Chapter 1.9, Problem 60E

If $A$ is a square matrix, we define the powers $A^2$, $A^3$, and so on, as follows: $A^2=AA$, $A^3=A\left(A^2\right)$, and so on. Suppose $A$ is an $\left(n\times n\right)$ matrix such that $A^3-2A^2+3A-I=0$. Show that $AB=I$, where $B=A^2-2A+3I$.