Solution:

(i) A + C = C + A

Taking Left Hand Side: A + C

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

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

= $\left[ \begin{matrix} 0 & 2 & 3 \\ 2 & 1 & 4 \\ 2 & 0 & 2 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: C + A

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

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

= $\left[ \begin{matrix} 0 & 2 & 3 \\ 2 & 1 & 4 \\ 2 & 0 & 2 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

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

(ii) A + B = B + A

Taking Left Hand Side: A + B

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

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

= $\left[ \begin{matrix} 2 & 1 & 4 \\ 4 & 1 & 3 \\ 4 & 0 & 3 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: B + A

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

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

= $\left[ \begin{matrix} 2 & 1 & 4 \\ 4 & 1 & 3 \\ 4 & 0 & 3 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

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

(iii) B + C = C + B

Taking Left Hand Side: B + C

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

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

= $\left[ \begin{matrix} 0 & -1 & 1 \\ 2 & -4 & 5 \\ 4 & 2 & 5 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: C + B

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

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

= $\left[ \begin{matrix} 0 & -1 & 1 \\ 2 & -4 & 5 \\ 4 & 2 & 5 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

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

(iv) A + (B + A) = 2A + B

Taking Left Hand Side: A + (B + A)

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

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

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

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

= $\left[ \begin{matrix} 3 & 3 & 7 \\ 6 & 4 & 4 \\ 5 & -1 & 3 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: 2A + B

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

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

= $\left[ \begin{matrix} 2 & 4 & 6 \\ 4 & 6 & 2 \\ 2 & -2 & 0 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2+1 & 4+\left( -1 \right) & 6+1 \\ 4+2 & 6+\left( -2 \right) & 2+2 \\ 2+3 & -2+1 & 0+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & 3 & 7 \\ 6 & 4 & 4 \\ 5 & -1 & 3 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

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

(v) (C$~-$ B) + A = C + (A $-$ B)

Taking Left Hand Side: (C$-$B) + A

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

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

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

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

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

Taking Right Hand Side: C + (A - B)

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

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

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

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

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

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

(vi) 2A + B = A + (A + B)

Taking Left Hand Side: 2A + B

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

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

= $\left[ \begin{matrix} 2 & 4 & 6 \\ 4 & 6 & 2 \\ 2 & -2 & 0 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2+1 & 4+\left( -1 \right) & 6+1 \\ 4+2 & 6+\left( -2 \right) & 2+2 \\ 2+3 & \left( -2 \right)+1 & 0+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & 3 & 7 \\ 6 & 4 & 4 \\ 5 & -1 & 3 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: A + (A + B)

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

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

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

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

= $\left[ \begin{matrix} 3 & 3 & 7 \\ 6 & 4 & 4 \\ 5 & -1 & 3 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

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

(vii) (C $-$ B) $-$A = (C$-$A) B

Taking Left Hand Side: (C – B) – A

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

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

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

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

= $\left[ \begin{matrix} -3 & -1 & -4 \\ -4 & -3 & 0 \\ -3 & 1 & -1 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: (C – A) – B

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

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

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

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

= $\left[ \begin{matrix} -3 & -1 & -4 \\ -4 & -3 & 0 \\ -3 & 1 & -1 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

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

(viii) (A + B) + C = A + (B + C)

Taking Left Hand Side: (A + B) + C

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

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

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

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

= $\left[ \begin{matrix} 1 & 1 & 4 \\ 4 & -1 & 6 \\ 5 & 1 & 5 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: A + (B + C)

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

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

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & -1 & 1 \\ 2 & -4 & 5 \\ 4 & 2 & 5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+0 & 2+\left( -1 \right) & 3+1 \\ 2+2 & 3+\left( -4 \right) & 1+5 \\ 1+4 & -1+2 & 0+5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 1 & 4 \\ 4 & -1 & 6 \\ 5 & 1 & 5 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

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

(ix) A + (B$-$ C) = (A$-$ C) + B

Taking Left Hand Side: A + (B $-$ C)

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

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

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

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

= $\left[ \begin{matrix} 3 & 1 & 4 \\ 4 & 3 & 0 \\ 3 & -1 & 1 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: (A$-$ C) + B

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

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

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

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

= $\left[ \begin{matrix} 3 & 1 & 4 \\ 4 & 3 & 0 \\ 3 & -1 & 1 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

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

(x) 2A + 2B = 2(A + B)

Taking Left Hand Side: 2A + 2B

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

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

= $\left[ \begin{matrix} 2 & 4 & 6 \\ 4 & 6 & 2 \\ 2 & -2 & 0 \\ \end{matrix} \right] + \left[ \begin{matrix} 2 & -2 & 2 \\ 4 & -4 & 4 \\ 6 & 2 & 6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2+2 & 4+\left( -2 \right) & 6+2 \\ 4+4 & 6+\left( -4 \right) & 2+4 \\ 2+6 & \left( -2 \right)+2 & 0+6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & 2 & 8 \\ 8 & 2 & 6 \\ 8 & 0 & 6 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

Taking Right Hand Side: 2(A + B)

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

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

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

= $\left[ \begin{matrix} 2\times2 & 2\times1 & 2\times4 \\ 2\times4 & 2\times1 & 2\times3 \\ 2\times4 & 2\times0 & 2\times3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & 2 & 8 \\ 8 & 2 & 6 \\ 8 & 0 & 6 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

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