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.

Question

Prove or disprove:

Let $A$ be the matrix:

$$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{pmatrix} $$

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?

Answer 1

Hint:

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

Secondly, write $$C=\begin{pmatrix} c_{11}&c_{12}&c_{13}\\c_{21}&c_{22}&c_{23}\\c_{31}&c_{32}&c_{33}\end{pmatrix}$$ and try multiplying this by the matrix $A$. What does this give you? From here, how can you construct $B$?

Answer 2

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 $\det B=1$.