Question

If $A=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right],$ show that $A^n=\left[\begin{array}{ccc}1& n& \frac{n(n+1)}{2}\\ 0& 1& n\\ 0& 0& 1\end{array}\right]\forall n\in N$.

Answer 1

$A^2=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]\times \left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]A^3=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]\times \left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 3& 6\\ 0& 1& 3\\ 0& 0& 1\end{array}\right]A^4=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]\times \left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 4& 10\\ 0& 1& 4\\ 0& 0& 1\end{array}\right]....A^n=\left[\begin{array}{ccc}1& n& \frac{n(n+1)}{2}\\ 0& 1& n\\ 0& 0& 1\end{array}\right]\forall n\in N$