prove or disprove invertible matrix with given equations Given a non-scalar matrix $A$ in size $n\times n$ over $\mathbb{R}$ that maintains the following equation 
$$A^2 + 2A = 3I$$
given matrix $B$ in size $n\times n$ too $$B = A^2 + A- 6I$$
Is $B$ an invertible matrix?
 A: An alternative version to the other excellent answers by Quark and user1551. 
Note first, as the others did, that $(A + 3 I) (A - I) = 0$, and $B = (A - 2 I)(A + 3 I)$.
Then
$$
B (A - I) = 0.
$$
So if $A \ne I$, then $B$ is definitely not invertible.
If $A = I$, as noted by Quark, then $B = -4 I$ is invertible. 
But Tharsis (thanks!) made me notice that $A$ is non-scalar by assumption, so the case $A = I$ is excluded (although easy to deal with).
A: It depends, If you factorize the first equation you get,
$$(A-I)*(A+3I)=0$$
This implies$\; \det(A-I)\;\; or\;\; \det(A-3I)\;$ must be zero.
Now factorizing B gives you (A-2I)*(A+3I)
So $\det(B) = \det(A-2I)*\det(A+3I)$
If $\det(A-3I)$ is zero from the first condition, then $\det(B)$ is also zero, so B is not invertible.
But, if $\det(A-I)$ is zero and not $\det(A-3I)$, I don't think anything can be said 
(not very sure, but I can give a example: If you take A=I, the first equation is satisfied, and B comes out to be -4I which is invertible...)
A: Hint: The first equation implies that $A^2+2A-3I=(A-I)(A+3I)=0$. Hence the minimal polynomial $m_A(x)$ of $A$ divides $(x-1)(x+3)$. What happens if $x+3$ is not a factor of $m_A(x)$?
