How to show $\text{rref }[\left.A\right|AB]=[\left.I_n\right|B]$?

Question

For invertible $A^{n\times n}, B^{n\times n}$, how do I show that $\text{rref }[\left.A\right|AB]=[\left.I_n\right|B]?$

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Tentatively: $\text{rref }[\left.A\right|AB]=[\left.I_n\right|ABA^{-1}],\text{where rref }[\left.A\right|I_n]=[\left.I_n\right|A^{-1}]\\\text{and }ABA^{-1}=AA^{-1}B=I_nB\\\Rightarrow\text{rref }[\left.A\right|AB]=[\left.I_n\right|ABA^{-1}]=[\left.I_n\right|I_nB]=[\left.B^{-1}\right|I_n^2]=[\left.B^{-1}\right|I_n]=[\left.I_n\right|B]\\\text{ where }\det A, \det B\neq0\ $

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I'm not confident about the third line, specifically the difference between $\text{rref }[\left.A\right|AB]$ and $[\left.A\right|AB]$ with augmented matrices. Can I still move things around and take inverses without $\text{rref}$ on the left?

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I'm given the following hint:

"The left part of $\text{rref }[\left.A\right|AB]$ is $\text{rref }(A)=I_n$. Write $\text{rref }[\left.A\right|AB]=[I_n|M]$; we have to show that $M=B$. To demonstrate this, note that the columns of matrix$$\begin{bmatrix}B\\-I_n\end{bmatrix}$$are in the kernel of $[\left.A\right|AB]$ and therefore in the kernel of $[\left.I_n\right|M]$."

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I didn't use the hint at all in formulating my proof, which raises doubts. Is the proof along the lines of $$\begin{bmatrix}B\\-I_n\end{bmatrix}\in\text{ker }([\left.A\right|AB],\begin{bmatrix}B\\-I_n\end{bmatrix}\in\text{ker }([\left.I_n\right|B])\\\Rightarrow\text{ker }[\left.I_n\right|B]\subseteq\text{ker }([\left.A\right|AB]\\\Rightarrow\text{rref }[\left.A\right|AB]=[\left.I_n\right|B]\,?$$

Answer 1

First proof: Write $A$ as a product of Elementary matrices, and use the fact that if you make the corresponding row operation you cancel an elementary matrix from left.

Second proof Let $b_j$ be the $j$-th column of $B$. Then the $jth$ column of $AB$ is $Ab_j$.

Now, when you row reduce $[A|Ab_j]$ you are solving the system of equations $$Ax=Ab_j$$ since $A$ is invertible, it has unique solution, which implies that $$rref[A|Ab_j] = [I_n|b_j]$$ for each $j$. Conclude from here that $$ rref[A|Ab_1 Ab_2 ... Ab_k] = [I_n|b_1 b_2 ... b_k]=[I | B]$$

Answer 2

suppose $[A|AB]$ is row equivalent to $[I|C].$ we are given $\begin{bmatrix}B\\-I\end{bmatrix}$ is in the null space of $[A|AB].$ therefore is in the null space of $[I|C].$ that implies $B-C = 0$ that is $$C = B. $$