3×3 Systems of Equations (AA HL)
Three equations in three unknowns arise whenever a problem gives you three independent pieces of information about three quantities - three prices, three currents, three intersecting planes. The working challenge is entirely organisational: eliminate one variable at a time until you're down to a single equation, then rebuild the rest by back-substitution. This page covers the elimination method, a worked walkthrough, and the mistakes that cost marks. It's part of the broader Systems of Equations topic.
22 questions on this sub-topic.
The method
Covered under IB syllabus reference AHL1.16 - solutions of systems of linear equations with up to three equations in three unknowns, including cases with a unique solution, infinitely many solutions, or no solution, and finding a general solution when there are infinitely many.
Row reduction / elimination
Systematically add or subtract multiples of the equations to eliminate variables, one at a time, until only one unknown remains.
Not a formula-booklet result - it's a method. You can do it by hand, or feed the augmented matrix into your GDC's row-reduction or simultaneous-equation tool.
Three outcomes
A 3×3 system has exactly one of: a unique solution (three planes meet at one point), infinitely many solutions (the planes share a common line), or no solution (a contradiction appears once you eliminate down).
Need the full syllabus wording and formula-booklet reference table? See Systems of Equations.
Worked examples
A shop sells three items. Buying 2 of A, 1 of B and 1 of C costs $13; 1 of A, 3 of B and 2 of C costs $21; and 3 of A, 1 of B and 2 of C costs $20.
Find the price of each item.
Worked solution
\(2a + b + c = 13,\ a + 3b + 2c = 21,\ 3a + b + 2c\) M1 \(= 20.\) A1
Eq3 − Eq1: \(a + c = 7\); substituting \(b = 13 - 2a - c\) into Eq2 gives \(5a + c\) M1 \(= 18.\) A1
\(4a = 11 \Rightarrow a = 2.75,\ c = 4.25,\ b\) M1 \(= 3.25.\) A1
Calculator questions like this can also be entered directly into your GDC's simultaneous-equation solver - see the parent topic's GDC guide for keystrokes on your model.
Three planes are given by \(x+y+z=1,\ 2x+y-z=2,\ 3x+2y=4.\) Determine the nature of their intersection.
Worked solution
(1)+(2): \(3x + 2y\) M1 \(= 3.\) A1
but (3) says \(3x + 2y\) M1 \(= 4.\) A1 Contradiction: the planes have no common point - no simultaneous solution. R1
Common mistakes
- Sign errors when subtracting equations. Subtracting one equation from another to eliminate a variable is a common source of arithmetic slips - double-check every term, including the constants, not just the variable you're cancelling.
- Introducing a parameter but only checking two of the three equations. A parametric general solution must satisfy all three original equations, not just the two you used to derive it - always verify with the third.
- Stopping as soon as a contradiction appears without stating the conclusion. Reaching \(3=4\) is not itself the final answer - you still need to state clearly that this means the system (or the three planes) has no solution.
Ready to practise properly?
22 three-variable-system questions, marked instantly like the real exam.
Quick answers
How do you solve a 3x3 system of equations by hand?
Eliminate one variable at a time by combining pairs of equations, reducing three equations in three unknowns to two in two, then one in one, before back-substituting to recover the rest.
What does it mean if a 3x3 system has no solution?
The three planes don't share a common point. Algebraically this shows up as a contradiction, such as \(3=4\), once you've eliminated down to a single equation.