Every elementary row operation has a unique inverse
Dependencies:
Let $A$ be a matrix over a field. Then every elementary row operation on $A$ has a unique inverse which is also an elementary row operation.
This means that if applying $R$ on $A$ gives $B$, then applying the inverse of $R$ (denoted as $R^{-1}$) on $B$ gives $A$.
- Inverse of $\langle i \rangle \leftarrow c\langle i \rangle$ is $\langle i \rangle \leftarrow (c^{-1}) \langle i \rangle$.
- Inverse of $\langle i \rangle \leftarrow \langle i \rangle + c\langle j \rangle$ $(i \neq j)$ is $\langle i \rangle \leftarrow \langle i \rangle - c\langle j \rangle$.
- $\langle i \rangle \leftrightarrow \langle j \rangle$ is its own inverse.
By the above means of finding inverse, it can be seen that the inverse of $R^{-1}$ is $R$.
Proof
Suppose $B$ is obtained from $A$ by applying the operation $R$. Let $a_i$ be the $i^{\textrm{th}}$ row of $A$ and $b_i$ be the $i^{\textrm{th}}$ row of $B$.
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$\langle p \rangle \leftarrow c\langle p \rangle$: \[ b_i = \begin{cases} a_i & i \neq p \\ ca_i & i = p \end{cases} \implies a_i = \begin{cases} b_i & i \neq p \\ (c^{-1})b_i & i = p \end{cases} \]
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$\langle p \rangle \leftarrow \langle p \rangle + c\langle q \rangle$ where $p \neq q$: \[ b_i = \begin{cases} a_i & i \neq p \\ a_p + ca_q & i = p \end{cases} \implies a_i = \begin{cases} b_i & i \neq p \\ b_p - ca_q & i = p \end{cases} = \begin{cases} b_i & i \neq p \\ b_p - cb_q & i = p \end{cases} \]
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$\langle p \rangle \leftrightarrow \langle q \rangle$: \[ b_i = \begin{cases} a_i & i \neq p \wedge i \neq q \\ a_q & i = p \\ a_p & i = q \end{cases} \implies a_i = \begin{cases} b_i & i \neq p \wedge i \neq q \\ b_p & i = q \\ b_q & i = p \end{cases} \]
Dependency for:
- Row equivalence of matrices
- Row equivalent matrices have the same row space
- Elementary row operation is matrix pre-multiplication
Info:
- Depth: 4
- Number of transitive dependencies: 6