9 min readConcept sheet10-mark problem: solve 2 or 3 equations by row operations

Solving systems by row reduction (Gauss-Jordan)

Write the equations as an augmented matrix, then use row operations to reach [I | solution]; the last column then gives the answers.

Augmented matrix[coefficients | constants]

Row operationsSwap; multiply by non-zero k; add k × another row

TargetReduced row-echelon form [I | x]

NotationR2 → R2 − 3R1

The three row operations (they never change the solution)

Swap two rowsR1 ↔ R2

Multiply a row by a non-zero numberR2 → (−1/7)R2

Add a multiple of one row to anotherR2 → R2 − 3R1

Row [0 0 | k] with k ≠ 0No solution (inconsistent)

A whole row of zeros [0 0 | 0]Infinitely many solutions, if no row is [0 0 | k] and there are fewer leading 1s than unknowns

Gauss-Jordan steps

Two equations

Problem

Solve x + 2y = 8 and 3x − y = 3 by row reduction.

Working

Augmented matrix: [1 2 | 8; 3 −1 | 3].
R2 → R2 − 3R1: [1 2 | 8; 0 −7 | −21].
R2 → (−1/7)R2: [1 2 | 8; 0 1 | 3].
R1 → R1 − 2R2: [1 0 | 2; 0 1 | 3].
Check: 2 + 2(3) = 8 ✓ and 3(2) − 3 = 3 ✓.

Answer

x = 2, y = 3

Three equations

Problem

Solve x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2 by Gauss-Jordan elimination.

Working

Augmented matrix: [1 1 1 | 6; 2 −1 1 | 3; 1 2 −1 | 2].
R2 → R2 − 2R1 and R3 → R3 − R1: [1 1 1 | 6; 0 −3 −1 | −9; 0 1 −2 | −4].
Swap R2 and R3 to get a leading 1: [1 1 1 | 6; 0 1 −2 | −4; 0 −3 −1 | −9].
R1 → R1 − R2 and R3 → R3 + 3R2: [1 0 3 | 10; 0 1 −2 | −4; 0 0 −7 | −21].
R3 → (−1/7)R3: [1 0 3 | 10; 0 1 −2 | −4; 0 0 1 | 3].
R1 → R1 − 3R3 and R2 → R2 + 2R3: [1 0 0 | 1; 0 1 0 | 2; 0 0 1 | 3].
Check: 1 + 2 + 3 = 6 ✓; 2 − 2 + 3 = 3 ✓; 1 + 4 − 3 = 2 ✓.

Answer

x = 1, y = 2, z = 3

A workshop production plan

Problem

A Moratuwa furniture workshop makes chairs (x) and tables (y). A chair needs 2 hours of carpentry and 1 hour of finishing; a table needs 3 hours of each. It has 120 carpentry hours and 90 finishing hours a week and wants to use them all. How many of each should it make?

Working

Carpentry: 2x + 3y = 120. Finishing: x + 3y = 90.
Augmented matrix [2 3 | 120; 1 3 | 90]. Swap rows for a leading 1: [1 3 | 90; 2 3 | 120].
R2 → R2 − 2R1: [1 3 | 90; 0 −3 | −60].
R2 → (−1/3)R2 then R1 → R1 − 3R2: [1 0 | 30; 0 1 | 20].
Check: 2(30) + 3(20) = 120 ✓ and 30 + 3(20) = 90 ✓.

Answer

30 chairs and 20 tables a week.

In Sri Lankan business

Moratuwa furniture makers and other Sri Lankan workshops that want to use all available hours in several departments face a system of equations. Row reduction solves it for any number of products, and it is what spreadsheet solvers do inside.

Exam angle

Short answer: “State the three elementary row operations.” Swap two rows; multiply a row by a non-zero constant; add a multiple of one row to another.

Structured question: “Solve x + 2y = 8 and 3x − y = 3 by row reduction.” [1 2 | 8; 3 −1 | 3] → [1 0 | 2; 0 1 | 3]: x = 2, y = 3.

Essay: “Solve x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2 by Gauss-Jordan elimination.” Reduce to [I | 1; 2; 3]: x = 1, y = 2, z = 3, with a check in all three equations.

Remember it

Work column by column: make the leading 1, then clear the column above and below it. Write the operation (e.g. R2 → R2 − 3R1) beside every step: examiners mark the method.