CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Row reducing $[A ∣ I]$ yields $[I ∣ A^{- 1}]$ exactly when $A$ is invertible

Statement

For $A \in M_{n} (F)$ , row reduction of $[A ∣ I_{n}]$ can reach $[I_{n} ∣ B]$ if and only if $A$ is invertible. Whenever it does, $B = A^{- 1}$ .

Facts & Assumptions

Given: A square matrix $A$ over a field and a finite row reduction of $[A ∣ I_{n}]$ .

[L1]

$A$ is invertible exactly when its RREF is $I_{n}$ (Invertible matrix theorem: invertibility, full pivot rank, RREF $I$ , trivial nullspace and unique solvability are equivalent).

[L2]

Each row operation is simultaneous left multiplication on both blocks by its elementary matrix (Applying an elementary row operation is left multiplication by its elementary matrix).

[L3]

A whole reduction is encoded by one product $E$ of elementary matrices (A finite row reduction from $A$ to $B$ is encoded by $B = E_{r} \dots E_{1} A$ ).

[L4]

An augmented matrix records simultaneous row operations on all its columns (Matrix equation $A x = b$ , its solution set, consistency, homogeneous systems and the augmented matrix $[A ∣ b]$ ).

Proof

technique · constructive

1.1

Let $E$ be the product encoding the row operations. The two output blocks are $E A$ and $E I_{n} = E$ .

L2L3L4construct

2.1

If the output is $[I_{n} ∣ B]$ , then $E A = I_{n}$ and $B = E$ . By [L1], $A$ is invertible, and multiplying $E A = I_{n}$ by $A^{- 1}$ gives $E = A^{- 1} = B$ .

step 1.1L1algebra

3.1

Conversely, if $A$ is invertible, [L1] supplies a reduction of $A$ to $I_{n}$ ; applying it to $[A ∣ I_{n}]$ gives $[I_{n} ∣ A^{- 1}]$ by step 2.1. If the left block cannot reach $I_{n}$ , [L1] says no inverse exists.

step 2.1L1discharge-construct∎

Depends on

Used by

A 3×3 inverse computed by row reducing [A∣ I] Example

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

D. Margalit and J. Rabinoff, Interactive Linear Algebra, §3.6 (standard reference, not scraped)

Row reducing [A∣I][A∣I] yields [I∣A−1][I∣A−1] exactly when AA is invertible