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.

[\begin{array}{cccc} 1 & 1 & 1 & 6 \\ 2 & - 1 & 1 & 3 \\ 1 & 2 & - 1 & 2 \end{array}]

Clear the first column

Replace R₂ with R₂ − 2R₁ and R₃ with R₃ − R₁ to put zeros below the first pivot.

[\begin{array}{cccc} 1 & 1 & 1 & 6 \\ 0 & - 3 & - 1 & - 9 \\ 0 & 1 & - 2 & - 4 \end{array}]

Clear the second column

Swap R₂ and R₃ so that the second pivot is 1, then replace the new R₃ with R₃ + 3R₂.

[\begin{array}{cccc} 1 & 1 & 1 & 6 \\ 0 & 1 & - 2 & - 4 \\ 0 & 0 & - 7 & - 21 \end{array}]

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.

\begin{array}{l} - 7 z = - 21 \\ z = 3 \end{array}

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.

[\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3 \end{array}]

Check

Substitute (1, 2, 3) into each original equation.

\begin{array}{l} 1 + 2 + 3 = 6 \\ 2 \cdot 1 - 2 + 3 = 3 \\ 1 + 2 \cdot 2 - 3 = 2 \end{array}

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.

\begin{array}{l} 2 - (- 1) + 2 \cdot 1 = 5 \\ 2 \cdot 2 + (- 1) - 1 = 2 \\ 2 + 2 \cdot (- 1) + 1 = 1 \end{array}

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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