Practicing Success
If $A =\begin{bmatrix}1&a\\0&1\end{bmatrix}$, then $A^n$ (where $n ∈ N$) equals |
$\begin{bmatrix}1&na\\0&1\end{bmatrix}$ $\begin{bmatrix}1&n^2a\\0&1\end{bmatrix}$ $\begin{bmatrix}1&na\\0&0\end{bmatrix}$ $\begin{bmatrix}n&na\\0&n\end{bmatrix}$ |
$\begin{bmatrix}1&na\\0&1\end{bmatrix}$ |
We have, $A =\begin{bmatrix}1&a\\0&1\end{bmatrix}$ $∴A^2 =\begin{bmatrix}1&a\\0&1\end{bmatrix}\begin{bmatrix}1&a\\0&1\end{bmatrix}=\begin{bmatrix}1&2a\\0&1\end{bmatrix}$ It can proved by the principle of mathematical induction that $A^n=\begin{bmatrix}1&na\\0&1\end{bmatrix}$ for all $n ∈ N$. |