0\times \left( -5 \right) \\ \end{matrix} \right] \right)\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-9 & -2-15 \\ 2+0 & 4+0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10 & -17 \\ 2 & 4 \\ \end{matrix} \right]~\times \left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10\times 2+\left( -17 \right)\times 1 & -10\times 1+\left( -17 \right)\times 3 \\ 2\times 2+4\times 1 & 2\times 1+4\times 3 \\ \end{matrix} \right]$ = $\left[ \begin{matrix} -20-17 & -10-51 \\ 4+4 & 2+12 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -37 & -61 \\ 8 & 14 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(2)$

 

From $(1)$ and $(2)$, it is proved that: A(BC) = (AB)C

 

(iii) A(B $+$ C) = AB $+$ AC

Taking Left Hand Side: A(B $+$ C)

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\left( \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]+~\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left( \left[ \begin{matrix} 1+2 & 2+1 \\ -3+1 & -5+3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]~\times \left[ \begin{matrix} 3 & 3 \\ -2 & -2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times 3+3\times \left( -2 \right) & -1\times 3+3\times \left( -2 \right) \\ 2\times 3+0\times \left( -2 \right) & 2\times 3+0\times \left( -2 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -3-6 & -3-6 \\ 6+0 & 6+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -9 & -9 \\ 6 & 6 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: AB $+$ AC

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]+\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times 1+3\times \left( -3 \right) & -1\times 2+3\times \left( -5 \right) \\ 2\times 1+0\times \left( -3 \right) & 2\times 2+0\times \left( -5 \right) \\ \end{matrix} \right]+\left[ \begin{matrix} -1\times 2+3\times 1 & -1\times 1+3\times 3 \\ 2\times 2+0\times 1 & 2\times 1+0\times 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-9 & -2-15 \\ 2+0 & 4+0 \\ \end{matrix} \right]+\left[ \begin{matrix} -2+3 & -1+9 \\ 4+0 & 2+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10 & -17 \\ 2 & 4 \\ \end{matrix} \right]+~\left[ \begin{matrix} 1 & 8 \\ 4 & 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10+1 & -17+8 \\ 2+4 & 4+2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -9 & -9 \\ -6 & 6 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(2)$

 

From $(1)$ and $(2)$, it is proved that: A(B $+$ C) = AB $+$ AC

 

(iv) A(B $-$ C) = AB $-$ AC

Taking Left Hand Side: A(B $-$ C)

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\left( \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]-~\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left( \left[ \begin{matrix} 1-2 & 2-1 \\ -3-1 & -5-3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]~\times \left[ \begin{matrix} -1 & 1 \\ -4 & -8 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times \left( -1 \right)+3\times \left( -4 \right) & -1\times 1+3\times \left( -8 \right) \\ 2\times \left( -1 \right)+0\times \left( -4 \right) & 2\times 1+0\times \left( -8 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1-12 & -1-24 \\ -2+0 & 2+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -11 & -25 \\ -2 & 2 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(1)$ 

 

Taking Right Hand Side: AB $-$ AC

= $\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & 2 \\ -3 & -5 \\ \end{matrix} \right]-\left[ \begin{matrix} -1 & 3 \\ 2 & 0 \\ \end{matrix} \right]\times \left[ \begin{matrix} 2 & 1 \\ 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1\times 1+3\times \left( -3 \right) & -1\times 2+3\times \left( -5 \right) \\ 2\times 1+0\times \left( -3 \right) & 2\times 2+0\times \left( -5 \right) \\ \end{matrix} \right]-\left[ \begin{matrix} -1\times 2+3\times 1 & -1\times 1+3\times 3 \\ 2\times 2+0\times 1 & 2\times 1+0\times 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-9 & -2-15 \\ 2+0 & 4+0 \\ \end{matrix} \right]-\left[ \begin{matrix} -2+3 & -1+9 \\ 4+0 & 2+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10 & -17 \\ 2 & 4 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 8 \\ 4 & 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -10-1 & -17-8 \\ 2-4 & 4-2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -11 & -25 \\ -2 & 2 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(2)$ 

 

From $(1)$ and $(2)$, it is proved that: A(B $-$ C) = AB $-$ AC