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(B) Given $A = \begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix}$.We can observe that $A = I + N$,where $I$ is the identity matrix and

Vedclass · Accepted Answer

$2020$

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(B) Given $A = \begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix}$.We can observe that $A = I + N$,where $I$ is the identity matrix and $N = \begin{bmatrix} 0 & 1 & 1 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix}$.Note that $N^2 = \begin{bmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$ and $N^3 = 0$.Using the binomial expansion,$A^n = (I+N)^n = I + nN + \frac{n(n-1)}{2}N^2$.Thus,$A^n = \begin{bmatrix} 1 & n & n + \frac{n(n-1)}{2} \ 0 & 1 & n \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & n & \frac{n^2+n}{2} \ 0 & 1 & n \ 0 & 0 & 1 \end{bmatrix}$.The sum of elements of $A^n$ is $S_n = 1 + n + \frac{n^2+n}{2} + 0 + 1 + n + 0 + 0 + 1 = 3 + 2n + \frac{n^2+n}{2} = 3 + \frac{5n+n^2}{2}$.Sum of elements of $M = \sum_{n=1}^{20} S_n = \sum_{n=1}^{20} (3 + \frac{5}{2}n + \frac{1}{2}n^2) = 3(20) + \frac{5}{2} \frac{20(21)}{2} + \frac{1}{2} \frac{20(21)(41)}{6}$.$= 60 + 525 + 1435 = 2020$.

If $A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ and $M = A + A^{2} + A^{3} + \dots + A^{20}$,then the sum of all the elements of the matrix $M$ is equal to $.....$

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