Elementary Row Matrix for Inverse of Elementary Row Operation is Inverse/Proof 2

Proof
We will demonstrate this for each of the $3$ types of elementary row operation.

In the below:
 * $e$ denotes a given elementary row operation
 * $\mathbf E$ denotes the elementary row matrix corresponding to $e$
 * $e'$ denotes the inverse of $e$
 * $\mathbf E'$ denotes the elementary row matrix corresponding to $e'$.

Let $n$ denote the order of $\mathbf E$ and $\mathbf E'$.

The strategy is to demonstrate that:
 * $\mathbf E \mathbf E' = \mathbf I$

where $\mathbf I$ denotes the unit matrix of order $n$.

Let $x_{i, j}$ and $y_{i, j}$ denote the elements of $\mathbf E$ and $\mathbf E'$ respectively at indices $\tuple {i, j}$.

Let $z_{i j}$ denote the element of $\mathbf E \mathbf E'$ at indices $\tuple {i, j}$.

$\text {ERO} 1$: Scalar Product of Row
Let $e$ be the elementary row operation:


 * $e := r_k \to \lambda r_k$

where $\lambda \ne 0$.

From Elementary Matrix corresponding to Elementary Row Operation, $\mathbf E$ is of the form:


 * $x_{a b} = \begin {cases} \delta_{a b} & : a \ne k \\ \lambda \cdot \delta_{a b} & : a = k \end {cases}$

where:
 * $\delta_{a b}$ is the Kronecker delta:
 * $\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$

From Existence of Inverse Elementary Row Operation: Scalar Product of Row, $e'$ is the elementary row operation:
 * $e' := r_k \to \dfrac 1 \lambda r_k$

From Elementary Matrix corresponding to Elementary Row Operation, $\mathbf E'$ is of the form:


 * $y_{a b} = \begin {cases} \delta_{a b} & : a \ne k \\ \dfrac 1 \lambda \cdot \delta_{a b} & : a = k \end {cases}$

By definition of matrix product:
 * $\displaystyle \forall a, b \in \set {1, 2, \ldots, n}: z_{a b} = \sum_{p \mathop = 1}^n x_{a p} y_{p b}$

Thus $z_{a b} \ne 0$ $a = p$ and $b = p$.

When $a \ne k$:
 * $x_{a a} = y_{a a} = 1$

and so:
 * $z_{a a} = 1 \times 1 = 1$

When $a = k$:
 * $x_{a a} = \lambda$, $y_{l b} = \dfrac 1 \lambda$

and so:
 * $z_{a a} = \lambda \times \dfrac 1 \lambda = 1$

and for all $z_{a b}$ where $a \ne b$:
 * $z_{a b} = 0$

That is:


 * $z_{a b} = \delta_{a b}$

and by definition:
 * $\mathbf E \mathbf E' = \mathbf I$

$\text {ERO} 2$: Add Scalar Product of Row to Another
Let $e$ be the elementary row operation:
 * $e := r_i \to r_i + \lambda r_j$

From Elementary Matrix corresponding to Elementary Row Operation, $\mathbf E$ is of the form:
 * $x_{a b} = \delta_{a b} + \lambda \cdot \delta_{a i} \cdot \delta_{j b}$

From Existence of Inverse Elementary Row Operation: Add Scalar Product of Row to Another, $e'$ is the elementary row operation:
 * $e' := r_i \to r_i - \lambda r_j$

From Elementary Matrix corresponding to Elementary Row Operation, $\mathbf E'$ is of the form:


 * $y_{a b} = \delta_{a b} - \lambda \cdot \delta_{a i} \cdot \delta_{j b}$

We have that:

But:
 * $\lambda \cdot \delta_{j b} \cdot \delta_{a i} - \lambda \cdot \delta_{j b} \cdot \delta_{a i} = 0$

So everything vanishes except $\delta_{a b}$, and so:


 * $z_{a b} = \delta_{a b}$

and by definition, again:
 * $\mathbf E \mathbf E' = \mathbf I$

$\text {ERO} 3$: Exchange Rows
Let $e$ be the elementary row operation:


 * $e := r_i \leftrightarrow r_j$

From Elementary Matrix corresponding to Elementary Row Operation, $\mathbf E$ is of the form:
 * $x_{a b} = \begin {cases} \delta_{a b} & : \text {if $a \ne i$ and $a \ne j$} \\ \delta_{j b} & : \text {if $a = i$} \\ \delta_{i b} & : \text {if $a = j$} \end {cases}$

From Existence of Inverse Elementary Row Operation: Exchange Rows, $e'$ is the elementary row operation:
 * $e' := r_i \leftrightarrow r_j$

From Elementary Matrix corresponding to Elementary Row Operation, $\mathbf E'$ is of the form:


 * $y_{a b} = \begin {cases} \delta_{a b} & : \text {if $a \ne i$ and $a \ne j$} \\ \delta_{j b} & : \text {if $a = i$} \\ \delta_{i b} & : \text {if $a = j$} \end {cases}$

By definition of matrix product:
 * $\displaystyle \forall a, b \in \set {1, 2, \ldots, n}: z_{a b} = \sum_{p \mathop = 1}^n x_{a p} y_{p b}$

When $a \ne i$ and $b \ne j$ we have:

When $a = i$ and $b \ne i$ we have:

When $a = j$ and $b \ne j$ we have:

When $a = b = i$ we have:

When $a = b = j$ we have:

Hence by definition, again:
 * $\mathbf E \mathbf E' = \mathbf I$

Thus in all cases:
 * $\mathbf E \mathbf E' = \mathbf I$

Hence the result.