Prove that every matrix $\in\mathbb{R}^{3\times3}$ with determinant equal 6 can be written as $AB$, when $|B|=1$ and $A$ is the given matrix. Prove or disprove:
  
  Let $A$ be the matrix:
  
  $$
    \begin{pmatrix}
    1 &amp; 0 &amp; 0 \\
    0 &amp; 2 &amp; 0 \\
    0 &amp; 0 &amp; 3 \\
    \end{pmatrix}
$$
  
  Then every matrix $\in\mathbb{R}^{3\times3}$ with determinant equal to $6$ can be written as $AB$, when $|B|=1$ and $A$ is the given matrix.


My way so far:

Let $C\in\mathbb{R}^{3\times3}$ such as $|C|=6$.
Assuming $C=AB$ then $A^{-1}C=B$. And indeed $|B|=1$.

But does it prove that $C$ can be every real matrix with determinant equal to $6$? I mean how can I even make this $(C=AB)$ assumption, if that's what I need to prove?
Hint:

We have that $$A=\begin{pmatrix}
    1 &amp; 0 &amp; 0 \\
    0 &amp; 2 &amp; 0 \\
    0 &amp; 0 &amp; 3 \\
    \end{pmatrix}$$
Suppose $B$ is some matrix with $\det B=1$. Then what is $\det AB$?

Secondly, write $$C=\begin{pmatrix} c_{11}&amp;c_{12}&amp;c_{13}\\c_{21}&amp;c_{22}&amp;c_{23}\\c_{31}&amp;c_{32}&amp;c_{33}\end{pmatrix}$$ and try multiplying this by the matrix $A$. What does this give you? From here, how can you construct $B$?
Too easy. Let $C\in \mathbb R^{3\times 3}$ with $\det C=6$. Then $C=AB$, where  $B=A^{-1}C$. We have $\det C=\det A\det B$, hence $\det B=1$.