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

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Aicharya · Accepted Answer

We have, $A=\begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix}$ and $M=A+A^2+A^3+\ldots+A^{20}$ Now, $A^2=A.A=\begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 2 & 3 \ 0 & 1 & 2 \ 0 & 0 & 1 \end{bmatrix}$ $\Rightarrow A^3=A^2.A=\begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \ 0 & 1 & 2 \ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 3 & 6 \ 0 & 1 & 3 \ 0 & 0 & 1 \end{bmatrix}$ $\ldots$ $A^n=\begin{bmatrix} 1 & n & \frac{n(n+1)}{2} \ 0 & 1 & n \ 0 & 0 & 1 \end{bmatrix}$ So, required sum $=20	imes3+2\sum_{r=1}^{20}r+\sum_{r=1}^{20}\frac{r(r+1)}{2}$ $=20	imes3+2\sum_{r=1}^{20}r+\frac{1}{2}\sum_{r=1}^{20}r^2+\frac{1}{2}\sum_{r=1}^{20}r$ $=20	imes3+$\frac{5}{2}$\sum_{r=1}^{20}r+$\frac{1}{2}$\sum_{r=1}^{20}r^2$ $=20	imes3+\frac{5}{2}\left(\frac{20	imes21}{2}ight)+\frac{1}{2}\left(\frac{20	imes21	imes41}{6}ight)$ $\left[ \Rightarrow \sum_{r=1}^{n}r=\frac{n(n+1)}{2},\ \sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6} ight]$ $=60+5(5	imes21)+5(7	imes41)$ $=60+525+1435$ $=2020$

If $A=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ and $M=A+A^{2}+A^{3}+\ldots+A^{20}$…

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