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Let $A = \,\left[ {\begin{array}{*{20}{c}}
1&0&0\\
1&1&0\\
1&1&1
\end{array}} \right]$ and $B = A^{20}$ . Then the sum of the elements of the first column of $B$ is?
$211$
$210$
$231$
$251$
Solution
Here $A = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
1&1&0\\
1&1&1
\end{array}} \right]$
$\therefore {A^2} = A.A = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
1&1&0\\
1&1&1
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
1&0&0\\
1&1&0\\
1&1&1
\end{array}} \right]$
$ = \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$
