The Further Mathematics Support Programme
Degree Topics in Mathematics
Gaussian Elimination and Row Echelon Form
Who was Gauss?
Johann Carl Friedrich Gauss was born in 1777 and showed academic
prowess from an early age. In primary school he amazed his teacher
by quickly answering the problem that he had been set:
“Add up the integers 1 to 1000”
Gauss listed the numbers in both ascending and descending order
noticing that each vertical pairing of numbers gave an equal sum of 1001 when added together.
1 +
2 + 3 + 4 + ……..+ 997 + 998 + 999 + 1000
1000 + 999 + 998 + 997 + …….+ 4
1001 1001 1001 1001
+ 3
+ 2
+ 1
1001 1001 1001 1001
Gauss realised there were 1000 pairs which each added to 1001 and this represented double
the total of the 1000 integers. So the total of the integers 1 to 1000 was:
(1000 x 1001) ÷ 2 = 500 500.
Can you use the same idea to prove the formula for the sum of an
arithmetic series?
{
(
) }
Gauss went on to write the influential number theory text ‘Disquisitiones Arithmeticae’ which
was influential in the development of this area of mathematics as we study it today. For
example, he proved the theorem:
“Every positive integer can be expressed as a sum of at most three triangular numbers”
Check this theorem for some positive integers and see if you can identify the triangular
numbers in the sum, e.g. 17 = 1 + 6 + 10 and 25 = 10 + 15.
Gauss also worked with astronomers and physicists to model the motion of planets and circuit
laws in electricity.
In the activities below some of Gauss’ work on linear algebra is discussed.
Image taken from http://www.thefamouspeople.com/profiles/carl-f-gauss-442.php
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Gaussian (or Gauss- Jordan) Elimination is a method used to solve linear simultaneous
equations.
It is particularly useful when there are 3 or more variables, as manipulating the algebra in these
cases can get quite complicated. The method uses matrices.
What is a matrix?
A matrix is an array (grid) of numbers. They are used in a number of ways within mathematics.
In this situation, the numbers correspond to coefficients and the solutions to the equations.
For example, consider the equations:
Solving these by elimination or substitution leads to the solution
.
Gaussian elimination could also be used. The process starts by setting up an augmented
matrix as follows:
[
]
The left hand side is formed from the coefficients of the two equations, with x coefficients first,
followed by y coefficients. The right hand side is formed from the solutions to the two
equations. A dotted line is used to separate the two parts.
To think about….
All systems of linear equations have either no solution, exactly one solution or an infinite
number of solutions. Thinking about two equations in two unknowns, can
you explain why?
We know from solving simultaneous equations by elimination that the following steps do not
change the solution to a system of linear simultaneous equations:


Multiplying through an equation by a constant;
Adding or subtracting the equations from each other.
These are known as row operations.
Gaussian elimination involves carrying out row operations to adapt the system of equations to a
format with 1s on the leading diagonal and zeroes elsewhere on the left hand side of the matrix.
i.e. [
] This set of equations would have the solution
.
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An example of Gaussian Elimination (2 equations in 2 unknowns)
Solving the system of equations shown above would proceed as follows:
[
Divide row 1 by 2 to get [
]
]. The top left hand entry is now correct.
Now subtract 4 x row 1 from row 2 to obtain [
]. The first column is now in the correct
format.
Next divide row 2 by -7 to get [
] which makes the bottom right entry in the left hand side
of the matrix correct.
Finally subtract 1.5 x row 2 from row 1 to get the final matrix [
We can now see that
].
as earlier.
In which order do I change the left hand side of the matrix?
Carry out row operations to make the following changes occur, in this order:




Make top left entry equal to 1
Make bottom left entry equal 0
Make bottom right entry equal 1
Make top right entry equal 0
A harder example of Gaussian Elimination (2 equations in 2 unknowns)
Use Gaussian elimination to solve the equations:
First form an augmented matrix [
] and think about how to form the correct format on the
left hand side of the line.
Divide row one by 5 to give [
] so that there is a 1 in the top left corner.
Now carry out the calculation R2 – 2xR1 (i.e. row two, minus double row one) to get [
Note that the first column is now in the correct format.
Next, carry out R2 ÷
to get [
].
]
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The final step is to make the
into a zero, so carry out R1 + R2. This results in [
and so we can conclude that
]
.
Remember to use the a b/c
button on a calculator here to make
the calculations quicker.
Task 1
Solve the simultaneous equations:
using Gaussian elimination and one other method – check that your solution is the same by both
methods.
Task 2
Solve the simultaneous equations:
using Gaussian elimination and one other method – check that your solution is the same by both
methods.
Task 3
Research how to carry out Gaussian elimination to solve a system of three linear equations in
three unknowns.
What is the difference between Gaussian elimination and Gauss-Jordan elimination?
Use your research to solve the equations:
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Solution: To think about…
Two equations in two unknowns will only have three possible scenarios:
No solution
If two lines are parallel they will not intersect and so the two
equations will have no solution.
e.g.
One unique solution
If two lines meet in a single point, the two equations will have a
unique solution.
e.g.
Infinite solutions
If two lines are coincident (i.e. on top of each other) the two
equations will have an infinite number of solutions.
e.g.
Note, these two equations are essentially the same – dividing
the second equation by two gives the first equation. An infinite
number of pairs of values for and satisfy the equation(s).
Solution 1
] divide R1 by 6 to obtain [
Starting with [
[
] then carry out R2 – R1 to get
].
Now divide R2 by
to obtain [
].
Finally carry out R1 - R2 to get to the final augmented matrix of [
Therefore the solution is
correct.
].
and checking these in the original equations shows they are
You could also use the standard methods of elimination or substitution that you learnt at GCSE to
check your answers. For two equations, Gaussian elimination may be more time consuming, but
for larger sets of equations it is can be a better approach – see Task 3.
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Solution 2
The equations
are solved as follows:
Start with [
] and divide R1 by 3 to get [
obtain [
].
Next divide R2 by
which gives [
[
] and then carry out R2 – 5R1 to
] and then finally carry out R1 - R2 to obtain
].
Therefore the solution to the equations is
.
Solution 3
The following websites might be useful for researching how to solve systems of three linear
equations in three unknowns:





Oregon State University explanation
Youtube video (Gauss Jordan elimination) – this video partially illustrates the difference
between Gaussian elimination and Gauss- Jordan elimination.
Purplemath.com explanation of Gaussian elimination – this demonstration is
particularly useful as it shows the technique of ‘back substitution’ which is needed when
using Gaussian elimination (but not when carrying out Gauss-Jordan elimination – see
below).
Step by step guide which includes clear definitions of the terminology being used.
Cliffsnotes.com – a more detailed and complex explanation which looks at more
complicated cases of three equations in three unknowns and at the various geometrical
scenarios which might arise. (This is similar to the table above which shows all the possible
cases for two variable simultaneous equations).
Gaussian elimination only requires zeroes below the leading diagonal of 1s in the augmented
matrix; Gauss-Jordan elimination requires zeroes above the leading diagonal of 1s as well.
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Why not try solving the system of three equations using both methods?
Using Gaussian elimination and back substitution on this system of equations:
] and carry out R2 + R1 and R3 – 3R1 to
we start with the augmented matrix [
obtain [
]. The first column is now in the correct format.
Next, multiplying R2 by -1 gives [
[
] and then carrying out R3 + 10R2 we obtain
]. The second column is now correctly formatted.
Finally, dividing R3 by -52 gives [
].
Using back substitution:
R3 tells us that
R1 tells us that
Checking
and R2 shows
and so
and so we can deduce that
.
.
in the original equations shows this solution to be correct.