Gaussian Elimination

Back Substitution

Solve a 3x3 system starting from a given upper-triangular row echelon form. Work upward from the bottom row: normalize each pivot to 1, then use that row to eliminate the same column from all rows above it. Each step isolates one more variable.

Example

Use the upper-triangular system to solve from the bottom row upward.

highlighted = computed this step

Step 1 — Set up

Start with the augmented matrix.

[12−1201390012]\left[\begin{array}{ccc|c}1&2&-1&2\\0&1&3&9\\0&0&1&2\end{array}\right]

Step 2 — Back-sub row 1

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

[120401390012]\left[\begin{array}{ccc|c}\hl{1}&\hl{2}&\hl{0}&\hl{4}\\0&1&3&9\\0&0&1&2\end{array}\right]

Step 3 — Back-sub row 2

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

[120401030012]\left[\begin{array}{ccc|c}1&2&0&4\\\hl{0}&\hl{1}&\hl{0}&\hl{3}\\0&0&1&2\end{array}\right]

Step 4 — Back-sub row 1

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

[100-201030012]\left[\begin{array}{ccc|c}\hl{1}&\hl{0}&\hl{0}&\hl{-2}\\0&1&0&3\\0&0&1&2\end{array}\right]

Step 5 — Result

Read the solution: x1 = -2, x2 = 3, x3 = 2.

[100−201030012],solution: x1=−2, x2=3, x3=2\left[\begin{array}{ccc|c}1&0&0&-2\\0&1&0&3\\0&0&1&2\end{array}\right],\quad \text{solution: }x1=-2,\ x2=3,\ x3=2
back-substitution After forward elimination produces an upper-triangular matrix, work upward row by row: first normalize the pivot to 1, then subtract multiples of that row from all rows above to zero out the pivot column above.
pivot-normalization Multiply a row by the reciprocal of its leading (pivot) entry so the pivot becomes exactly 1. This enables clean elimination of that column in subsequent steps.