Chapter 1.6, Problem 28E

Let $A$ and $B$ be as in Exercise 27. Find the flaw in the following proof that $A=B$. Since $A^2=AB$, $A^2-AB=0$. Factoring yields $A\left(A-B\right)=0$. Since $A\ne 0$, it follows that $A-B=0$. Therefore, $A=B$.

Exercise 27:

Let $A$ and $B$ be $\left(2\times 2\right)$ matrices such that $A^2=AB$ and $A\ne 0$. Can we assert that, by cancellation, $A=B$? Explain.