Question
If $A=\left[\begin{array}{ll}2 & -1 \\ 3 & -2\end{array}\right]$, find $A^3$.

Answer

$\begin{aligned} A^2=A \cdot A & =\left[\begin{array}{ll}2 & -1 \\ 3 & -2\end{array}\right]\left[\begin{array}{ll}2 & -1 \\ 3 & -2\end{array}\right] \\ & =\left[\begin{array}{ll}4-3 & -2+2 \\ 6-6 & -3+4\end{array}\right] \\ A^2 & =\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\end{aligned}$

$\therefore \mathrm{A}^2=1$

Multiplying throughout by A, we get

$\begin{aligned} & A^3=A .1 \\ & \therefore A^3=A\end{aligned}$

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