Let $A$ be an $n \times n$ matrix satisfying
\[A^2+c_1A+c_0I=O,\]
where $c_0, c_1$ are scalars, $I$ is the $n\times n$ identity matrix, and $O$ is the $n\times n$ zero matrix.

Prove that if $c_0\neq 0$, then the matrix $A$ is invertible (nonsingular).
How about the converse? Namely, is it true that if $c_0=0$, then the matrix $A$ is not invertible?

Suppose first that $c_0\neq 0$.
Then we have
\begin{align*}
A^2+c_1A=-c_0I\\
\Leftrightarrow A(A+c_1I)=-c_0 I\\
\Leftrightarrow A\left(\frac{-1}{c_0}(A+c_1I) \right)=I.
\end{align*}
It is in the last step that we needed to assume $c_0\neq0$.

Thus, if we put
\[B=\frac{-1}{c_0}(A+c_1I),\]
then we have proved that
\[AB=I.\]

Similarly, one can check that $BA=A$. Hence $B$ is the inverse matrix of $A$.
Namely,
\[A^{-1}=\frac{-1}{c_0}(A+c_1I).\]
This proves that when $c_0\neq 0$ the matrix $A$ is invertible.

Is it true that if $c_0=0$, then the matrix $A$ is not invertible?

Next, let us consider the case $c_0=0$.
We claim that the matrix $A$ can be invertible even $c_0=0$.

For example, if $A=I$, then $A$ satisfies
\[A^2-A=O.\]
(Thus, $c_1=-1$ and $c_0=0$.)

Since the identity matrix is invertible, the condition $c_0=0$ does not force the matrix $A$ to be non-invertible.

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