Determinant of Elementary Row Matrix/Scale Row and Add

Theorem
Let $e_2$ be the elementary row operation $\text {ERO} 2$:

which is to operate on some arbitrary matrix space.

Let $\mathbf E_2$ be the elementary row matrix corresponding to $e_2$.

The determinant of $\mathbf E_2$ is:
 * $\map \det {\mathbf E_2} = 1$

Proof
By Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add, $\mathbf E_2$ is of the form:


 * $E_{a b} = \delta_{a b} + \lambda \cdot \delta_{a i} \cdot \delta_{j b}$

where:
 * $E_{a b}$ denotes the element of $\mathbf E$ whose indices are $\tuple {a, b}$
 * $\delta_{a b}$ is the Kronecker delta:
 * $\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$

Because $i \ne j$ it follows that:
 * if $a = i$ and $b = j$ then $a \ne b$

Hence when $a = b$ we have that:
 * $\delta_{a i} \cdot \delta_{j b} = 0$

Hence the diagonal elements of $\mathbf E_2$ are all equal to $1$.

We also have that $\delta_{a i} \cdot \delta_{j b} = 1$ $a = i$ and $b = j$.

Hence, all elements of $\mathbf E_2$ apart from the diagonal elements and $a_{i j}$ are equal to $0$.

Thus $\mathbf E_2$ is a triangular matrix (either upper or lower).

From Determinant of Triangular Matrix, $\map \det {\mathbf E_2}$ is equal to the product of all the diagonal elements of $\mathbf E_2$.

But as we have seen, these are all equal to $1$.

Hence the result.