<s>[INST] Given a $2\times 2$ matrix $A$, compute $A^7$. Let $A =
        \begin{pmatrix}
        e^{2x} & -1  \\
        0 & e^{2x}-1  \\
        \end{pmatrix}
$. Compute $A^7$. 
I've tried the obvious way of multiplying $A$ with $A$, then $A^2$ with $A^2$, but I arrived at a messy result in the top right member of the matrix. Is there a general form to be noticed here? 
 [/INST] You have $A=B+C$, where 
$$
B=\begin{bmatrix} e^{2x}&0\\0&e^{2x}-1\end{bmatrix},\ \ \ C=\begin{bmatrix}0&-1\\0&0\end{bmatrix}. 
$$
The key is that $C^2=0$, and that $CB^kC=0$. Also, 
\begin{align}
B^kCB^m&=\begin{bmatrix} e^{2kx}&0\\0&(e^{2x}-1)^k\end{bmatrix}\begin{bmatrix}0&-1\\0&0\end{bmatrix}
\begin{bmatrix} e^{2mx}&0\\0&(e^{2x}-1)^m\end{bmatrix}
=\begin{bmatrix} 0&-(e^{2x})^k(e^{2x}-1)^m\\0&0\end{bmatrix}\\ \ \\
&=(e^{2k})^k(e^{2x}-1)^m\,C.
\end{align}
Then, after cancelling all the zero terms, 
\begin{align}
(B+C)^7&=B^7+\sum_{k=0}^7B^kCB^{7-k}=B^7+\sum_{k=0}^7(e^{2x})^k(e^{2x}-1)^{7-k}C
\end{align}
So
$$
A^7=(B+C)^7=\begin{bmatrix} e^{14x}&-\displaystyle\sum_{k=0}^7(e^{2x})^k(e^{2x}-1)^{7-k}\\0&(e^{2x}-1)^7\end{bmatrix}.
$$
The sum can be simplified by noting that 
$$
\sum_{k=0}^7a^kb^{7-k}=\frac{b^7-a^7}{b-a}.
$$
Thus
$$
A^7=(B+C)^7=\begin{bmatrix} e^{14x}&\displaystyle (e^{2x}-1)^{7}-e^{14x}\\0&(e^{2x}-1)^7\end{bmatrix}.
$$
</s>