Gaussian Elimination

Reduced Row Echelon Form

Reduce a 3x3 augmented matrix to reduced row echelon form (RREF). Forward elimination creates the upper-triangular REF; back- substitution then normalizes each pivot to 1 and zeros entries above each pivot, yielding a unique solution directly.

Example

Continue row operations until each pivot column is reduced.

highlighted = computed this step

Step 1 — Set up

Start with the augmented matrix.

[111323481024]\left[\begin{array}{ccc|c}1&1&1&3\\2&3&4&8\\1&0&2&4\end{array}\right]

Step 2 — Update row 2

Update row 2 using row 1: r2 <- r2 - 2*r1.

[111301221024]\left[\begin{array}{ccc|c}1&1&1&3\\\hl{0}&\hl{1}&\hl{2}&\hl{2}\\1&0&2&4\end{array}\right]

Step 3 — Update row 3

Update row 3 using row 1: r3 <- r3 - r1.

[111301220-111]\left[\begin{array}{ccc|c}1&1&1&3\\0&1&2&2\\\hl{0}&\hl{-1}&\hl{1}&\hl{1}\end{array}\right]

Step 4 — Update row 3

Update row 3 using row 2: r3 <- r3 + r2.

[111301220033]\left[\begin{array}{ccc|c}1&1&1&3\\0&1&2&2\\\hl{0}&\hl{0}&\hl{3}&\hl{3}\end{array}\right]

Step 5 — Normalize row 3

Scale row 3: r3 <- 1/3*r3.

[111301220011]\left[\begin{array}{ccc|c}1&1&1&3\\0&1&2&2\\\hl{0}&\hl{0}&\hl{1}&\hl{1}\end{array}\right]

Step 6 — Back-sub row 1

Update row 1 using row 3: r1 <- r1 - r3.

[110201220011]\left[\begin{array}{ccc|c}\hl{1}&\hl{1}&\hl{0}&\hl{2}\\0&1&2&2\\0&0&1&1\end{array}\right]

Step 7 — Back-sub row 2

Update row 2 using row 3: r2 <- r2 - 2*r3.

[110201000011]\left[\begin{array}{ccc|c}1&1&0&2\\\hl{0}&\hl{1}&\hl{0}&\hl{0}\\0&0&1&1\end{array}\right]

Step 8 — Back-sub row 1

Update row 1 using row 2: r1 <- r1 - r2.

[100201000011]\left[\begin{array}{ccc|c}\hl{1}&\hl{0}&\hl{0}&\hl{2}\\0&1&0&0\\0&0&1&1\end{array}\right]

Step 9 — Result

Read the solution: x1 = 2, x2 = 0, x3 = 1.

[100201000011],solution: x1=2, x2=0, x3=1\left[\begin{array}{ccc|c}1&0&0&2\\0&1&0&0\\0&0&1&1\end{array}\right],\quad \text{solution: }x1=2,\ x2=0,\ x3=1
rref A matrix is in reduced row echelon form when it is in REF and every pivot equals 1 with zeros both above and below it.
back-substitution Starting from the bottom-right pivot, divide each pivot row to make the pivot 1, then subtract multiples from rows above to eliminate that column entry everywhere else.