Proof.

1.

Write \(e\) for the row operation and \(E=e(I)\) for the elementary matrix it produces. The key observation is that row \(i\) of a product \(MA\) depends only on row \(i\) of \(M\): explicitly, \((MA)_{i\bullet }=\sum _{j}m_{ij}A_{j\bullet }\), a combination of the rows of \(A\) with the coefficients taken from row \(i\) of \(M\).

Now each of the three elementary operations changes the rows of \(I\) in a particular way, and applying \(E\) to \(A\) reproduces exactly that change on the rows of \(A\). For instance, if \(e\) is \(r_i\rightarrow r_i+kr_j\), then row \(i\) of \(E\) is \(\text {(row $i$ of $I$)}+k\,\text {(row $j$ of $I$)}\), so row \(i\) of \(EA\) is \(A_{i\bullet }+kA_{j\bullet }\), while every other row of \(E\) is a row of \(I\) and leaves the corresponding row of \(A\) unchanged. That is precisely \(e(A)\). The other two operations are checked the same way. Hence \(B=e(A)=EA\).

2.

Every elementary row operation can be undone by an elementary row operation of the same type: \[r_i\leftrightarrow r_j \ \text { by }\ r_i\leftrightarrow r_j,\qquad r_i\rightarrow kr_i \ \text { by }\ r_i\rightarrow \tfrac {1}{k}r_i \ (k\neq 0), \qquad r_i\rightarrow r_i+kr_j \ \text { by }\ r_i\rightarrow r_i-kr_j.\] Let \(e'\) be the operation undoing \(e\), and \(E'=e'(I)\) the corresponding elementary matrix. Applying part (1) twice, \(E'E=e'(E)=e'(e(I))=I\) and similarly \(EE'=I\). So \(E\) is invertible with \(E^{-1}=E'\), which is elementary.

3.

Induction on the number of operations. If \(B\) is obtained from \(A\) by operations \(e_1,\dots ,e_k\) applied in that order, let \(A_0=A\) and \(A_t=e_t(A_{t-1})\), so \(B=A_k\). By part (1), \(A_t=E_tA_{t-1}\) for each \(t\). Composing, \[B=A_k=E_kA_{k-1}=E_kE_{k-1}A_{k-2}=\cdots =E_kE_{k-1}\cdots E_1A.\]

4.

By definition \(R\) is obtained from \(A\) by a finite sequence of elementary row operations, so part (3) gives \(R=E_kE_{k-1}\cdots E_1A\). By part (2) each \(E_t\) is invertible, so we may multiply on the left successively by \(E_k^{-1},E_{k-1}^{-1},\dots \) to obtain \[A=E_1^{-1}E_2^{-1}\cdots E_k^{-1}R,\] and each \(E_t^{-1}\) is elementary, again by part (2). Renaming these inverses \(E_1,\dots ,E_k\) gives the statement.

Solution. \begin {align*} & \begin {pmatrix} -1&1&2&&|&1&0&0\\ 1&1&0&&|&0&1&0\\0&-2&-1&&|&0&0&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2+r_1\\ \longrightarrow \\ r_3\rightarrow -r_3\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&1&2&&|&1&0&0\\0&2&2&&|&1&1&0\\0&2&1&&|&0&0&-1\\ \end {pmatrix}\\\\ &\hspace {0.3cm} \begin {matrix} r_3\rightarrow r_3-r_2\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&1&2&&|&1&0&0\\0&2&2&&|&1&1&0\\0&0&-1&&|&-1&-1&-1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_2\rightarrow \frac {1}{2}r_2\\ \longrightarrow \\ r_3\rightarrow -r_3\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&1&2&&|&1&0&0\\0&1&1&&|&1/2&1/2&0\\0&0&1&&|&1&1&1\\ \end {pmatrix}\\\\ &\hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2-r_3\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&1&2&&|&1&0&0\\0&1&0&&|&-1/2&-1/2&-1\\0&0&1&&|&1&1&1\\ \end {pmatrix}\\\\ \hspace {0.3cm}& \begin {matrix} r_1\rightarrow -r_1\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&-1&-2&&|&-1&0&0\\0&1&0&&|&-1/2&-1/2&-1\\0&0&1&&|&1&1&1\\ \end {pmatrix}\\\\ &\hspace {0.3cm} \begin {matrix} r_1\rightarrow r_1+r_2\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&-2&&|&-3/2&-1/2&-1\\0&1&0&&|&-1/2&-1/2&-1\\0&0&1&&|&1&1&1\\ \end {pmatrix}\\\\ &\hspace {0.3cm} \begin {matrix} r_1\rightarrow r_1+2r_3\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&0&&|&1/2&3/2&1\\0&1&0&&|&-1/2&-1/2&-1\\0&0&1&&|&1&1&1\\ \end {pmatrix}\\ \end {align*}

\[\therefore \hspace {0.3cm} A^{-1}= \begin {pmatrix} 1/2&3/2&1\\-1/2&-1/2&-1\\1&1&1\\ \end {pmatrix} \]