1.16 Systems of linear equations (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

Three types of solution

A system of up to three linear equations in three unknowns has exactly one of:

a unique solution — one value of each unknown (three planes meeting at a point);
infinitely many solutions — the equations are consistent but not independent (planes meeting in a line, or all the same plane);
no solution — the system is inconsistent (for example, reduction gives $0 = 5$ ).

Each equation $ax + by + cz = d$ represents a plane, which gives a geometric way to picture the three cases.

Key termsunique solutioninconsistent

Common mistake

A row $0 = 0$ does not mean 'no solution' — it means one equation depended on the others, giving infinitely many solutions (if nothing else is contradictory).

Section 2

Solving algebraically by elimination (row reduction)

Use one equation to eliminate $x$ from the other two, then eliminate $y$ from the resulting pair. This is row reduction: in the augmented matrix $\left(\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & -1 & 1 & 3 \\ 1 & 2 & -1 & 2 \end{array}\right)$ you aim for zeros below the leading diagonal, then back-substitute.

Example: (2) $-$ 2×(1) gives $3y + z = 9$ ; (3) $-$ (1) gives $y - 2z = -4$ ; eliminating $y$ gives $7z = 21$ , so $z = 3$ , $y = 2$ , $x = 1$ . Always check in all three original equations.

Key termsrow reductionback-substitution

Exam tip

Label your equations (1), (2), (3) and write the operation each time, e.g. (2) − 2×(1). It earns method marks and prevents sign slips.

Section 3

Using technology

On Paper 2 you are expected to solve systems with a GDC (simultaneous equation solver or matrix row reduction). Write down the system you entered, then the answer — for example, for $2c + t + 3j = 62$ , $c + 3t + j = 44$ , $3c + 2t + 2j = 74$ : $c = 14$ , $t = 7$ , $j = 9$ .

A GDC will report an error or give a 'free' variable when there is no unique solution — you must then work algebraically to decide between infinitely many and none.

Key termsGDC

Section 4

General solution for infinitely many solutions

When reduction leaves two independent equations in three unknowns, set one unknown equal to a parameter, $z = \lambda$ , and express the others in terms of it.

Example: $x + 2y - z = 3$ and $y + 3z = 4$ give $y = 4 - 3\lambda$ and $x = -5 + 7\lambda$ . The general solution $(x, y, z) = (-5 + 7\lambda,\ 4 - 3\lambda,\ \lambda)$ describes a line: every point on it satisfies all the equations. Different choices of parameter give equivalent forms.

Key termsparametergeneral solution

Exam tip

Check the general solution by substituting it into every original equation — the parameter should cancel completely.

Section 5

Systems with an unknown coefficient

When a coefficient or constant is unknown, reduce until one equation has the form $(a - 4)z = b - 7$ . Then:

$a \ne 4$ : unique solution;
$a = 4$ , $b = 7$ : $0 = 0$ , infinitely many solutions;
$a = 4$ , $b \ne 7$ : $0 = b - 7 \ne 0$ , inconsistent — no solution.

Common mistake

Dividing by $(a - 4)$ without first dealing with the case $a = 4$ separately.

Must know

Up to three equations in three unknowns: unique, infinitely many, or no solution.
Solve by elimination/row reduction (algebraically) and by GDC.
No solution ⇔ inconsistent (reduction gives $0 = k$ , $k \ne 0$ ).
Infinitely many: set a parameter and write the general solution.
With unknown coefficients, reduce to $(\ldots)z = \ldots$ and split into cases.

That's the notes covered.

Carry on to the next subtopic.