Elementary Row Matrix for Inverse of Elementary Row Operation is Inverse/Proof 1

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.

Let $\sim$ denote row equivalence.

Let $\mathbf I$ denote the unit matrix.

$\text {ERO} 1$: Scalar Product of Row

Let $e_1$ be the elementary row operation:

$e_1 := r_i \to \lambda r_i$

for an arbitrary row $r_i$ of $\mathbf I$.

Let $\mathbf E_1$ be the elementary row matrix created by applying $e_1$ to $\mathbf I$.

From Existence of Inverse Elementary Row Operation: Scalar Product of Row, the inverse $e_1'$ of $e_1$ is given by:

$e_1' := r_i \to \dfrac 1 \lambda r_i$

Thus applying $e_1'$ to $\mathbf E_1$ transforms $\mathbf E_1$ back to $\mathbf I$.

From Elementary Row Operations as Matrix Multiplications, for every elementary row operation there exists a corresponding elementary row matrix.

Let $\mathbf E_1'$ be the elementary row matrix created by applying $e_1'$ to $\mathbf I$.

Then:

$\mathbf E_1 \mathbf E_1' = \mathbf E_1' \mathbf E_1 = \mathbf I$

$\text {ERO} 2$: Add Scalar Product of Row to Another

Let $e_2$ be the elementary row operation:

$e_2 := r_i \to r_i + \lambda r_j$

for arbitrary row $r_i$ and $r_j$ of $\mathbf I$ such that $i \ne j$.

Let $\mathbf E_2$ be the elementary row matrix created by applying $e_2$ to $\mathbf I$.

From Existence of Inverse Elementary Row Operation: Add Scalar Product of Row to Another, the inverse $e_2'$ of $e_2$ is given by:

$e_2' := r_i \to r_i - \lambda r_j$

Thus applying $e_2'$ to $\mathbf E_2$ transforms $\mathbf E_2$ back to $\mathbf I$.

From Elementary Row Operations as Matrix Multiplications, for every elementary row operation there exists a corresponding elementary row matrix.

Let $\mathbf E_2'$ be the elementary row matrix created by applying $e_2'$ to $\mathbf I$.

Then:

$\mathbf E_2 \mathbf E_2' = \mathbf E_2' \mathbf E_2 = \mathbf I$

$\text {ERO} 3$: Exchange Rows

Let $e_3$ be the elementary row operation:

$e_3 := r_i \leftrightarrow r_j$

for arbitrary row $r_i$ and $r_j$ of $\mathbf I$ such that $i \ne j$.

Let $\mathbf E_3$ be the elementary row matrix created by applying $e_3$ to $\mathbf I$.

From Existence of Inverse Elementary Row Operation: Exchange Rows, the inverse $e_3'$ of $e_3$ is $e_3$ itself:

$e_3' := r_i \leftrightarrow r_j = e_3$

Thus applying $e_3$ to $\mathbf E_3$ transforms $\mathbf E_3$ back to $\mathbf I$.

$\mathbf E_3 \mathbf E_3 = \mathbf I$

Hence the result, from Proof by Cases.

$\blacksquare$

Sources

• 1995: John B. Fraleigh and Raymond A. Beauregard: Linear Algebra (3rd ed.) $\S 1.5$