"When you perform elementary row operations on the augmented matrix $[A|I_n]$, you are essentially applying the same operations simultaneously to both $A$ and $I_n$."

"From an algebraic perspective, this means you are left-multiplying the entire augmented matrix by the same sequence of elementary matrices:"

"According to the definition of matrix multiplication (specifically, how it applies to block matrices), this is equivalent to:"

How to prove this step:

This step is crucial for understanding how the augmented matrix method for finding the inverse works. It is not a direct application of the distributive property (as matrix multiplication over matrices does not distribute in this sense for blocks), but rather based on the definition of block matrix multiplication.

Let's first prove the case for a single elementary matrix $E$.

Proof of $E[A|I_n]=[EA|E{I}_n]$:

Let $A$ be an $n\times n$ matrix, and $I_n$ be the $n\times n$ identity matrix.
The augmented matrix $[A|I_n]$ is an $n\times (n+n)$, i.e., $n\times (2n)$ matrix.
Let $E$ be an $n\times n$ elementary matrix (representing a single row operation).

We can consider the augmented matrix $[A|I_n]$ as a single $n\times (2n)$ matrix, whose columns are split into two blocks:

• The first $n$ columns form matrix $A$.
• The second $n$ columns form matrix $I_n$.

Let $C=[A|I_n]$. We can write $C$ in terms of its column vectors:
$C=[{\mathbf{c}}_1{\mathbf{c}}_2\cdots {\mathbf{c}}_n|{\mathbf{c}}_{n+1}\cdots {\mathbf{c}}_{2n}]$
Where ${\mathbf{c}}_1,\dots ,{\mathbf{c}}_n$ are the column vectors of $A$, and ${\mathbf{c}}_{n+1},\dots ,{\mathbf{c}}_{2n}$ are the column vectors of $I_n$.

Recall the fundamental definition of matrix multiplication: when a matrix $E$ left-multiplies another matrix $C$, each column of the resulting matrix $EC$ is $E$ multiplied by the corresponding column of $C$.
So,

Now, let's examine the blocks of the resulting matrix $EC$:

1. The first $n$ columns of $EC$: These are $E{\mathbf{c}}_1,E{\mathbf{c}}_2,\dots ,E{\mathbf{c}}_n$. By the definition of matrix multiplication, these are precisely the column vectors that form the matrix product $EA$.
   Therefore, the left block of the result is $EA$.

2. The last $n$ columns of $EC$: These are $E{\mathbf{c}}_{n+1},E{\mathbf{c}}_{n+2},\dots ,E{\mathbf{c}}_{2n}$. Similarly, these are precisely the column vectors that form the matrix product $E{I}_n$.
   Therefore, the right block of the result is $E{I}_n$.

Combining these two observations, we have demonstrated that:

Generalization to a Sequence of Elementary Matrices:

This principle extends straightforwardly to a sequence of elementary matrices $E_1,E_2,\dots ,E_k$. We can apply the proven principle iteratively:

1. Start with $E_1$: $E_1[A|I_n]=[E_{1}A|E_1{I}_n]$
2. Apply $E_2$ to the result: $E_2(E_1[A|I_n])=E_2[E_{1}A|E_1{I}_n]=[E_2(E_{1}A)|E_2(E_1{I}_n)]$
3. Continue this process for all $k$ elementary matrices:$$(E_k\cdots E_2{E}_1)[A|I_n]=[(E_k\cdots E_2{E}_1)A|(E_k\cdots E_2{E}_1)I_n]$$

This is why, when you perform a sequence of elementary row operations on the augmented matrix $[A|I_n]$, the left block $A$ is transformed into the product $(E_k\cdots E_1)A$, and the right block $I_n$ is simultaneously transformed into $(E_k\cdots E_1)I_n$. If the process successfully transforms $A$ into $I_n$ (i.e., $(E_k\cdots E_1)A=I_n$), then the product $(E_k\cdots E_1)$ must be $A^{-1}$, and the right side becomes $A^{-1}{I}_n=A^{-1}$.