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Let $A=\left(\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)$ Then, for positive integer
$n, A^{n}$ is
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$n, A^{n}$ is
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The correct answer is:
$\left(\begin{array}{lll}1 & n & n\left(\frac{n+1}{2}\right) \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right)$
We have, $A=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
$\therefore \quad A^{2}=A \cdot A$
$=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{lll}1+0+0 & 1+1+0 & 1+1+1 \\ 0+0+0 & 0+1+0 & 0+1+1 \\ 0+0+0 & 0+0+0 & 0+0+1\end{array}\right]$
$=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 2 & \frac{2(2+1)}{2} \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]$
Again, $A^{3}=A^{2} \cdot A$
$=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
$\begin{aligned} &=\left[\begin{array}{lll}1+0+0 & 1+2+0 & 1+2+3 \\ 0+0+0 & 0+1+0 & 0+1+2 \\ 0+0+0 & 0+0+0 & 0+0+1\end{array}\right] \\=\left[\begin{array}{lll}1 & 3 & 6 \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 3 & \frac{3(3+1)}{2} \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{array}\right] \\\therefore A^{n} =\left[\begin{array}{lll}1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right] \end{aligned}$
$\therefore \quad A^{2}=A \cdot A$
$=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{lll}1+0+0 & 1+1+0 & 1+1+1 \\ 0+0+0 & 0+1+0 & 0+1+1 \\ 0+0+0 & 0+0+0 & 0+0+1\end{array}\right]$
$=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 2 & \frac{2(2+1)}{2} \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]$
Again, $A^{3}=A^{2} \cdot A$
$=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
$\begin{aligned} &=\left[\begin{array}{lll}1+0+0 & 1+2+0 & 1+2+3 \\ 0+0+0 & 0+1+0 & 0+1+2 \\ 0+0+0 & 0+0+0 & 0+0+1\end{array}\right] \\=\left[\begin{array}{lll}1 & 3 & 6 \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 3 & \frac{3(3+1)}{2} \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{array}\right] \\\therefore A^{n} =\left[\begin{array}{lll}1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right] \end{aligned}$
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