First note that \(T\) takes a \(3\)-tuple and outpus a \(2\)-tuple. Thus \(A\) must be \(2 \times 3\). We need the first column of \(A\) to contain the coefficients of \(u\), the second column to contain the coefficients of \(v\), and the third column to contain the coefficients of \(w\). Thus \(A = \begin{bmatrix} 1 & 1 & -1 \\ 2 & 0 & 1\end{bmatrix}\).

Note that \begin{eqnarray*} S\left(T\left(\begin{bmatrix} u\\ v\\ w\end{bmatrix}\right)\right) & = & S\left( \begin{bmatrix} u - v + w \\ 3u + v - w\end{bmatrix}\right) \\ & = & \begin{bmatrix} 2(u - v + w)+ (3u + v - w) \\ -(u-v+w) + 2(3u + v- w)\end{bmatrix} \\ & = & \begin{bmatrix} 5u - v + w \\ 5u+3v-3w\end{bmatrix} \\ & = & \begin{bmatrix} 5 & -1 & 1\\ 5 & 3 & -3 \end{bmatrix} \begin{bmatrix} u\\ v \\ w\end{bmatrix} \end{eqnarray*} Hence, the required matrix is \(\begin{bmatrix} 5 & -1 & 1\\ 5 & 3 & -3 \end{bmatrix} \).

\begin{eqnarray*} \begin{bmatrix} 1 & 2 \\ -1 & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix} 3 & 1 & 2 \\ -1 & 0 & 4\end{bmatrix} & = & \begin{bmatrix} 1\cdot 3 + 2 \cdot (-1) & 1 \cdot 1 + 2\cdot 0 & 1\cdot 2 + 2\cdot 4 \\ (-1)\cdot 3 + 1 \cdot (-1) & (-1) \cdot 1 + 1\cdot 0 & (-1)\cdot 2 + 1\cdot 4 \\ 1\cdot 3 + 0 \cdot (-1) & 1 \cdot 1 + 0\cdot 0 & 1\cdot 2 + 0\cdot 4 \end{bmatrix} & = & \begin{bmatrix} 1 & 1 & 10 \\ -4 & -1 & 2 \\ 3 & 1 & 2 \end{bmatrix} \end{eqnarray*}

Performing \(R_2 \leftarrow 3 R_2\) to \(A\), we get \(\left[\begin{array}{rrr} 1 & -1 & 1 \\ 0 & 3 & 9 \\ -2 & 1 & 0\\ \end{array}\right]\). Then performing \(R_3 \leftarrow R_3 + 2 R_1\), we get \(\left[\begin{array}{rrr} 1 & -1 & 1 \\ 0 & 3 & 9 \\ 0 & -1 & 2\\ \end{array}\right]\). This matrix is precisely \(B\).

Now, the elementary matrices corresponding the first and second operations are, respectively, \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1\\ \end{bmatrix}\) and \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1\\ \end{bmatrix}\).

Hence, \(M =\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1\\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1\\ \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 2 & 0 & 1\\ \end{bmatrix}\). One can check that \(M A = B\).

Remark. Observe that one could obtain \(M\) by applying the same sequence of elementary row operations to the \(3\times 3\) identity matrix. The reason is that each multiplication (on the left) by an elementary matrix is the same as performing the elementary row operation associated with the matrix. Noting this can save some writing and unnecessarily calculations when the number of elementary row operations is large.

The extended augmented matrix capturing the three systems is \[\left[\begin{array}{rrr|rrr} 1 & 0 & 2 & 1 & 0 & 0 \\ 1 & 3 & 3 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}\right].\] Performing \(R_2 \leftarrow R_2 - R_1\) gives \[\left[\begin{array}{rrr|rrr} 1 & 0 & 2 & 1 & 0 & 0 \\ 0 & 3 & 1 & -1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}\right].\] Performing \(R_2 \leftarrow R_2 - 3R_3\) gives \[\left[\begin{array}{rrr|rrr} 1 & 0 & 2 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1 & 1 & -3 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}\right].\] Performing \(R_1 \leftarrow R_1 - 2R_2\) gives \[\left[\begin{array}{rrr|rrr} 1 & 0 & 0 & 3 & -2 & 6 \\ 0 & 0 & 1 & -1 & 1 & -3 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}\right].\] Performing \(R_2 \leftrightarrow R_3\) gives \[\left[\begin{array}{rrr|rrr} 1 & 0 & 0 & 3 & -2 & 6 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & -1 & 1 & -3 \end{array}\right].\]

Hence, \(\begin{bmatrix} x_1 \\x_2\\ x_3\end{bmatrix} = \begin{bmatrix} 3 \\ 0 \\ -1 \end{bmatrix}\) is the unique solution to \(Ax = e_1\), \(\begin{bmatrix} x_1 \\x_2\\ x_3\end{bmatrix} = \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}\) is the unique solution to \(Ax = e_2\), and \(\begin{bmatrix} x_1 \\x_2\\ x_3\end{bmatrix} = \begin{bmatrix} 6 \\ 1 \\ -3 \end{bmatrix}\) is the unique solution to \(Ax = e_3\).