Determinant of Elementary Row Matrix/Scale Row

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Theorem

Let $e_1$ be the elementary row operation $\text {ERO} 1$:

\((\text {ERO} 1)\)   $:$   \(\displaystyle r_k \to \lambda r_k \)    For some $\lambda \ne 0$, multiply row $k$ by $\lambda$             

which is to operate on some arbitrary matrix space.


Let $\mathbf E_1$ be the elementary row matrix corresponding to $e_1$.

The determinant of $\mathbf E_1$ is:

$\map \det {\mathbf E_1} = \lambda$


Proof

By Elementary Matrix corresponding to Elementary Row Operation: Scale Row, the elementary row matrix corresponding to $e_1$ is of the form:

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

where:

$E_{a b}$ denotes the element of $\mathbf E_1$ 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}$

Thus when $a \ne b$, $E_{a b} = 0$.

This means that $\mathbf E_1$ is a diagonal matrix.


\(\ds \displaystyle \map \det {\mathbf E_1}\) \(=\) \(\ds \prod_i E_{i i}\) Determinant of Diagonal Matrix where the index variable $i$ ranges over the order of $\mathbf E_1$
\(\ds \) \(=\) \(\ds \prod_i \paren {\begin {cases} 1 & : i \ne k \\ \lambda & : a = k \end{cases} }\)
\(\ds \) \(=\) \(\ds \prod_{i \mathop \ne k} 1 \times \prod_{i \mathop = k} \lambda\)
\(\ds \) \(=\) \(\ds 1 \times \lambda\)
\(\ds \) \(=\) \(\ds \lambda\)

$\blacksquare$


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