How to use Gauss–Jordan elimination to reach reduced row-echelon form

Gauss–Jordan elimination uses row operations to make a system’s solution easier to read. We build pivot columns with a leading 1 and zeros above and below it. Let’s follow that process through a three-variable system.

Recognize echelon and reduced echelon forms

In row-echelon form, zero rows are at the bottom and each nonzero row’s leading entry is farther right than the leading entry in the row above. In reduced row-echelon form, each leading entry is 1 and is the only nonzero entry in its column. Entries in columns without pivots may still be nonzero.

Write the system as an augmented matrix

\begin{array}{l} - x - 5 y + z = 17 \\ - 5 x - 5 y + 5 z = 5 \\ 2 x + 5 y - 3 z = - 10 \end{array}

[\begin{array}{r} - 1 & - 5 & 1 & | & 17 \\ - 5 & - 5 & 5 & | & 5 \\ 2 & 5 & - 3 & | & - 10 \end{array}]

At each pivot stage, look for a nonzero entry at or below the current row. If the current entry is zero, swap a suitable row into that position. If no such entry exists in the column, move to the next column.

Create the first pivot and clear its column

First apply $\frac{1}{5} R_{2} \to R_{2}$ and $- R_{1} \to R_{1}$ .

[\begin{array}{r} 1 & 5 & - 1 & | & - 17 \\ - 1 & - 1 & 1 & | & 1 \\ 2 & 5 & - 3 & | & - 10 \end{array}]

Now apply $R_{2} + R_{1} \to R_{2}$ and $R_{3} - 2 R_{1} \to R_{3}$ .

[\begin{array}{r} 1 & 5 & - 1 & | & - 17 \\ 0 & 4 & 0 & | & - 16 \\ 0 & - 5 & - 1 & | & 24 \end{array}]

Create the second pivot and clear its column

Apply $\frac{1}{4} R_{2} \to R_{2}$ , giving row 2 equal to $(0, 1, 0, - 4)$ . Then apply $R_{1} - 5 R_{2} \to R_{1}$ and $R_{3} + 5 R_{2} \to R_{3}$ .

[\begin{array}{r} 1 & 0 & - 1 & | & 3 \\ 0 & 1 & 0 & | & - 4 \\ 0 & 0 & - 1 & | & 4 \end{array}]

Create the third pivot and finish reducing

Apply $- R_{3} \to R_{3}$ , followed by $R_{1} + R_{3} \to R_{1}$ .

[\begin{array}{r} 1 & 0 & 0 & | & - 1 \\ 0 & 1 & 0 & | & - 4 \\ 0 & 0 & 1 & | & - 4 \end{array}]

The reduced matrix gives $x = - 1$ , $y = - 4$ , and $z = - 4$ . Substituting these values into the original equations gives $17$ , $5$ , and $- 10$ on their left sides, matching all three constants.

Interpret other possible reduced forms

A row with zero coefficients and a nonzero constant means the system has no solution. If a consistent system has a variable column without a pivot, that variable is free. The remaining variables can be written in terms of the free variables.

Learn mathKrista KingAugust 18, 2021math, learn online, online course, online math, equilibrium solutions