1The idea in one paragraph

A system of linear equations is a set of equations, each of the form ax + by + cz = d, that must all hold at once. The method is elimination: combine equations to remove one unknown at a time until the last equation has only one unknown, then substitute back. Writing only the numbers in an array and applying row operations is the same method, tidier. The last reduced row tells you everything. If it reads, say, −2z = −8, there is a unique solution. If it reads 0 = 0, one equation added nothing new and there are infinitely many solutions, described with a parameter. If it reads 0 = 5, the equations contradict each other and there is no solution: the system is inconsistent. Geometrically the equations are planes, and the three cases are planes meeting at a point, in a line, or not all together at all.

2Two unknowns: the three possibilities

Start where the pictures are flat. Each equation in x and y is a straight line, and a solution is a point on both. Figure 1 shows the only three things that can happen.

• One solution. x + y = 3 and 2x − y = 0 cross at (1, 2).
• No solution. x + y = 3 and x + y = 1 are parallel. Subtracting one from the other gives 0 = 2, which is impossible. The system is inconsistent.
• Infinitely many solutions. 2x + 2y = 6 is just x + y = 3 doubled. Both equations describe the same line, and every point on it is a solution.

Three unknowns work the same way, except that the lines become planes and there are more ways for them to fail to meet.

3Three unknowns: elimination and row reduction

The plan: use the first equation to remove x from the other two, then use the new second equation to remove y from the third. The system becomes triangular, and you solve from the bottom up.

Worked example 1. Paper 1. Solve the system

Check in all three original equations: 1 + 4 − 4 = 1; 2 + 2 + 4 = 8; 3 − 2 + 8 = 9. Checking in the equation you did not use last is the one that catches slips.

The same thing with the numbers only. Writing x, y, z and = on every line is slow. Keep just the coefficients and the right-hand sides in an array, the augmented matrix, with a line where the = signs were. The allowed row operations are exactly the moves you made with equations:

• swap two rows;
• multiply a row by a non-zero number;
• add a multiple of one row to another.

None of these changes the set of solutions, because each can be undone. Figure 2 does Worked example 1 this way. The shaded entries below the diagonal are the zeros you are aiming for; this is row-echelon or triangular form.

Two habits keep row reduction accurate on Paper 1. Label every operation (R2 − 2R1), because the marks are for method and a marker cannot follow unlabelled numbers. And choose the row you work from so that its leading coefficient is 1, swapping rows first if needed; dividing by 3 in the first step creates fractions you then carry to the end.

4Reading the last row

After reduction, the last row has zeros in the x and y columns. What remains decides the case, as Figure 3 summarises.

Last row a z = b with a ≠ 0: unique solution. Last row 0 = 0: infinitely many solutions. Last row 0 = b with b ≠ 0: no solution (the system is inconsistent).

The middle case needs a closer look. 0 = 0 is true, but it says nothing: one of the three equations was a combination of the other two, so there are really only two independent equations for three unknowns. One unknown can then take any value, and the others follow from it.

5Infinitely many solutions: the general solution

When the reduction gives 0 = 0, let one unknown be a parameter, usually λ, and express the other two in terms of it. The result is the general solution: one formula that gives every solution as λ runs through the real numbers.

Worked example 2. Paper 1. Show that the system below has infinitely many solutions and find the general solution.

Check with any value of λ. λ = 1 gives (1, 2, 1): 1 + 2 + 1 = 4, 1 − 2 + 2 = 1, 2 + 3 = 5. All three hold.

The general solution can be written as a vector, (x, y, z) = (5/2, 3/2, 0) + λ(−3/2, 1/2, 1), which is the vector equation of a line (AHL 3.14). So geometrically the three planes all pass through one common line, like pages of a book meeting at the spine: a sheaf. Figure 4(b) draws it.

Different choices of parameter give general solutions that look different but describe the same line. Letting z = 2μ − 3 instead would give x = 7 − 3μ, y = μ, z = 2μ − 3. Either is correct; a mark scheme accepts any valid parametrisation, and your check will confirm yours.

6No solution: an inconsistent system

Change one number in Worked example 2 and the picture changes completely.

Worked example 3. Paper 1. Show that the following system has no solution.

The two reduced equations ask −2y + z to be −3 and −1 at the same time. No values of x, y and z can do that, so the system is inconsistent and has no solution. The conclusion must be stated in words: "0 = 2 is impossible, so there is no solution".

Geometrically there are two ways three planes can have no common point, and Figure 4 shows both. In (c) each pair of planes meets in a line, but the three lines are parallel, forming a triangular prism: this is what happens in Worked example 3, where no two of the planes are parallel. In (d) two of the planes are parallel and never meet at all. AHL 3.18 treats these arrangements in detail.

7Systems with a parameter

The favourite Paper 1 question puts an unknown constant into the system and asks when the solution stops being unique. Reduce as usual, carrying the constant along; the last row will have a coefficient that depends on it.

Worked example 4. Paper 1. Consider the system

(a) Find the value of k for which the system does not have a unique solution. (b) For that value of k, find the value of m for which the system has infinitely many solutions, and find the general solution.

Figure 5 lays out the decision. The logic generalises: a coefficient that can be zero splits the question into "unique" (coefficient not zero) and "not unique" (coefficient zero), and the not-unique case splits again on whether the right-hand side is also zero.

When k ≠ −2 the last row gives z = (m + 3)/(k + 2) directly, and back-substitution finds x and y. A question may ask for that unique solution in terms of k and m; the method is the same, only the algebra is heavier.

8Solving with technology

On Paper 2 the GDC does the arithmetic, and the guide expects you to use it.

• A simultaneous equation solver (or matrix mode) finds a unique solution directly: enter the coefficients and right-hand sides row by row. When there is no unique solution, most solvers report "no solution" or "infinite solutions" rather than the general solution.
• The rref function (reduced row-echelon form) applied to the augmented matrix gives the fully reduced array. A last row of 0 0 0 | 1 means inconsistent; a last row of zeros means infinitely many, and the other rows give the general solution once you set the free unknown equal to λ.

Worked example 5. Paper 2. A café sells small, medium and large coffees. On Monday it sold 50 small, 80 medium and 40 large for €533. On Tuesday it sold 60 small, 70 medium and 50 large for €564. On Wednesday it sold 40 small, 50 medium and 60 large for €488. Find the price of each size. (Invented data.)

A small coffee costs €2.50, a medium €3.20 and a large €3.80. Check one day by hand: 50 × 2.5 + 80 × 3.2 + 40 × 3.8 = 125 + 256 + 152 = 533.

If a solver reports "no solution" or "infinite solutions" for a question that must have one answer, the likeliest cause is a coefficient typed wrongly. Re-enter the matrix before concluding anything.

Write down the equations you entered on Paper 2, not just the answer: the equations carry the method marks, and a typo in one coefficient is then a lost A mark, not the whole question.