if $(B^2+AB)B=3I$, how come B is invertible matrix

Question

I know that B is invertible but I don't understand why..

The question goes like that:

Let $A,B$ be $3x3$ matrix so that:

$B^2A = -2B^3$ and $B^3+AB^2 = 3I$

Prove that $A$ and $B$ invertible and express $A^{-1}$ and $B^{-1}$ with $B$

Two things I don't understand.

1. They answered the question like this:

   $$B^3+AB^2=3I \Rightarrow (B^2+AB)B=3I$$ Because 3I is invertible, so does $(B^2+AB)B$, and because (*) B is invertible

The (*) part is just a sentence that say's:

Let A,B be square matrix, if $AB=I$ then both are invertible and each of them are the inverse of the other.

The question: Why is $(B^2+AB)B$ invertible because of 3I? And how did they use the sentence above if they didn't say that $(B^2+AB)$ is invertible instead of $(B^2+AB)B$?

($(B^2+AB)B$ is a matrix by itself..)

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2. The second part of the question is that they just said:

   (After saying B is invertible) Because of that $-2B^3$ is invertible as well and because of that $B^2A$ is invertible.

The question: How did they came to the conclusion that both $-2B^3$ and $B^2A$ is invertible based on $B$?

Thanks in advance! If something is not understood please let me know so I can edit the question.

Answer 1

Consider the matrix $\frac13(B^2+AB)$.

Answer 2

These remarks are specific to matrix theory: $AB=I$ implies that the linear application associated to $A$: $$\begin{aligned} k^n & \longrightarrow k^n\\ X & \longmapsto AX \end{aligned}$$ is surjective, so that it is an isomorphism, thus implying that $A$ is invertible.

In the same way you prove that the linear application associated to $B$ is injective so that $B$ is also invertible. Each time you use the rank theorem (this is why it is specific to matrix theory and linear algebra in finite dimension).

Since $A$ and $B$ are invertible and $AB=I$ you get $A=B^{-1}$.

So basically in matrix algebra, a matrix admits a left inverse $\iff$ it admits a right inverse. This is not true in general.

Thus if $M=UV$ is invertible of inverse $M^{-1}=N$ then you have $I=NM=(NU)V$ and $I=MN=U(VN)$ so both $U$ and $V$ are invertible.

These remarks should help