If $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$, show that $A^{2}=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$ and $A^{3}=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$.
Given : $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$
Now,
$A^{2}=A A$
$\Rightarrow A^{2}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{ll}1+0 & 1+1 \\ 0+0 & 0+1\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$
$A^{3}=A^{2} A$
$\Rightarrow A^{3}=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$
$\Rightarrow A^{3}=\left[\begin{array}{ll}1+0 & 1+2 \\ 0+0 & 0+1\end{array}\right]$
$\Rightarrow A^{3}=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$
Hence proved.
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