$(A -3I)(A + 2I) = 0$ implies $A = 3I$ or $A = -2I$? I know that square matrices' multiplication $XY=0$ does not imply $X=0$ or $Y=0$.
However, what if $X = A - 3I $, $Y = A + 2I $, given $I$ the identity matrix.
$$(A - 3I)(A + 2I) = 0$$
If it goes with the same rule of $XY=0$, then $A$ can be one matrix other than $-3I$ and $2I$.
But, we have the restriction that we can find an $A$ such that  $A - 3I$ and $A + 2I$ are the multiplication that produce $0$.
Can I still find a matrix $A$ which is not equal to $3I$ or $-2I$?
 A: To find all solutions in the $2 \times 2$-case, write $A= \begin{pmatrix}a & b\\ c & d\end{pmatrix}$. The condition $$(A-3I)(A+2I) = 0$$
leads to the system
$$\begin{cases}(a-3)(a+2) + bc = 0 \\ (a-3)b + b(d-2) = 0 \\ c(a+2) + c(d-3) = 0 \\ bc + (d-3)(d+2) = 0\end{cases}$$
In particular, $a= 3, b = 0, c = 0, d = -2$ satisfies this, so the matrix
$$A:=\begin{pmatrix} 3  & 0  \\ 0 & -2\end{pmatrix}$$
presents a counterexample.
There are even more solutions: see https://www.wolframalpha.com/input/?i=solve+for+a%2Cb%2Cc%2Cd%3A+%28a-3%29%28a%2B2%29%2Bbc+%3D+0%2C+%28a-3%29b+%2B+b%28d-2%29+%3D+0%2C+c%28a%2B2%29+%2B+c%28d-3%29+%3D+0%2C+bc+%2B+%28d-3%29%28d%2B2%29+%3D+0+
A: $X$ and $Y$ are not null, then
$$XY=O \implies \det|X| =0 ~and~ \det|Y|=0$$
Also $$(A-3I)(A-2I)=O \implies A^2-A-6I=0$$ is the characteristic eq. satisfied by the matrix $A$, where $3,-2$ are the  eigenvalues. The qudratic can be compared with the monic quadratoc satisfied by a $2 \times 2$ matrix, namely: $$A^2-Tr(A) A+|A|I=0$$
This means the matrix $A$ has trace $Tr(A)=1$ and $\det|A|=-6$. So the simple choice is $$A=\begin{pmatrix} 3 & 0 \\0 & -2 \end{pmatrix}$$
Also one may check that $$\det|A-3I|=0=\det|A+2I|$$
A: hint
By Cayley-Hamilton Theorem, if the charactetistic polynomial is
$$P(x)=(x-3)(x+2)=(3-x)(-2-x)$$ then matrix will satisfy $$(A-3I)(A+2I)=0$$
