RREF (Row-Reduced Echelon Form)
Linear Algebra
Intro: We display each elementary row operation (swap, scale, replace) and the resulting matrix in echelon or reduced row-echelon form. Augmented matrices are supported.
Worked example
- Reduce to RREF: $\left[\begin{array}{cc|c} 1 & 2 & 5 \\ 3 & 4 & 11 \end{array}\right]$
- Start with the augmented matrix (vertical bar separates coefficients from constants): $$\left[\begin{array}{cc|c} 1 & 2 & 5 \\ 3 & 4 & 11 \end{array}\right].$$
- Eliminate below the leading 1 in row 1: $$R_2 \leftarrow R_2 - 3R_1 \;\Rightarrow\; \left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & -2 & -4 \end{array}\right].$$
- Scale row 2 to make the pivot 1: $$R_2 \leftarrow (-\tfrac{1}{2})R_2 \;\Rightarrow\; \left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & 1 & 2 \end{array}\right].$$
- Clear above the pivot in column 2: $$R_1 \leftarrow R_1 - 2R_2 \;\Rightarrow\; \left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right].$$
- This is RREF. Read off the solution: $$x=1,\; y=2.$$
- Check (optional): $$\begin{cases} 1(1)+2(2)=5 \\ 3(1)+4(2)=11 \end{cases} \;\Rightarrow\; 5=5,\; 11=11\;\checkmark$$
- Solve [ [1,2,−1|4], [2,4,−2|8], [0,1,1|3] ] to rref.
- Matrix: $$\left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 2 & 4 & -2 & 8 \\ 0 & 1 & 1 & 3 \end{array}\right].$$
- Eliminate below the leading 1 in row 1: $$R_2 \leftarrow R_2 - 2R_1 \Rightarrow \left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 3 \end{array}\right].$$
- Swap rows 2 and 3 to expose the next pivot: $$R_2 \leftrightarrow R_3 \Rightarrow \left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 0 & 1 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right].$$
- Clear above the pivot in column 2: $$R_1 \leftarrow R_1 - 2R_2 \Rightarrow \left[\begin{array}{ccc|c} 1 & 0 & -3 & -2 \\ 0 & 1 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right].$$
- This is RREF. Let the free variable be $z=t$. From row 2: $y + z = 3 \Rightarrow y = 3 - t$. From row 1: $x - 3z = -2 \Rightarrow x = -2 + 3t$.
- Parametric solution set: $$\boxed{(x,y,z)=(-2,\,3,\,0) + t\,(3,\,-1,\,1),\; t\in\mathbb{R}}.$$
- Solve [ [1,1|2], [2,2|5] ] to rref.
- Matrix: $$\left[\begin{array}{cc|c} 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right].$$
- Eliminate: $$R_2 \leftarrow R_2 - 2R_1 \Rightarrow \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right].$$
- Row 2 reads $0x+0y=1$ which is impossible. Therefore the system is $$\boxed{\text{inconsistent (no solution)}}.$$
FAQs
Fractions?
Yes—row scaling uses rational numbers so pivots become 1; we keep exact fractions whenever possible.
How do I spot free variables?
Columns without pivots correspond to free variables; pivot columns correspond to leading variables solved in terms of the free ones.
What about inconsistency?
If a row reduces to [0 0 … 0 | c] with c ≠ 0, the system has no solution.
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