Solution:

(i) (A $+$ B)$^{t}$ = A$^{t}+$ B$^{t}$

Taking Left Hand Side

= (A $+$ B)$^{t}$

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

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

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

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

Taking Right Hand Side

= A$^{t}+$ B$^{t}$

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

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

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

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

From (1) and (2), it is proved that: (A $+$ B)$^{t}$ = A$^{t}+$ B$^{t}$

(ii) (A $-$ B)$^{t}$ = A$^{t}-$ B$^{t}$ 

Taking Left Hand Side

= (A $-$ B)$^{t}$

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

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

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

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

Taking Right Hand Side

= A$^{t}-$ B$^{t}$

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

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

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

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

From $(1)$ and $(2)$, it is proved that: (A $-$ B)$^{t}$ = A$^{t}-$ B$^{t}$

(iii) To prove A $+$ A$^{t}$ is symmetric

A $+$ A$^{t}$

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

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

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

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

Now we will take transpose of A$ + $A$^{t}$

(A $+$ A$^{t}$)$^{t}$

= ${{\left[ \begin{matrix} 2 & 2 \\ 2 & 2 \\ \end{matrix} \right]}^{t}}$

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

From $1$ and $2$, it is proved that: A $+$ A$^{t}$ = (A $+$ A$^{t}$)$^{t}$

So, it is symmetric.

(iv) To prove A $-$ A$^{t}$ is skew-symmetric

A $-$ A$^{t}$

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

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

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

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

Now we will take transpose of A $-$ A$^{t}$

(A $-$ A$^{t}$)$^{t}$

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

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

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

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

= $-$(A $-$ A$^{t}$)

From $(1)$ and $(2)$, it is proved that: (A $-$ A$^{t}$)$^{t}$ = $-$(A $-$ A$^{t}$)

So A $-$ A$^{t}$ is skew-symmetric

(v) To prove B $+$ B$^{t}$ is symmetric

B $+$ B$^{t}$

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

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

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

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

Now we will take transpose of B $+$ B$^{t}$

(B $+$ B$^{t}$)$^{t}$

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

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

From $(1)$ and $(2)$, it is proved that: B $+$ B$^{t}$ = (B $+$ B$^{t}$)$^{t}$

So, it is symmetric.

(vi) To prove B $-$ B$^{t}$ is skew-symmetric

B $-$ B$^{t}$

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

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

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

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

Now we will take transpose of B $-$ B$^{t}$

(B $-$B $^{t}$)$^{t}$

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

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

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

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

= $-$(B $-$ B$^{t}$)

From $(1)$ and $(2)$, it is proved that: (B $-$ B$^{t}$)$^{t}$ = $-$(B $-$ B$^{t}$)

So, it is skew-symmetric.