Q:

Is $B = A^2 + A - 6I$ invertible when $A^2 + 2A = 3I$?

Given:
$$A \in M_{nxn} (\mathbb C), \; A \neq \lambda I, \; A^2 + 2A = 3I$$
Now we define:
$$B = A^2 + A - 6I$$
The question:

Is $B$ inversable?

Now, what I did is this:

$A^2 + 2A = 3I \rightarrow \lambda^2v + 2\lambda v = 3v \rightarrow \lambda_1 = 1, \lambda_2 = -3$

Is what I suggested correct? I know that if so, I just do the same to B and calculate the determinant.

A:

Note that
$$
\begin{align}
(A-I)B
&=(A-I)(A^2+A-6I)\\
&=(A-2I)(A^2+2A-3I)\\[4pt]
&=0
\end{align}
$$
Thus, unless $A=I$, and therefore, $B=-4I$, $(A-I)B=0$ implies that $B$ is not invertible.
Clarification: Suppose that $B^{-1}$ exists, then
$$
\begin{align}
A-I
&=(A-I)BB^{-1}\\
&=0B^{-1}\\[6pt]
&=0
\end{align}
$$
Thus, if $B^{-1}$ exists, then $A=I$. This is the contapositive of "if $A\ne I$, then $B$ is not invertible".

A bit of explanation
I used the Euclid-Wallis Algorithm to try and write $(A^2+A-6I)x+(A^2+2A-3I)y=I$ to compute an inverse for $A^2+A-6I\bmod A^2+2A-3I$:
$$
\begin{array}{c|c}
&&1&-A+I\\\hline
1&0&1&A-I\\
0&1&-1&-A+2I\\
A^2+A-6I&A^2+2A-3I&-A-3I&0
\end{array}
$$
Unfortunately, this showed that $A+3I$ was the GCD of $A^2+A-6I$ and $A^2+2A-3I$. However, this did show that
$$
(A-I)(A^2+A-6I)-(A-2I)(A^2+2A-3I)=0
$$
which was used in the answer above.

A:

$B=A^{2}+A-6I$ 
$B=(A+3I)(A-2I)$ 
$det(B)=det(A+3I)det(A-2I)$
now attention :
$A^{2}+2A=3I$
$A^2+2A-3I=0$    $\to$  $det(A+3I)det(A-I)=0$
$A^2+2A-3I-4A+4I=-4A+4I$
$(A-I)^2=-4(A-I)$ $\to$ $if det(A-I)=-4$ then B willnot be inversable
in other word if -3 or 2 be eigen value of A then B will not be inversable

