$\therefore {A^2} = A.A = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
1&1&0\\
1&1&1
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
1&0&0\\
1&1&0\\
1&1&1
\end{array}} \right]$

          $ = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
2&1&0\\
3&2&1
\end{array}} \right]$

also ${A^3} = {A^2}.A = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
2&1&0\\
3&2&1
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
1&0&0\\
1&1&0\\
1&1&1
\end{array}} \right]$

          $ = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
3&1&0\\
6&3&1
\end{array}} \right]$

and, ${A^4} = {A^3}.A = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
3&1&0\\
6&3&1
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
1&0&0\\
1&1&0\\
1&1&1
\end{array}} \right]$

           $ = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
4&1&0\\
{10}&4&1
\end{array}} \right]$

On observing the pattern, we come to a conclusion that,

$A = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
n&1&0\\
{\frac{{n\left( {n + 1} \right)}}{2}}&n&1
\end{array}} \right]$

${A^{20}} = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
{20}&1&0\\
{210}&{20}&1
\end{array}} \right]$

Therefore, sum of first column of 

${A^{20}} = \left[ {1 + 20 + 210} \right]$

$ = 231$