Let \(A \in \mathbb{F}^{m\times n}\) for some field \(\mathbb{F}\). One of the advantages of having a matrix \(M \in \mathbb{F}^{m\times m}\) such that \(MA\) is a matrix in RREF is that you can now easily find solutions to the system \(Ax = b\) for any \(b \in \mathbb{F}^m\) because the matrix \(MA\) is in RREF and we can readily read off the solutions from the system \(MA = Mb\).
However, if we are interested in solving the system \(Ax = b\) for just a few different tuples \(b\), we can simply form an extended augmented matrix and performing row reduction as illustrated in the example below.
Let \(A = \begin{bmatrix} 1 & 0 & -1 \\ -1 & 2 & 1\\ 1 & 2 & -1 \end{bmatrix}\), \(b = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}\), and \(b' = \begin{bmatrix} 3 \\ 2 \\ 1\end{bmatrix}\) be defined over the real numbers. Find all solutions to each of the systems \(Ax = b\) and \(Ax = b'\) where \(x = \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix}\).
If we row reduce \([A\mid b]\) to \([R \mid d]\) and \([A\mid b']\) to \([R'\mid d']\), we must have \(R = R'\) since they are both the RREF of \(A\), which is unique. Hence, the same sequence of elementary row operations applied to \(Ax = b\) will bring \([A \mid b']\) to \([R'\mid d']\). As a result, we can row reduce \([A\mid b~b']\) to solve both systems: \[ \begin{array}{rl} & \left[\begin{array}{rrr|rr} 1 & 0 & -1 & 0 & 3 \\ -1 & 2 & 1 & 1 & 2\\ 1 & 2 & -1 & 1 & 1 \end{array}\right] \\ \stackrel{R_2 \leftarrow R_2 + R_1}{\longrightarrow} & \left[\begin{array}{rrr|rr} 1 & 0 & -1 & 0 & 3 \\ 0 & 2 & 0 & 1 & 5\\ 1 & 2 & -1 & 1 & 1 \end{array}\right] \\ \stackrel{R_3 \leftarrow R_3 - R_1}{\longrightarrow} & \left[\begin{array}{rrr|rr} 1 & 0 & -1 & 0 & 3 \\ 0 & 2 & 0 & 1 & 5\\ 0 & 2 & 0 & 1 &-2 \end{array}\right]\\ \stackrel{R_3 \leftarrow R_3 - R_2}{\longrightarrow} & \left[\begin{array}{rrr|rr} 1 & 0 & -1 & 0 & 3 \\ 0 & 2 & 0 & 1 & 5\\ 0 & 0 & 0 & 0 &-7 \end{array}\right] \\ \stackrel{R_2 \leftarrow \frac{1}{2} R_2}{\longrightarrow} & \left[\begin{array}{rrr|rr} 1 & 0 & -1 & 0 & 3 \\ 0 & 1 & 0 & \frac{1}{2} & \frac{5}{2}\\ 0 & 0 & 0 & 0 &-7 \end{array}\right] \end{array} \] In the final matrix above, the first three columns together with the second right-most column gives a matrix in RREF row-equivalent to \([A\mid b]\) whereas the first three columns together with the right-most column gives a matrix in RREF row-equivalent to \([A\mid b']\). Thus, \(Ax = b\) has solutions \(x = \begin{bmatrix} t\\\frac{1}{2}\\t \end{bmatrix}\) for all \(t \in \mathbb{R}\) whereas \(Ax = b'\) has no solutions. Another way to say this is that there are infinitely many ways to write \(b\) as a linear combination of the columns of \(A\) but there is no way to write \(b'\) as a linear combination of the columns of \(A\). This interpretation will be revisited when we discuss the column space of a matrix.
Let \(A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 1\\ 1 & -1 & 3 \end{bmatrix}\), \(b = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\), \(b' = \begin{bmatrix} 1 \\ -2 \\ 3\end{bmatrix}\), and \(b'' = \begin{bmatrix} 2 \\ -1 \\ -1\end{bmatrix}\) be defined over the real numbers. Find all solutions to each of the systems \(Ax = b\), \(Ax = b'\), and \(Ax = b''\).