## Solving systems of linear equations with the same coefficient matrix

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.

### Example

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.

## Exercises

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 $$Ax = b''$$.