I am reading this passage in my linear algebra textbook:

To prove the theorem in one direction, assume A is invertible. From Theorem 2.11 you know that the system of linear equations represented by Ax = O has only the trivial solution. But this implies that the augmented matrix [A O] can be rewritten in the form [I O] (using elementary row operations corresponding to $E_1$, $E_2$, . . . , and $E_k$). So, $E_k$... $E_3E_2E_1A = I$ and it follows that $A=E_1^{-1}E_2^{-1}E_3^{-1}...E_k^{-1}$. A can be written as the product of elementary matrices.

Few questions:

What is the shorthand [A 0]. I only have seen this notation when adjoining matrices (like adjoining the identity matrix to A for the purposes of finding the inverse).

Why does this imply that the augmented matrix [A O] can be rewritten in the form [I O]?

Can all invertible matrices be converted to identity matrices using row operations?

Theorem 11 is this btw:

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  1. the short hand is what you have seen .
  2. Refer to Thm. 2.11 [maybe you could add this theorem in your post as a reference for us].
  3. Yes.
  • $\begingroup$ edited. But when I've seen adjoining it didn't have an equal sign? In this context, doesn't [A 0] mean matrix A = 0? $\endgroup$ – Jwan622 Aug 3 '18 at 1:04
  • $\begingroup$ why is 3 true?? $\endgroup$ – Jwan622 Aug 3 '18 at 1:04
  • $\begingroup$ @Jwan622 No. This is only a notation, where $[\boldsymbol A\; \boldsymbol O]$ is a matrix formed by putting $\boldsymbol A$ and $\boldsymbol O$ horizontally [or say that $\boldsymbol O$ is put at the right of $\boldsymbol A$]. $\endgroup$ – xbh Aug 3 '18 at 1:16
  • $\begingroup$ So no equals sign right? @xbh $\endgroup$ – Jwan622 Aug 3 '18 at 1:17
  • $\begingroup$ @Jwan622 3 is true according to the proof. Think about Gauss elimination. $\endgroup$ – xbh Aug 3 '18 at 1:17

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