A) 211
B) 210
C) 231
D) 251
Correct Answer: C
Solution :
Given, \[A=\left[ \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{matrix} \right].\]Computing higher powers of A, \[{{A}^{2}}=A.A=\left[ \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \\ \end{matrix} \right]\] \[{{A}^{3}}={{A}^{2}}.A=\left[ \begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 6 & 3 & 1 \\ \end{matrix} \right]\] \[{{A}^{4}}={{A}^{3}}.A=\left[ \begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 6 & 3 & 1 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 10 & 4 & 1 \\ \end{matrix} \right].\] On observing the pattern, we come to a conclusion that, \[{{A}^{k}}=\left[ \begin{matrix} 1 & 0 & 0 \\ k & 1 & 0 \\ \frac{k(k+1)}{2} & k & 1 \\ \end{matrix} \right].\] Therefore, sum of first column of \[{{A}^{20}}\]is \[(1+20+\frac{20\times 21}{2})=231.\]Option C is correct.
You need to login to perform this action.
You will be redirected in
3 sec
