Question

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?
(A) $211$
(B) $210$
(C) $231$
(D) $251$

Answer 1

Firstly, find the value of ${A^2},{A^3},...$ and so on. Then make a relation between the elements of the first column by finding the general term. For example- The general term for the sequence $2,4,6,8,.....$ is given by $n\left( {n + 1} \right)$.

Complete step-by-step answer:
Given: A = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 1&1&0 \\\ 1&1&1 \end{array}} \right] and $B = {A^{20}}$
Now, ${A^2} = A \times A$
{A^2} = \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]
{A^2} = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 2&1&0 \\\ 3&2&1 \end{array}} \right]
Similarly, ${A^3} = {A^2} \times A$
{A^3} = \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]
{A^3} = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 3&1&0 \\\ 6&3&1 \end{array}} \right]
Now from above calculation in first column, we observed that:
In every multiplication, first elements of the column is $1$.
Second element increases by $1$.
Third element being in a series: $1,3,6,10,....$; whose general term is $\dfrac{{n\left( {n + 1} \right)}}{2}$.
Therefore, sum of first column in ${A^{20}} = 1 + n + \dfrac{{n\left( {n + 1} \right)}}{2}$
Put n=20,
Sum of first column in ${A^{20}} = 1 + 20 + \dfrac{{20\left( {20 + 1} \right)}}{2}$
$= 1 + 20 + 10 \times 21$
$= 1 + 20 + 210$
$= 231$

Hence, option (C) is the correct answer.

Note: Another method to solve this question is find the value of ${A^2},{A^3},.....{A^{20}}$.
B = {A^{20}} = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ {20}&1&0 \\\ {210}&{20}&1 \end{array}} \right]
Then, the sum of first column in $B$ $= 1 + 20 + 210$ $= 231$