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 $z_{a b} \ne 0$ {{iff}} $a = p$ 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: :$\mathbf E \mathbf E' = \mathbf I$ {{qed|lemma}} === $\text {ERO} 2$: 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: :$y_{a b} = \delta_{a b} - \lambda \cdot \delta_{a i} \cdot \delta_{j b}$ {{begin-eqn}} {{eqn | lo= \forall a, b \in \set {1, 2, \ldots, n}:       | l = z_{a b}       | r = \sum_{p \mathop = 1}^n x_{a p} y_{p b}       | c =  }} {{eqn | r = \sum_{p \mathop = 1}^n \paren {\delta_{a p} + \lambda \cdot \delta_{a i} \cdot \delta_{j p} } \cdot \paren {\delta_{p b} - \lambda \cdot \delta_{p i} \cdot \delta_{j b} }       | c =  }} {{eqn | r = \sum_{p \mathop = 1}^n \paren {\delta_{a p} \cdot \delta_{p b} + \lambda \cdot \delta_{a i} \cdot \delta_{j p} \cdot \delta_{p b} - \lambda \cdot \delta_{p i} \cdot \delta_{j b} \cdot \delta_{a p} - \lambda \cdot \delta_{a i} \cdot \delta_{j p} \cdot \lambda \cdot \delta_{p i} \cdot \delta_{j b} }       | c =  }} {{end-eqn}} We have that: {{begin-eqn}} {{eqn | l = \sum_{p \mathop = 1}^n \delta_{a p} \cdot \delta_{p b}       | r = \delta_{a b}       | c =  }} {{eqn | l = \sum_{p \mathop = 1}^n \lambda \cdot \delta_{p i} \cdot \delta_{j b} \cdot \delta_{a p}       | r = \lambda \cdot \delta_{a i} \cdot \delta_{j b}       | c =  }} {{eqn | l = -\sum_{p \mathop = 1}^n \lambda \cdot \delta_{a i} \cdot \delta_{j p} \cdot \delta_{p b}       | r = -\lambda \cdot \delta_{j b} \cdot \delta_{a i}       | c =  }} {{eqn | l = \sum_{p \mathop = 1}^n \lambda \cdot \delta_{a i} \cdot \delta_{j p} \cdot \lambda \cdot \delta_{p i} \cdot \delta_{j b}       | r = \lambda^2 \sum_{p \mathop = 1}^n \delta_{a i} \cdot \delta_{j p} \cdot \delta_{p i} \cdot \delta_{j b}       | c =  }} {{eqn | r = \lambda^2 \delta_{a i} \cdot \delta_{j i} \cdot \delta_{j b}       | c =  }} {{eqn | r = 0       | c = as $i \ne j$ }} {{end-eqn}} 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: :$\mathbf E \mathbf E' = \mathbf I$ {{qed|lemma}} === $\text {ERO} 3$: 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: {{begin-eqn}} {{eqn | l = z_{a b}       | r = \sum_{p \mathop = 1}^n \delta_{a p} \delta_{p b}       | c =  }} {{eqn | r = \delta_{a b}       | c =  }} {{end-eqn}} When $a = i$ and $b \ne i$ we have: {{begin-eqn}} {{eqn | l = z_{a b}       | r = \sum_{p \mathop = 1}^n \delta_{j p} \delta_{p b}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} When $a = j$ and $b \ne j$ we have: {{begin-eqn}} {{eqn | l = z_{a b}       | r = \sum_{p \mathop = 1}^n \delta_{i p} \delta_{p b}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} When $a = b = i$ we have: {{begin-eqn}} {{eqn | l = z_{i i}       | r = \sum_{p \mathop = 1}^n \delta_{j p} \delta_{p j}       | c =  }} {{eqn | r = \delta_{j j}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} When $a = b = j$ we have: {{begin-eqn}} {{eqn | l = z_{j j}       | r = \sum_{p \mathop = 1}^n \delta_{i p} \delta_{p i}       | c =  }} {{eqn | r = \delta_{i i}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} Hence by definition, again: :$\mathbf E \mathbf E' = \mathbf I$ {{qed|lemma}} Thus in all cases: :$\mathbf E \mathbf E' = \mathbf I$ Hence the result. {{qed}}