How to solve a linear system from a row echelon form matrix

Key concept: When a matrix is already in row echelon form, the work is mostly back-substitution. You read from the bottom row upward and solve one variable at a time.

Step by step

We are given the augmented matrix

1 & -1 & 4 & 1\\ 0 & 2 & -2 & 3 \end{bmatrix}.$$ This corresponds to the system $$\begin{cases} x-y+4z=1\\ 2y-2z=3 \end{cases}$$ ### Step 1: Solve the second equation $$2y-2z=3$$ Divide by 2: $$y-z=\frac32$$ so $$y=z+\frac32.$$ ### Step 2: Substitute into the first equation $$x-y+4z=1$$ Substitute $y=z+\frac32$: $$x-(z+\frac32)+4z=1$$ $$x+3z-\frac32=1$$ $$x+3z=\frac52$$ so $$x=\frac52-3z.$$ ### Step 3: Parametric form Let $z=t$. Then $$(x,y,z)=\left(\frac52-3t,\ t+\frac32,\ t\right), \quad t\in\mathbb{R}.$$ So there are infinitely many solutions. ### Pitfall alert Don’t treat REF as if it already gives the final answer. The bottom row usually gives one variable in terms of a free variable, and then the row above must be updated. A small sign slip in back-substitution is the most common error here. ### Try different conditions If a third row were added with a pivot in the last variable, the system could become a single exact solution instead of infinitely many. If the second row were $[0\ 2\ -2\ 0]$, then the same process would still work, but the constants would change the final parameterization. ### Further reading row echelon form, back-substitution, parametric solution