Solution:

(i) To prove: $(\rm{A}^{t})^{t} = \rm{A}$

Given

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

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

Taking transpose of $\rm{A}^{t}$, we will get

$(\rm{A}^{t})^{t} = \left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]= \rm{A}$

Hence proved: $(\rm{A}^{t})^{t} = \rm{A}$

 

(ii) To prove: $(\rm{B}^{t})^{t} = \rm{B}$

Given

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

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

Taking transpose of $\rm{B}^{t}$, we will get

$(\rm{B}^{t})^{t} = \left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right] = \rm{B}$

Hence proved: $(\rm{B}^{t})^{t} = \rm{B}$