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
A: Hint: Using Jordan normal form for $A$.
A: We have 
$$B=-3I-A$$
and we can see easily that the spectrum of $B$ is
$$\mathrm{sp}(B)=\{-3-\lambda_1,-3-\lambda_2\}=\{-4,0\}$$
so $B$ is not invertible.
Added Since $A^2+2A-3I=(A+3I)(A-I)=0$ and since $A\not= \lambda I$ then the minimal polynomial of is $x^2+2x-3$ which is  a product of distinct linear factors over $\mathbb{R}$ then $A$ is diagonalisable and $-3$ and $1$ are it's eigenvalues hence there's $P\in\mathrm{GL}_n(\mathbb{R})$ s.t. $A=PDP^{-1}$ where $D$ is a diagonal matrix and then we have $$B=-3I-A=-3PP^{-1}-PDP^{-1}=P(-3I-D)P^{-1}$$
so we can deduce  the spectrum of $B$ from the spectrum of $A$ as explained above.
A: $A$ is a root of the polynomial $x^2 + 2x - 3=(x-1)(x+3).$ 
Since $A$ is not of the form $\lambda I,$ this is the minimal polynomial of $A.$ 
Consequently $-3$ is an eigenvalue of $A.$
$$\color{red}{B=A^2 + A - 6I=A^2+2A-3I-A-3I=-A-3I}$$
$-3$ is an eigenvalue of $A\implies\det(A+3I)=0\implies\det(-B)=0\implies\det B=0.$
