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Δβα\Delta _{\beta \alpha}
The Δelta βeta αrchivesThe \ \Delta elta \ \beta eta \ \alpha rchives
CHAPTER 01
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Linear functions

CHAPTER 02
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Quadratic functions

under construction
CHAPTER 03
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Functions

under construction
CHAPTER 04
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Trigonometry

under construction
CHAPTER 05
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Trigonometric functions

under construction
CHAPTER 06
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Linear algebra

under construction
CHAPTER 07
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Vectors

under construction
CHAPTER 08
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Logarithms

under construction
CHAPTER 09
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Complex numbers

under construction
CHAPTER 10
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Differentiation

under construction
CHAPTER 11
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Integration

under construction

A deeper dive into Gaussian elimination06_01A

From Chapter 1, you know that there's different forms of an augmented matrix that lets you find solutions to systems involving 2 equations, and those rules almost exactly apply to systems with 3 (the only difference being that we are now working in 3D space, hence the additional variable
zz
):

Row forms

1 solution:
(1□□□01□□001□)\begin{pmatrix} 1 & \square & \square & \square \\ 0 & 1 & \square & \square \\ 0 & 0 & 1 & \square \\ \end{pmatrix}
0 solutions:
(1□□□01□□000□)\begin{pmatrix} 1 & \square & \square & \square \\ 0 & 1 & \square & \square \\ 0 & 0 & 0 & \square \\ \end{pmatrix}
Infinite solutions:
(1□□□01□□0000)\begin{pmatrix} 1 & \square & \square & \square \\ 0 & 1 & \square & \square \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}
Now, with three equations, we can find points of intersection in 3D space. When you solve these, the best way to achieve the zeros are as follows: get the zero in the center left first, then the bottom left, then the bottom middle.

Solving systems with Gaussian elimination

Consider the system
{x−2y+5z=12x−y+8z=2−3x−11z=−3\left\{ \begin{array}{l} x-2y+5z=1 \\ 2x-y+8z=2 \\ -3x-11z=-3 \end{array} \right.
.
As usual, transition the system into a matrix to get
(1−2512−182−30−11−3)\begin{pmatrix} 1 & -2 & 5 & 1 \\ 2 & -1 & 8 & 2 \\ -3 & 0 & -11 & -3 \\ \end{pmatrix}
(note how, since there's no
yy
-variable for that last function, I just subbed in a 0)
Then perform the operation
R2=2R1−R2R_2=2R_1-R_2
to get that middle left zero:
(1−2510−320−30−11−3)\begin{pmatrix} 1 & -2 & 5 & 1 \\ 0 & -3 & 2 & 0 \\ -3 & 0 & -11 & -3 \\ \end{pmatrix}
Then perform the operation
R3=−3R1−R3R_3=-3R_1-R_3
to get that bottom left zero:
(1−2510−32006−40)\begin{pmatrix} 1 & -2 & 5 & 1 \\ 0 & -3 & 2 & 0 \\ 0 & 6 & -4 & 0 \\ \end{pmatrix}
Finally perform the operation
R3=2R2+R3R_3=2R_2+R_3
to get those bottom zeros:
(1−2510−3200000)\begin{pmatrix} 1 & -2 & 5 & 1 \\ 0 & -3 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}
You can see however, we get infinite solutions (because the bottom row is all zeros). From here, we assign
z=tz=t
and sub that
tt
into the equations we extract from the matrices above. So,
−3y+2(t)=0-3y+2(t)=0
which re-arranges to produce
y=2t3y=\frac{2t}{3}
and then sub that and
tt
into
x−2(2t3)+5t=1x-2\left(\frac{2t}{3} \right)+5t=1
which re-arranges to get
x=1−11t3x=1-\frac{11t}{3}
.
These are therefore your solutions:
z=tz=t
,
y=2t3y=\frac{2t}{3}
, and
x=1−11t3x=1-\frac{11t}{3}
The example above, shows how you solve it when a) a variable is missing, and b) there are infinite solutions. All questions related to simultaneous equations will take any of these forms:
- Find the solutions of a system
- Find the value of a variable in a system (like
aa
) which makes the system have no, one, or infinite solutions
So, you should only really focus on the reduced row forms for this topic, and from there, you can find any answer you need! Below are some practice questions.

Solving systems using Gaussian elimination

1) Consider the system
{x−2y−z=−12x+y+3z=13x+8y+9z=a\left\{ \begin{array}{l} x-2y-z=-1 \\ 2x+y+3z=13 \\ x+8y+9z=a \end{array} \right.
. For what values of
aa
does the system have no solutions?
2) Consider the system
{x+2y+z=32x−y+4z=1x+7y−z=k\left\{ \begin{array}{l} x+2y+z=3 \\ 2x-y+4z=1 \\ x+7y-z=k \end{array} \right.
. For what values of
kk
does the system have no solutions? How about infinite solutions? Find the solutions in this case. Finally, explain why the system never has a unique solution.
3) Consider the system
{x+4y+mz=−m(m+1)x+4y+z=14x+4y+z=1\left\{ \begin{array}{l} x+4y+mz=-m \\ (m+1)x+4y+z=1 \\ 4x+4y+z=1 \end{array} \right.
. Show that for one value of
mm
, the system has no solutions. Show that for a different value of
mm
, the system has infinitely many solutions and find the solutions in this case.
4) Consider the system
{x+4y−z=k−2x+z=35x+4y+(k−6)z=−3\left\{ \begin{array}{l} x+4y-z=k \\ -2x+z=3 \\ 5x+4y+(k-6)z=-3 \end{array} \right.
. Show that for one value of
kk
, the system has infinitely many solutions and find the solutions in this case. Show that there is a unique solution for all other values of
kk
. Write this solution in terms of
kk
.
That is the extent to which the IB includes linear algebra in their course, hence why this page is quite short!

Practical applications06_02A

Gaussian elimination and 3D simultaneous equations have wide-ranging practical applications in various fields, including engineering, economics, physics, and computer science, where they are used to solve complex systems of linear equations for tasks that have more than two variables that need to be taken into account.

Practical applications

1) At a busy intersection, traffic engineers are studying the flow of vehicles. They have the following data, where
xx
,
yy
, and
zz
represent the number of cards from three different directions. Determine the traffic flow from each direction.
Road A:
x+y+z=1200x+y+z=1200
(cars per hour)
Road B:
2x+y+3z=21002x+y+3z=2100
(cars per hour)
Road C:
x+3y+2z=1800x+3y+2z=1800
(cars per hour)
2) An economist is analyzing the relationship between three economic factors: GDP growth (
xx
), unemployment rate (
yy
), and inflation rate (
zz
). Based on historical data, they have derived the following equations. Find the values of
xx
,
yy
, and
zz
.
2x+3y−z=42x+3y-z=4
x−y+2z=3x-y+2z=3
3x+2y+z=73x+2y+z=7
3) An electrical engineer is analyzing a circuit with three loops. Using Kirchhoff's laws, they derive the following systems of equations where
I1I_1
,
I2I_2
,
I3I_3
are currents in each loop. Solve for the currents.
5I1−2I2+I3=105I_1-2I_2+I_3=10
−2I1+7I2−I3=5-2I_1+7I_2-I_3=5
I1−I2+4I3=3I_1-I_2+4I_3=3
4) A company produces three types of products (A, B, and C) using three resources (labor, materials, and machinery). The constraints are the following, where
xx
,
yy
, and
zz
represent the number of units of products
AA
,
BB
, and
CC
respectively. Find the maximum possible production of each product.
2x+y+3z≤10002x+y+3z \leq 1000
(labor hours)
x+3y+2z≤1500x+3y+2z \leq 1500
(material units)
3x+2y+z≤12003x+2y+z \leq 1200
(machine hours)

Topics coming soon

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