Math · Precalculus · Worked example
Solve a 3 × 3 system with row reduction
Solve the system x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2 by row reduction.
Clear the first column
Replace R₂ with R₂ − 2R₁ and R₃ with R₃ − R₁ to put zeros below the first pivot.
Clear the second column
Swap R₂ and R₃ so that the second pivot is 1, then replace the new R₃ with R₃ + 3R₂.
Back-substitute
The last row says −7z = −21, so z = 3. Then y − 2z = −4 gives y = 2, and x + y + z = 6 gives x = 1.
Or finish to RREF
Scaling R₃ by −1/7 and clearing the entries above each pivot leaves the identity matrix on the left; the last column is the solution.
Check
Substitute (1, 2, 3) into each original equation.
Result
x = 1, y = 2, z = 3.
Your turn
Solve x − y + 2z = 5, 2x + y − z = 2, x + 2y + z = 1 with an augmented matrix.
Show the answer and explanation
x = 2, y = −1, z = 1.
Row reduction leaves the identity matrix beside the column (2, −1, 1). Substituting checks each equation: 2 + 1 + 2 = 5, 4 − 1 − 1 = 2 and 2 − 2 + 1 = 1.
Keep exploring
In Matrices & linear systems, the operation trail shows each step to the RREF. Remove the last column and switch the task to Determinant: it is 7, not 0, which is why the system has exactly one solution.
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