IB Analysis and Approaches HL · Systems of linear equations · Section 2 of 3

Gaussian elimination method

3 worked examples · By Álvaro López

Key idea

The Gaussian elimination method transforms a system into an equivalent one (with the same solutions) but in echelon form, where it is solved from the bottom up. Besides giving the solution, the last row tells you whether the system has one solution, infinitely many or none. In the IB exam it is the tool for justifying by hand what the calculator does not explain.

Equivalent systems

Two systems are equivalent if they have exactly the same set of solutions. The idea of the method is to replace the system with an equivalent one that is simpler each time, until the solution can be read almost by itself.

To avoid losing or inventing solutions, only three transformations are allowed. The rows are labelled F1, F2, F3 (the first, second and third equation).

The intuition: if two equalities are true at the same time, so is their sum, and from the sum you recover the original by subtracting. Multiplying by $0$ would destroy information (the equation would become $0 = 0$ regardless of anything), which is why it is forbidden. When an operation mixes rows, the row that changes is the one on the left of the arrow; the other one stays as it was.

What we are aiming for: echelon form

The goal is for each equation to start with an unknown later than the previous one, so that the zeros form a staircase:

$⎩ ⎨ ⎧ a_{1} x + b_{1} y + c_{1} z b_{2} y + c_{2} z c_{3} z = d_{1} = d_{2} = d_{3}$

To achieve this, you work column by column: first use an equation with a convenient coefficient of $x$ (ideally $1$ ) to eliminate $x$ from the other two; then use the second one to eliminate $y$ from the third.

Write down each operation next to the row that changes (for example, " $F 2 \to F 2 - 2 F 1$ "). In the IB exam, steps with clear operations earn method marks even if there is an arithmetic slip later.

How each type is recognised in echelon form

A row of the form $0 x + 0 y + 0 z = k$ (all coefficients zero) summarises the two extreme possibilities: it is true for any values of the unknowns only if $k = 0$ .

Last row obtained	What it means	Type of system
$c z = d$ with $c \neq = 0$	$z$ is fixed and, working backwards, so are $y$ and $x$	Consistent, unique solution
$0 = k$ with $k \neq = 0$	An impossible equality	Inconsistent: no solution
$0 = 0$	The equation adds no condition	Consistent, infinitely many solutions

If $0 = k$ with $k \neq = 0$ appears in any row, the system is inconsistent, whatever the other equations say.

In the third case there are fewer useful equations than unknowns, so some unknown is free and is given a parameter. The number of parameters depends on how many rows vanish completely:

one row $0 = 0$ (two useful equations): one parameter, $t$ . The solutions form a line of points $(x, y, z)$ .
two rows $0 = 0$ (only one useful equation): two parameters, $t$ and $s$ . The solutions form a plane.
three rows $0 = 0$ : all three unknowns are free (this happens only if all the coefficients and constant terms are zero).

In the $3 \times 3$ system of this course, the usual case in the exam is the one with one parameter.

Abbreviated notation with the augmented matrix

As the unknowns only mark positions, the system can be written without them, using a table of coefficients and a bar before the constant terms. The row operations are identical:

$⎩ ⎨ ⎧ x + 2 y - z 2 x + y + z x - y + 2 z = 3 = 1 = 4 ⟷ 121 21 - 1 - 1 12 314$

It is just a faster way of writing. The column of $x$ holds its coefficients, that of $y$ its own, that of $z$ its own, and what lies to the right of the bar are the constant terms. A row $(000 ∣ k)$ is equivalent to $0 = k$ .

To see how the procedure works with the different types of system, use the three examples below. To recognise systems and their geometric interpretation, review linear equations and systems. When an unknown letter appears in the coefficients, move on to systems with a parameter.

Worked examples

Example 1 Basic

A system with a unique solution

Solve the following system by Gaussian elimination:

$⎩ ⎨ ⎧ x + y + z 2 x - y + z x + 2 y - z = 6 = 3 = 2$

Solution

The first row already has coefficient $1$ for $x$ . We eliminate $x$ from the other two rows:

$F 2 F 3 \to F 2 - 2 F 1 : \to F 3 - F 1 : - 3 y - z y - 2 z = - 9 = - 4$

The equivalent system is

$⎩ ⎨ ⎧ x + y + z - 3 y - z y - 2 z = 6 = - 9 = - 4$

We eliminate $y$ from F3 without fractions, with $F 3 \to 3 F 3 + F 2$ :

$3 (y - 2 z) + (- 3 y - z) = - 7 z 3 (- 4) + (- 9) = - 21$

The row becomes $- 7 z = - 21$ . The echelon system is

$⎩ ⎨ ⎧ x + y + z - 3 y - z - 7 z = 6 = - 9 = - 21$

Back substitution:

From $- 7 z = - 21$ : $z = 3$ .
From $- 3 y - z = - 9$ : $- 3 y - 3 = - 9$ , so $y = 2$ .
From $x + y + z = 6$ : $x + 2 + 3 = 6$ , so $x = 1$ .

Check in the original F2: $2 \cdot 1 - 2 + 3 = 3$ . Correct. In the original F3: $1 + 4 - 3 = 2$ . Correct.

The last coefficient is non-zero, so the system is consistent with a unique solution, namely $(x, y, z) = (1, 2, 3)$ .

Example 2 Medium

A system with no solution

Investigate the following system and solve it, if possible:

$⎩ ⎨ ⎧ x + 2 y - z 2 x + y + z x - y + 2 z = 3 = 1 = 4$

Classify it as consistent with a unique solution, consistent with infinitely many solutions, or inconsistent.

Solution

We write the augmented matrix and eliminate $x$ from F2 and F3:

$121 21 - 1 - 1 12 314 F 2 \to F 2 - 2 F 1 F 3 \to F 3 - F 1 100 2 - 3 - 3 - 1 33 3 - 5 1$

The rows F2 and F3 have the same coefficients in $y$ and $z$ but different constant terms. Subtracting: $F 3 \to F 3 - F 2$ :

$100 2 - 3 0 - 1 30 3 - 5 6$

The last row says $0 x + 0 y + 0 z = 6$ , that is, $0 = 6$ , which is impossible. No triple $(x, y, z)$ satisfies the three equations at once.

The system is inconsistent: it has no solution.

Example 3 Hard

A system with infinitely many solutions

Consider the system

$⎩ ⎨ ⎧ x - y + z 2 x + y - z 5 x + y - z = 3 = 3 = 9$

a) Reduce the system to echelon form and classify it as SCD, SCI or SI.

b) Find the general solution, using $t$ as the parameter and $z = t$ .

c) Among all the solutions, find the one that satisfies $x + y + z = 7$ .

Solution

a) We eliminate $x$ from F2 and F3 using the first row:

$F 2 F 3 \to F 2 - 2 F 1 : \to F 3 - 5 F 1 : 3 y - 3 z 6 y - 6 z = - 3 = - 6$

The third row is twice the second. We do $F 3 \to F 3 - 2 F 2$ :

$6 y - 6 z - 2 (3 y - 3 z) = 0 - 6 - 2 (- 3) = 0$

The row becomes $0 x + 0 y + 0 z = 0$ . The echelon system is

$⎩ ⎨ ⎧ x - y + z 3 y - 3 z 0 = 3 = - 3 = 0$

There is a row $0 = 0$ and none of the form $0 = k$ with $k \neq = 0$ : the system is consistent with infinitely many solutions, with $3 - 1 = 2$ useful equations and one free parameter.

b) We take $z = t$ . From the second row, $3 y - 3 t = - 3$ , so $y = t - 1$ . From the first, $x - (t - 1) + t = 3$ , that is $x + 1 = 3$ , so $x = 2$ . The general solution is

$(x, y, z) = (2, t - 1, t), t \in R$

Check in the original F3: $5 \cdot 2 + (t - 1) - t = 9$ . Correct for any $t$ .

c) We impose $x + y + z = 7$ :

$2 + (t - 1) + t = 7 \Rightarrow 2 t + 1 = 7 \Rightarrow t = 3$

Then $(x, y, z) = (2, 2, 3)$ .

In the exam: when you obtain the row $0 = 0$ , write the conclusion "consistent system with infinitely many solutions" and give the general solution with its parameter; giving only one particular solution does not count as solving the system. If instead of $0 = 0$ you had $0 = k$ with $k \neq = 0$ , the answer would be "inconsistent".

Common mistakes

Summary

Two systems are equivalent if they have the same solutions. You move from one to another with three operations: swapping rows, multiplying a row by a non-zero number, and adding to a row a multiple of another ( $F 2 \to F 2 - 2 F 1$ ).
Goal: echelon form (the zeros form a staircase). Eliminate $x$ from F2 and F3, then $y$ from F3; then solve by back substitution.
Last row $cz = d$ with $c \neq = 0$ : unique solution. Row $0 = k$ with $k \neq = 0$ : inconsistent. Row $0 = 0$ : consistent with infinitely many solutions.
One row $0 = 0$ gives one parameter; two rows $0 = 0$ give two parameters ( $t$ and $s$ ).
The augmented matrix is a shorthand: you operate on its rows exactly as on the equations.
Paper 2: the GDC solves systems with a unique solution; with infinitely many solutions or none you must reason by hand. Paper 1: everything by hand, writing down the operations.

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