IB Maths AA HLTopic 1 — Number & AlgebraPaper 1 & 2HL only~10 min read
Solving Systems Using Row Reduction
In the last note you learned how to set up a system of linear equations and how to solve one using your GDC. But on Paper 1 you don’t have a calculator — and even on Paper 2, examiners sometimes specifically ask for an algebraic method. Row reduction is that algebraic method, and it’s astonishingly systematic. The idea: write the system as a grid of numbers (an augmented matrix), apply a few simple operations to whittle it down to a “staircase” pattern, then read off the answer line by line. Once you trust the algorithm, even messy 3×3 systems become surprisingly mechanical — every step has a reason, and there’s no guessing.
📘 What you need to know
An augmented matrix packs a system of linear equations into a grid of just the coefficients and constants — no variables.
A matrix is in row-reduced (echelon) form when it has a “staircase” of leading 1’s on the diagonal and zeros below them.
There are three row operations you may use, none of which change the solution: swap two rows, multiply a row by a non-zero constant, or add a multiple of one row to another.
The standard row-reduction algorithm works column by column from the left: get a 1 at the top of the column, create zeros below it, ignore that column and row, repeat.
Once a 3×3 matrix is in row-reduced form, the last row gives z directly. Then back-substitute upwards to find y, then x.
Row reduction is the only algebraic method on Paper 1 for non-trivial 3×3 systems — your GDC isn’t allowed there.
The notation r1 ↔ r2 means swap rows; kr1 → r1 means multiply row 1 by k; r2 + 3r1 → r2 means replace row 2 with row 2 plus three times row 1.
The augmented matrix — packing a system into a grid
The first move is purely cosmetic but very useful: rewrite the system without the variables. Just keep the coefficients (and the constants on the right-hand side) in a rectangular grid called an augmented matrix:
The vertical bar separates the coefficients from the constants — it’s just a visual reminder that the right-hand side comes from after the equals sign. Important: if a variable is missing from any equation, write a 0 in its place — don’t leave a gap.
Pre-flight check: before converting to a matrix, make sure every equation is in standard form (ax + by + cz = d) with the variables in the same order across all rows. Missing variables → put 0; constants on the right of equals → after the bar.
What is row-reduced (echelon) form?
A matrix is in row-reduced echelon form when its non-zero entries form a “staircase” pattern: leading 1’s down the diagonal, with zeros below them. Above the diagonal, anything goes.
The “staircase” pattern of row-reduced echelon form
If your matrix looks like this at the end of the algorithm, the system is solved — well, almost. The last row gives z directly; the row above gives y in terms of z, which you’ve just found; and the top row gives x in terms of y and z. This last sweep is called back-substitution.
🤔 Why a “staircase” specifically?
Because the staircase shape makes back-substitution work in one direction. The bottom row has only one variable, so it’s solved directly. The next row up adds one new variable — but the previous one is already known, so it’s still a one-variable problem. And so on up. Each row introduces exactly one new unknown — that’s what the staircase shape encodes.
The three row operations
To get from your starting matrix to the staircase form, you may use only three operations. Each one is reversible, and crucially, none of them change the solution set — that’s what makes row reduction valid.
Operation 1
swap two rows
r1 ↔ r2 useful when the top-left entry is 0 or awkward
Operation 2
multiply a row by k
kr2 → r2 (k ≠ 0) useful for getting a leading 1
Operation 3
add a multiple of one row to another
r2 + kr1 → r2 useful for creating zeros below a leading 1
Each operation has a clean equation-level interpretation. Swapping rows = writing your equations in a different order. Multiplying a row by 5 = multiplying both sides of an equation by 5. Adding 3 times row 1 to row 2 = adding 3 copies of equation 1 to equation 2. None of these change which (x, y, z) satisfies the system, which is why they’re allowed.
The 4-step row reduction algorithm
Here’s the standard “going down the diagonal” recipe. Work column by column, from left to right.
🧭 Recipe — row reduction step by step
STEP 1 — Get a 1 in the top-left corner. Either divide the first row by its leading entry (if non-zero) or swap with another row that has a friendlier leading entry.
STEP 2 — Create zeros below that 1. For each row below, subtract a suitable multiple of row 1 to kill the entry in column 1.
STEP 3 — Repeat for the sub-matrix. Now ignore the first row and column (they’re done). Apply Steps 1–2 to what remains: get a 1 in the new top-left, then zeros below it.
STEP 4 — Get a 1 in the bottom row. If the bottom row’s only non-zero entry isn’t already 1, divide the row by its value. The matrix is now in echelon form.
After reaching echelon form, convert the matrix back into equations and back-substitute upwards.
Bookkeeping advice: write the operation you used next to the row in the new matrix — for example “r2 − 3r1 → r2“. This earns you method marks and helps you spot any mistakes when you check your working.
Back-substitution — recovering x, y, z
Once the matrix is in echelon form, switch back to equations and solve from the bottom up:
Back-substitution from echelon form
bottom row gives z = D3
middle row gives y + C2z = D2 ⟹ y = D2 − C2z
top row gives x + B1y + C1z = D1 ⟹ x = D1 − B1y − C1z
Each step gives you one new variable. By the time you get to the top, you’ve got the whole solution. The order is rigid: you cannot find x first because the top equation has three unknowns. Always start at the bottom and work up.
Worked examples
WE 1
Convert a system to an augmented matrix
Write the following system as an augmented matrix:
2x + y − 3z = 4 x + z = 6
3x − 2y + z = 9
Step 1: Make sure every equation is in standard form ax + by + cz = d2x + y − 3z = 4 ✓x + 0y + z = 6 (insert 0 for missing y!)3x − 2y + z = 9 ✓Step 2: Strip out the variables, write coefficients in a grid[2 1 −3 | 4] / [1 0 1 | 6] / [3 −2 1 | 9]missing variable → 0, never blank. The y coefficient in the second row is 0.
WE 2
Apply a single row operation
Given the matrix
[1 −1 2 | 5]
[2 1 −1 | 3]
[3 −2 1 | 6]
apply the row operation r2 − 2r1 → r2 and write the new matrix.
Step 1: Identify what we’re computingreplace row 2 with (row 2 − 2 × row 1)Step 2: Compute element by elementcol 1: 2 − 2(1) = 0col 2: 1 − 2(−1) = 1 + 2 = 3col 3: −1 − 2(2) = −1 − 4 = −5RHS: 3 − 2(5) = 3 − 10 = −7Step 3: Replace row 2 with the new values; rows 1 and 3 unchanged[1 −1 2 | 5] / [0 3 −5 | −7] / [3 −2 1 | 6]notice the new row 2 has a 0 in column 1 — exactly the goal of this operation
WE 3
Solve a 2×2 system using row reduction
Solve the system using row reduction: x + 2y = 8
3x − y = 3
Step 1: Convert to augmented matrix[1 2 | 8][3 −1 | 3]Step 2: Already 1 in top-left ✓ Make zero below: r₂ − 3r₁ → r₂row 2: [3−3, −1−6, 3−24] = [0, −7, −21][1 2 | 8][0 −7 | −21]Step 3: Get a 1 in the bottom row: r₂ ÷ (−7) → r₂[1 2 | 8][0 1 | 3]Step 4: Back-substitute. Bottom row: y = 3Top row: x + 2y = 8 → x + 6 = 8 → x = 2x = 2, y = 3verify: 2 + 6 = 8 ✓ 3(2) − 3 = 3 ✓
WE 4
Solve a 3×3 system using row reduction
Solve using row reduction: x + 2y + z = 7
2x + 5y − z = 6
3x + y + 2z = 13
Step 1: Convert to augmented matrix[1 2 1 | 7][2 5 −1 | 6][3 1 2 | 13]Step 2: Already 1 in top-left ✓ Create zeros below itr₂ − 2r₁ → r₂: [0, 1, −3, −8]r₃ − 3r₁ → r₃: [0, −5, −1, −8][1 2 1 | 7][0 1 −3 | −8][0 −5 −1 | −8]Step 3: Already 1 in row 2, col 2 ✓ Create zero below in row 3r₃ + 5r₂ → r₃: [0, 0, −16, −48][1 2 1 | 7][0 1 −3 | −8][0 0 −16 | −48]Step 4: Get a 1 in the bottom row: r₃ ÷ (−16) → r₃[1 2 1 | 7][0 1 −3 | −8][0 0 1 | 3]Step 5: Back-substitutez = 3y − 3z = −8 → y − 9 = −8 → y = 1x + 2y + z = 7 → x + 2 + 3 = 7 → x = 2x = 2, y = 1, z = 3verify in original equation 2: 2(2) + 5(1) − 3 = 4 + 5 − 3 = 6 ✓ — always check at least one
WE 5
3×3 system requiring row swaps
Solve the system using row reduction:
3x − 2y + z = 1 x + y − z = 4
2x + 3y − 2z = 11
Step 1: Convert to matrix; the top-left entry is 3, so swap to get a 1[3 −2 1 | 1][1 1 −1 | 4][2 3 −2 | 11]apply r₁ ↔ r₂:[1 1 −1 | 4][3 −2 1 | 1][2 3 −2 | 11]Step 2: Create zeros below the leading 1r₂ − 3r₁ → r₂: [0, −5, 4, −11]r₃ − 2r₁ → r₃: [0, 1, 0, 3][1 1 −1 | 4][0 −5 4 | −11][0 1 0 | 3]Step 3: Swap rows 2 and 3 to get a 1 in the (2,2) position[1 1 −1 | 4][0 1 0 | 3][0 −5 4 | −11]Step 4: Zero below the new (2,2) entry: r₃ + 5r₂ → r₃[1 1 −1 | 4][0 1 0 | 3][0 0 4 | 4]Step 5: Get a 1 in the bottom row: r₃ ÷ 4 → r₃[1 1 −1 | 4][0 1 0 | 3][0 0 1 | 1]Step 6: Back-substitutez = 1y = 3 (directly from middle row)x + y − z = 4 → x + 3 − 1 = 4 → x = 2x = 2, y = 3, z = 1two strategic swaps made the rest of the algebra clean — always swap when it gets you a friendlier leading entry
WE 6
Word problem solved using row reduction
Three friends — Yara, Zayn, and Wei — buy concert tickets at three price tiers (standard, premium, VIP).
• Yara buys 2 standard, 1 premium, 1 VIP for £110.
• Zayn buys 1 standard, 2 premium, 1 VIP for £120.
• Wei buys 1 standard, 1 premium, 2 VIP for £130.
Find the price of each tier using row reduction.
Step 1: Define variables and form equationsLet s, p, v be the prices in £2s + p + v = 110s + 2p + v = 120s + p + 2v = 130Step 2: Augmented matrix; swap r₁ ↔ r₂ to get 1 in top-left[1 2 1 | 120][2 1 1 | 110][1 1 2 | 130]Step 3: Zero below the leading 1r₂ − 2r₁ → r₂: [0, −3, −1, −130]r₃ − r₁ → r₃: [0, −1, 1, 10][1 2 1 | 120][0 −3 −1 | −130][0 −1 1 | 10]Step 4: Swap r₂ ↔ r₃ for a friendlier middle entry, then negate r₂[0 −1 1 | 10] → after −r₂ → r₂: [0, 1, −1, −10][1 2 1 | 120][0 1 −1 | −10][0 −3 −1 | −130]Step 5: Zero below: r₃ + 3r₂ → r₃[0, 0, −4, −160]divide by −4: [0, 0, 1, 40][1 2 1 | 120][0 1 −1 | −10][0 0 1 | 40]Step 6: Back-substitutev = 40p − v = −10 → p = −10 + 40 = 30s + 2p + v = 120 → s + 60 + 40 = 120 → s = 20standard = £20, premium = £30, VIP = £40always interpret in the context — “standard tickets cost £20” not just “s = 20”
💡 Top tips
Always rearrange to standard form first. Row reduction only works if every equation is in ax + by + cz = d form, with variables in the same order across rows.
Use 0 for missing variables. Don’t leave gaps in the matrix.
If an equation already starts with a 1 in x, put it as row 1. This saves you from dividing or multiplying to make a leading 1 — straight to step 2.
Annotate each row operation on your working (“r2 − 3r1 → r2“). Examiners give method marks for this.
Swap rows freely whenever it gets you a cleaner leading entry. Two swaps cost nothing in marks.
Check your work by plugging back in. A 3×3 system has three equations — verify your (x, y, z) satisfies all three.
For Paper 2, you can use your GDC to verify your row-reduction answer — even if the question demanded an algebraic method.
⚠ Common mistakes
Forgetting to update the constant column when applying a row operation. The vertical bar’s right side gets the same treatment as the coefficients.
Sign errors when computing r2 − kr1. Distribute the minus sign carefully.
Multiplying a row by zero. Operation 2 specifically requires a non-zero constant — multiplying by 0 destroys information.
Leaving the matrix without leading 1’s. The echelon form requires 1’s, not just any non-zero numbers, on the diagonal.
Trying to find x first. Back-substitution must start from the bottom row (z) and work upwards.
Confusing rows and columns. Operations always act on rows, never on columns. Mixing them up changes the solution.
Skipping step 4. If the bottom row reads “[0 0 k | d]” with k ≠ 1, you’re not yet in echelon form — divide by k first.
Row reduction is the workhorse algebraic method for systems of linear equations — and once you’ve practised the algorithm a few times, it becomes almost automatic. The next note covers what happens when row reduction doesn’t end up with a clean staircase: when the system has no solutions (an inconsistent system) or infinitely many solutions (a dependent system). The matrix tells you straight away which case you’re in — you just need to know what to look for.
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