Remark: Parametric representation of infinitely many solutions
When a linear system has infinitely many solutions, we express the solution set using parameters (typically denoted by , , , etc.).
For example, consider the system: \begin{align} x + 2y - z &= 1 \\ 2x + 4y - 2z &= 2 \end{align}
Since the second equation is just twice the first, we have infinitely many solutions. We can express the general solution as:
\begin{align} x &= 1 - 2t + s \\ y &= t \\ z &= s \end{align}$$ where $t$ and $s$ are **free parameters** that can take any real values. ^def-free-parameters ## Exercise ### Solving linear system with elementary row operations > Solve the following system of linear equations: > $$\begin{align} > 2x + y - z &= 8 \\ > -3x - y + 2z &= -11 \\ > -2x + y + 2z &= -3 > \end{align}$$ 1. Write as augmented matrix $$\left[\begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array}\right]$$ 2. $R_1 \leftarrow \frac{1}{2}R_1$ (multiply row 1 by $\frac{1}{2}$) $$\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array}\right]$$ 3. $R_2 \leftarrow R_2 + 3R_1$ (add 3 times row 1 to row 2) $$\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \\ -2 & 1 & 2 & -3 \end{array}\right]$$ 4. $R_3 \leftarrow R_3 + 2R_1$ (add 2 times row 1 to row 3) $$\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \\ 0 & 2 & 1 & 5 \end{array}\right]$$ 5. $R_2 \leftarrow 2R_2$ (multiply row 2 by 2) $$\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & 1 & 1 & 2 \\ 0 & 2 & 1 & 5 \end{array}\right]$$ 6. $R_3 \leftarrow R_3 - 2R_2$ (subtract 2 times row 2 from row 3) $$\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & -1 & 1 \end{array}\right]$$ 7. $R_3 \leftarrow -R_3$ (multiply row 3 by -1) $$\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & -1 \end{array}\right]$$ 8. $R_2 \leftarrow R_2 - R_3$ (subtract row 3 from row 2) $$\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1 \end{array}\right]$$ 9. $R_1 \leftarrow R_1 + \frac{1}{2}R_3$ (add $\frac{1}{2}$ times row 3 to row 1) $$\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & 0 & \frac{7}{2} \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1 \end{array}\right]$$ 10. $R_1 \leftarrow R_1 - \frac{1}{2}R_2$ (subtract $\frac{1}{2}$ times row 2 from row 1) $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1 \end{array}\right]$$ The matrix is now in **reduced row echelon form (RREF)**. The solution is. $$\boxed{x = 2, \quad y = 3, \quad z = -1}$$