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Let A = $$\left[ {\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{matrix} } \right]$$ and B = A20. Then the sum of the elements of the first column of B is :

JEE · Math · previous-year question

  1. A.210
  2. B.211
  3. C.231correct
  4. D.251

Answer

C. 231

Explanation

A = $$\left[ {\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{matrix} } \right]$$ A2 = A.A = $$\left[ {\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{matrix} } \right] \times \left[ {\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{matrix} } \right]$$ = $$\left[ {\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \\ \end{matrix} } \right]$$ A3 = A2.A = $$\left[ {\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \\ \end{matrix} } \right] \times \left[ {\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ \end{matrix} } \right]$$ = $$\left[ {\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 6 & 3 & 1 \\ \end{matrix} } \right]$$ Similarly A4 = $$\left[ {\begin{matrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ {10} & 4 & 1 \\ \end{matrix} } \right]$$ From this we can say, An = $$\left[ {\begin{matrix} 1 & 0 & 0 \\ n & 1 & 0 \\ {{{n\left( {n + 1} \right)} \over 2}} & n & 1 \\ \end{matrix} } \right]$$ $$\therefore\,\,\,$$ A20 = $$\left[ {\begin{matrix} 1 & 0 & 0 \\ {20} & 1 & 0 \\ {210} & {20} & 1 \\ \end{matrix} } \right]$$ $$\therefore\,\,\,$$ Sum of the first column = 1 + 20 + 210 = 231

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