Question

Let $A=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]$. Then for positive integer $n,A^n$ is

• A. $\left[\begin{array}{ccc}1& n& n^2\\ 0& n^2& n\\ 0& 0& n\end{array}\right]$
• B. $\left[\begin{array}{ccc}1& n& n\left(\frac{n+1}{2}\right)\\ 0& 1& n\\ 0& 0& 1\end{array}\right]$
• C. $\left[\begin{array}{ccc}1& n^2& n\\ 0& n& n^2\\ 0& 0& n^2\end{array}\right]$
• D. $\left[\begin{array}{ccc}1& n& 2n-1\\ 0& \frac{n+1}{2}& n^2\\ 0& 0& \frac{n+1}{2}\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{ccc}1& n& n\left(\frac{n+1}{2}\right)\\ 0& 1& n\\ 0& 0& 1\end{array}\right]$

$A=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]$

$\Rightarrow A^3=\left[\begin{array}{ccc}1& 3& 6\\ 0& 1& 3\\ 0& 0& 1\end{array}\right]$

Observing the pattern, we can conclude that
$\Rightarrow A^n=\left[\begin{array}{ccc}1& n& \frac{n(n+1)}{2}\\ 0& 1& n\\ 0& 0& 1\end{array}\right]$