Is there a good intuitive way to understand why matrix B is inverse of A when matrix A|I is turned into I|B

Question

I'm looking for some help with my intuition of basic matrix operations, specifically finding a matrix's inverse (as per my subject line). I have no problems with the steps. The basic row operations are relatively simple. I'd like to understand why/ how this solves the system of linear equations.

I know my question is asking more (or arguably less) than a concrete sequence of steps, a theorem, etc. But I think someone who understands linear algebra much better than I can get through to me better than my texts' treatment, which is little more than a worked example.

Thanks in advance..

Answer 1

Think of it directly - multiply the combined matrix $[A|I]$ by B on the left.

$$ B[A|I] = [BA|BI] = [I|B] $$ Where $BA=I$ because $B$ is the inverse of $A$. And of course, the basic matrix operations are just the equivalent of multiplying on the left by various matrices.

Answer 2

By doing the row operations that change $[A|I]$ to $[I|B]$ stores up the row operations necessary to "untangle" $A$. This compiles those operation into the matrix $B$. You can think of $B$ as an executable program that undoes $A$.

Answer 3

What you are doing is solving the linear systems given by each column of $I$ at the same time. Multipling the resulting columns (of $B$) by $A$ gives the original columns (of $I$). So the bunch of columns ($B$) as a matrix are the inverse of $A$.

If you are really interested in $A^{-1} \cdot C$, you can start with $[A | C]$ and you end up with $[I | A^{-1} C]$.

Answer 4

Take any vector $v$ that your invertible matrix $B$ transforms to $w$: $Bv=w$. Additionally to thinking of the inverse matrix $B^{-1}$ as the one such that $BB^{-1}=I$, keep the vectors in mind as this can provide a sublte chance for a bit more intuition.

If $B$ transforms $v$ to $w$, $B^{-1}$ intuitively transforms $w$ back to $v$, and we write $B^{-1}w=v$, $B^{-1}w=Iv$, and therefore $$B^{-1}w=(B^{-1}B)v.$$ The row operations luckily provide a reliable method for 'peeling of' the effect of $B$ on $v$ step by step. Why this method works you have probably already intuitively understood. So then after a number $k$ of 'peel-off' steps we necessarily reach $$(P_k...P_2P_1)w=(P_k...P_2P_1)Bv,$$ such that $(P_k...P_2P_1)B$ is the identity matrix. You have intuitively peeled $B$ and found a matrix that can recover $v$ from $w$.

A fun thing to notice is that anywhere you split the matrix product $(P_k...P_2P_1)B$, e.g: $(P_k..P_3)(P_2P_1B)$ you get two matrices that are the inverse of each other.