Let A=pmatrix1&1510&1pmatrix.

Question

Let $A=\begin{pmatrix}1&\frac1{51}\0&1\end{pmatrix}$. If $B=\begin{pmatrix}1&2\-1&-1\end{pmatrix}A\begin{pmatrix}-1&-2\1&1\end{pmatrix}$, then the sum of all the elements of the matrix $\displaystyle\sum_{n=1}^{50}B^n$ is equal to

Accepted Answer

Let $C=\begin{pmatrix}1&2\-1&-1\end{pmatrix}$, $D=\begin{pmatrix}-1&-2\1&1\end{pmatrix}$. Direct computation gives $DC=I$, so $B=CAD$, and $B^n=CA^nD$ for all $n$ (since $DC=I$ cancels the middle terms).

$A^n=\begin{pmatrix}1&\frac n{51}\0&1\end{pmatrix}$, so
$$B^n=C\begin{pmatrix}1&\frac n{51}\0&1\end{pmatrix}D=\begin{pmatrix}\frac n{51}+1 & \frac n{51}\ -\frac n{51} & 1-\frac n{51}\end{pmatrix}.$$

Summing $n=1$ to $50$: $\sum n=1275$, so $\sum\frac n{51}=25$.

$$\sum_{n=1}^{50}B^n=\begin{pmatrix}25+50 & 25\-25 & 50-25\end{pmatrix}=\begin{pmatrix}75&25\-25&25\end{pmatrix}.$$

Sum of all elements $=75+25-25+25=100$.

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