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

## Theorem

Let $e$ be an elementary row operation.

Let $\mathbf E$ be the elementary row matrix corresponding to $e$.

Let $e'$ be the inverse of $e$.

Then the elementary row matrix corresponding to $e'$ is the inverse of $\mathbf E$.

## 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$} \end {cases}$
$e' := r_k \to \dfrac 1 \lambda r_k$

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 $z_{a b} \ne 0$ if and only if $a = p$ 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:

$\mathbf E \mathbf E' = \mathbf I$

$\Box$

### $\text {ERO} 2$: 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:

$x_{a b} = \delta_{a b} + \lambda \cdot \delta_{a i} \cdot \delta_{j b}$
$e' := r_i \to r_i - \lambda r_j$

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

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

 $\, \ds \forall a, b \in \set {1, 2, \ldots, n}: \,$ $\ds z_{a b}$ $=$ $\ds \sum_{p \mathop = 1}^n x_{a p} y_{p b}$ $\ds$ $=$ $\ds \sum_{p \mathop = 1}^n \paren {\delta_{a p} + \lambda \cdot \delta_{a i} \cdot \delta_{j p} } \cdot \paren {\delta_{p b} - \lambda \cdot \delta_{p i} \cdot \delta_{j b} }$ $\ds$ $=$ $\ds \sum_{p \mathop = 1}^n \paren {\delta_{a p} \cdot \delta_{p b} + \lambda \cdot \delta_{a i} \cdot \delta_{j p} \cdot \delta_{p b} - \lambda \cdot \delta_{p i} \cdot \delta_{j b} \cdot \delta_{a p} - \lambda \cdot \delta_{a i} \cdot \delta_{j p} \cdot \lambda \cdot \delta_{p i} \cdot \delta_{j b} }$

We have that:

 $\ds \sum_{p \mathop = 1}^n \delta_{a p} \cdot \delta_{p b}$ $=$ $\ds \delta_{a b}$ $\ds \sum_{p \mathop = 1}^n \lambda \cdot \delta_{p i} \cdot \delta_{j b} \cdot \delta_{a p}$ $=$ $\ds \lambda \cdot \delta_{a i} \cdot \delta_{j b}$ $\ds -\sum_{p \mathop = 1}^n \lambda \cdot \delta_{a i} \cdot \delta_{j p} \cdot \delta_{p b}$ $=$ $\ds -\lambda \cdot \delta_{j b} \cdot \delta_{a i}$ $\ds \sum_{p \mathop = 1}^n \lambda \cdot \delta_{a i} \cdot \delta_{j p} \cdot \lambda \cdot \delta_{p i} \cdot \delta_{j b}$ $=$ $\ds \lambda^2 \sum_{p \mathop = 1}^n \delta_{a i} \cdot \delta_{j p} \cdot \delta_{p i} \cdot \delta_{j b}$ $\ds$ $=$ $\ds \lambda^2 \delta_{a i} \cdot \delta_{j i} \cdot \delta_{j b}$ $\ds$ $=$ $\ds 0$ as $i \ne j$

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:

$\mathbf E \mathbf E' = \mathbf I$

$\Box$

### $\text {ERO} 3$: 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$} \end {cases}$
$e' := r_i \leftrightarrow r_j$

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:

 $\ds z_{a b}$ $=$ $\ds \sum_{p \mathop = 1}^n \delta_{a p} \delta_{p b}$ $\ds$ $=$ $\ds \delta_{a b}$

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

 $\ds z_{a b}$ $=$ $\ds \sum_{p \mathop = 1}^n \delta_{j p} \delta_{p b}$ $\ds$ $=$ $\ds 0$

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

 $\ds z_{a b}$ $=$ $\ds \sum_{p \mathop = 1}^n \delta_{i p} \delta_{p b}$ $\ds$ $=$ $\ds 0$

When $a = b = i$ we have:

 $\ds z_{i i}$ $=$ $\ds \sum_{p \mathop = 1}^n \delta_{j p} \delta_{p j}$ $\ds$ $=$ $\ds \delta_{j j}$ $\ds$ $=$ $\ds 1$

When $a = b = j$ we have:

 $\ds z_{j j}$ $=$ $\ds \sum_{p \mathop = 1}^n \delta_{i p} \delta_{p i}$ $\ds$ $=$ $\ds \delta_{i i}$ $\ds$ $=$ $\ds 1$

Hence by definition, again:

$\mathbf E \mathbf E' = \mathbf I$

$\Box$

Thus in all cases:

$\mathbf E \mathbf E' = \mathbf I$

Hence the result.

$\blacksquare$