Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Chapter 1 - System of linear equations
Manuel Navarro García
(mnavarrogarcia@faculty.ie.edu)
IE University
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
1 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
1
Motivation
2
Gaussian elimination method
3
Matrices and systems of linear equations
4
Rouché-Frobenius theorem
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
Rouché-Frobenius theorem
IE University
2 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
1
Motivation
2
Gaussian elimination method
3
Matrices and systems of linear equations
4
Rouché-Frobenius theorem
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
Rouché-Frobenius theorem
IE University
3 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Initial example
A farmer has two types of milk, M1 with 24% fat and M2 with 18% fat. How many
liters of each should be mixed to obtain 150 liters of milk containing 20% fat?
How can we design a mixture that produces exactly 150 liters at 20% fat?
Any problem in this form has two key ingredients:
• Unknown quantities: decision variables that we need to determine.
• Constraints: restrictions that must be satisfied.
We must determine the liters of milk M1 and M2 . Thus, the unknown quantities of the
problem are
x1 = Liters of M1 , x2 = Liters of M2 .
The constraints are obtained by translating the conditions of the problem into equations:
1
Total volume. The sum of both amounts must be the final 150 liters:
x1 + x2 = 150.
2
Total fat content. Fat percentage is per liter, but in the equation we must count
liters of fat:
0.24x1 + 0.18x2 = 0.20 × 150 = 30.
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
4 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Geometric interpretation of the constraints
x2
• Each constraint describes a line in the
x1 x2 −plane.
x1 +
x2 = 150,
(50, 100)
0.24x1 + 0.18x2 = 30.
• The solution must satisfy both at the
same time, i.e., it is the intersection
point of the two lines:
(x1∗ , x2∗ ) = (50, 100).
x1 + x2 = 150
4x1 + 3x2 = 500
x1
The geometric method for solving systems of linear equations has important limitations:
• In R2 we can represent each equation as a line, and in R3 as a plane, but in higher
dimensions visualization is no longer possible.
• Even in R3 , accurately determining the exact coordinates of the solution from a
plot is often impractical.
We need a general, reliable method for higher dimensions.
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
5 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
1
Motivation
2
Gaussian elimination method
3
Matrices and systems of linear equations
4
Rouché-Frobenius theorem
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
Rouché-Frobenius theorem
IE University
6 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Definition of system of linear equations
A system of m linear equations in n unknowns x1 , . . . , xn is a family of the form
ai1 x1 + ai2 x2 + . . . + ain xn = bi , i = 1, . . . , m,
where the coefficients aij and the constants bi are given numbers. If all bi are zero,
the system is called homogeneous.
We are interested in the set of points that simultaneously satisfy all these equations.
First, we must define what a solution is.
A tuple (x1∗ , . . . , xn∗ ) ∈ Rn is a solution if it satisfies
ai1 x1∗ + ai2 x2∗ + . . . + ain xn = bi ,
for all i = 1, . . . , m. The solution set of the system is the set of solutions:
S = {(x1 , . . . , xn ) ∈ Rn : ai1 x1 + . . . + ain xn = bi , i = 1, . . . , m}.
Two systems of linear equations are equivalent if they have the same solution set.
In the remainder of the chapter, we pursue two main goals:
1
Design a procedure that transforms any system into an easier equivalent system
(with the same solutions) and that works in any dimension.
2
Characterize the solutions by determining when there is a unique solution, no
solution, or infinitely many solutions.
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
7 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Motivation of Gaussian elimination method
Consider the system of linear equations:

a11 x1 + a12 x2 + · · · + a1n xn = b1 ,






 a21 x1 + a22 x2 + · · · + a2n xn = b2 ,







.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
am1 x1 + am2 x2 + · · · + amn xn = bm .
Suppose we can transform the previous system into an equivalent system of the form

a11 x1 + a12 x2 + · · · + a1n xn = b1 ,





a22 x2 + · · · + a2n xn = b2 ,









.
.
.
.
.
.
amn xn = bm .
• This new system is triangular: each equation involves only one new variable in
addition to those already determined above it.
• Solving is now straightforward: start from the last equation and proceed upwards
by substitution.
• This is exactly what the Gaussian elimination method achieves.
To apply Gaussian elimination, we must clarify which transformations of a system
preserve its solution set.
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
8 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Elementary row transformations
To transform a system into an easier, equivalent one, we use the following elementary
row transformations:
1 Row exchange. Swap the position of two equations: (Rk ) ↔ (Rℓ ).
2
Row scaling. Multiply an equation by a constant λ ̸= 0: (Rk ) ← λ(Rk ).
Row replacement. Replace an equation by itself plus a multiple of another:
(Rk ) ← (Rk ) + λ(Rℓ ), k ̸= ℓ.
By repeatedly applying these transformations, we can convert any system into a triangular form, ready for back-substitution.
3
Applying any elementary row transformation to a system of linear equations does not
change its solution set.
Let (R1 ), . . . , (Rm ) be the equations of a given system.
1
2
Since the order of the equations is irrelevant, this is trivial.
Let k ∈ {1, . . . , m} and λ ̸= 0. Suppose (x1∗ , . . . , xn∗ ) is a solution of the original
system. In particular, it satisfies (Rk ):
ak1 x1∗ + · · · + akn xn∗ = bk .
Multiplying both sides by λ gives
(λak1 )x1∗ + · · · + (λakn )xn∗ = λbk ,
which is exactly λ(Rk ).The converse follows by dividing through by λ.
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
9 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Elementary row transformations
3
Let k, ℓ ∈ {1, . . . , m} with k ̸= ℓ. In the modified system, the k−th equation
becomes
(ak1 + λaℓ1 )x1 + · · · + (akn + λaℓn )xn = bk + λbℓ .
All other equations are unchanged.
(⊂) Suppose (x1∗ , . . . , xn∗ ) is a solution of the original system. Then
ak1 x1∗ + · · · + akn xn∗ = bk ,
aℓ1 x1∗ + · · · + aℓn xn∗ = bℓ .
Multiplying the second equality by λ and adding it to the first yields exactly the
new k−th equation. Thus, (x1∗ , . . . , xn∗ ) also solves the modified system.
(⊃) Conversely, suppose (x1∗ , . . . , xn∗ ) is a solution of the modified system. Then
(ak1 + λaℓ1 )x1∗ + · · · + (akn + λaℓn )xn∗ = bk + λbℓ .
Since (x1∗ , . . . , xn∗ ) also satisfies (Rℓ ) in both systems, subtracting λ(Rℓ ) from the
above equality recovers (Rk ). Hence, (x1∗ , . . . , xn∗ ) also solves the original system.
■
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
10 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Gaussian elimination method in action
Input: a system of m linear equations in n unknowns
ai1 x1 + ai2 x2 + · · · + ain xn = bi ,
i = 1, . . . , m.
Goal: transform it into an equivalent triangular system, then solve by back-substitution.
1
Pivot for x1 : choose an equation with a non-zero coefficient of x1 ; if needed, swap
it to the first position and scale it so that the coefficient of x1 is 1.
In the motivating example, this is already satisfied:
x1 +
x2 = 150,
0.24x1 + 0.18x2 = 30.
2
Eliminate x1 below: for each lower equation (Ei ) with coefficient ai1 , perform
(Ri ) ← (Ri ) −
ai1
(R1 ),
a11
so that x1 is removed from (Ri ).
Applying the transformation (R2 ) ← (R2 ) − 0.24 (R1 ), we get:
x1 +
x2 = 150,
0.06x2 =
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
6.
IE University
11 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Gaussian elimination method in action
3
Repeat for x2 , x3 , . . .: in the reduced subsystem, choose the next pivot with a nonzero coefficient, scale it to 1, and eliminate entries below. Continue until no more
elimination is possible.
1
In our example, scale the second equation by 0.06
:
x1 + x2 = 150,
x2 = 100.
At this stage, the system is in row echelon form: all pivot elements are 1, zeros appear
below each pivot, and each pivot is to the right of the pivot in the row above.
4
Back-substitution: solve the last equation, then substitute upward to find the
preceding variables.
Propagating the value of x2 in the first equation, we get:
x1 = 50,
x2 = 100.
This matches the geometric method but extends to any dimension.
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
12 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
1
Motivation
2
Gaussian elimination method
3
Matrices and systems of linear equations
4
Rouché-Frobenius theorem
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
Rouché-Frobenius theorem
IE University
13 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
From systems to matrices
So far, we have written systems of equations in expanded form:

a11 x1 + a12 x2 + · · · + a1n xn = b1 ,





a

 21 x1 + a22 x2 + · · · + a2n xn = b2 ,







.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
am1 x1 + am2 x2 + · · · + amn xn = bm .
• In Gaussian elimination, we modify the coefficients in a systematic way.
• The variables x1 , . . . , xn do not change order and appear in every equation.
• Writing all variables every time is repetitive and slow.
For our system, we define:


a11
a12
···
a1n
a22
···
a2n 
 a21


A= .
.
.  (coefficient matrix),
.
. 
 ..
.
.
am1 am2 · · ·
amn
 
 
x1
b1
x2 
 b2 
 
 
x =  .  (vector of unknowns), b =  .  (vector of constants).
 .. 
 .. 
xn
bm
If we define a meaningful matrix-vector product, the system becomes Ax = b.
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
14 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Matrix-vector product
Let A be an m × n matrix and x ∈ Rn a column vector. The product Ax is defined by:

a11
 a21

Ax =  .
 ..
am1
a12
a22
..
.
am2
···
···
···

  
a11 x1 + a12 x2 + · · · + a1n xn
x1
a1n




a x + a22 x2 + · · · + a2n xn 
x
a2n

  2   21 1
 ∈ Rm .
..
..   ..  = 

.
.  .  
xn
amn
am1 x1 + am2 x2 + · · · + amn xn
• The equation Ax = b means:
ai1 x1 + · · · + ain xn = bi ,
i = 1, . . . , m.
• This notation turns m equations into one.
For example, in our motivating problem:
(
x1 +
x2 = 150,
=⇒
0.24x1 + 0.18x2 = 30,
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
1
0.24
1
0.18
x1
x2
!
=
!
150
30
.
IE University
15 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Gaussian elimination with matrix form
We can apply elementary row operations directly to the augmented matrix
A b ,
using the notation (Ri ) for the i−th row. The initial system is written as:
1
1
150
.
0.24 0.18
30
Eliminate below the pivot with (R2 ) ← (R2 ) − 0.24 (R1 ):
1
1
150
.
6
0 0.06
1
Scale the pivot with (R2 ) ← − 0.06
(R2 ):
1
0
1
1
150
.
100
Eliminate above the pivot with (R1 ) ← (R1 ) − (R2 ):
1 0
50
.
0 1 100
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
16 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
1
Motivation
2
Gaussian elimination method
3
Matrices and systems of linear equations
4
Rouché-Frobenius theorem
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
Rouché-Frobenius theorem
IE University
17 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Rouché-Frobenius theorem
When applying Gaussian elimination, we transform a system into an equivalent one in
row echelon form:
• Some equations remain non-zero and contain essential information.
• Others may become zero rows (0 = 0), meaning they are redundant.
The number of non-zero rows tells us how many independent equations we have.
The rank of a matrix A, denoted rank(A), is the number of non-zero rows in any row
echelon form of A.
• The rank is invariant under elementary row operations.
• The inequality rank(A) ≤ min{m, n} always hold.
The system Ax = b is consistent (has at least one solution) if and only if
rank(A) = rank([A | b]) .
If the system is consistent:
• If rank(A) = n, the solution is unique.
• If rank(A) < n, there are infinitely many solutions, with n − rank(A) free variables.
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
18 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Proof of the Rouché-Frobenius theorem
Perform Gaussian elimination on the augmented matrix [A | b] to reach a row echelon
form [A′ | b′ ] that is equivalent to [A | b]. Let
r = rank(A) and R = rank([A | b]) .
By invariance of rank under elementary row operations, r (resp. R) is the number of
non-zero rows in any row echelon form of A (resp. [A | b]).
(⇒) Suppose Ax = b is consistent. Then [A′ | b′ ] cannot contain a row of the form
0, . . . , 0 α
with α ̸= 0,
because that row would encode the contradiction 0 = α. Hence every non-zero row in
[A′ | b′ ] has its pivot in the A′ block, so the number of non-zero rows in A′ equals the
number of non-zero rows in [A′ | b′ ]. Therefore r = R.
(⇐) Conversely, suppose r = R. This means that, after Gaussian elimination, the
augmented matrix has no row of the form
0, . . . , 0 α
with α ̸= 0.
Therefore, the reduced system is consistent: we can choose values for the free variables
(those without pivots) and then determine the pivot variables by back-substitution.
Multiplicity of solutions. If the system is consistent:
• If r = rank(A) = n, then every column of A contains a pivot, so there are no free
variables and the solution is unique.
• If r < n, then there are n−r free variables, which yield to infinitely many solutions.
■
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
19 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Case I: Unique solution
We start with the system of equations represented by the augmented matrix:
1
1
150
.
0.24 0.18
30
We transform it into the row echelon form
aplying elementary row operations:
1 0
50
.
0 1 100
(50, 100)
x1 + x2 = 150
Observe that:
rank(A) = 2,
x2
rank([A | b]) = 2.
By the Rouché-Frobenius theorem, the system is consistent and it has a unique solution:
4x1 + 3x2 = 500
x1
(x1∗ , x2∗ ) = (50, 100).
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
20 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Case II: Infinitely many solutions
We start with the system of equations represented by the augmented matrix:
1 1 150
.
2 2 300
x2
We transform it into the row echelon form applying elementary row operations:
1 1 150
.
0
0 0
infinitely many solutions
Observe that:
rank(A) = 1,
rank([A | b]) = 1.
By the Rouché-Frobenius theorem, the system is consistent and has infinitely many solutions:
(
x1 = µ,
x1 + x2 = 150 ⇒
µ ∈ R.
x2 = 150 − µ,
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
x1 + x2 = 150
2x1 + 2x2 = 300
x1
IE University
21 / 22
Motivation
Gaussian elimination method
Matrices and systems of linear equations
Rouché-Frobenius theorem
Case III: No solution (two parallel lines)
We start with the system of equations represented by the augmented matrix:
1 1 150
.
1 1 120
x2
We transform it into the row echelon form
aplying elementary row operations:
1 1 150
.
0 0 −30
Observe that
rank(A) = 1,
x1 + x2 = 150
rank([A | b]) = 2.
By the Rouché-Frobenius theorem, the system is inconsistent and it has no solutions.
x1 + x2 = 120
no intersection
x1
This is the empty intersection of two parallel
lines.
Manuel Navarro García (mnavarrogarcia@faculty.ie.edu)
Chapter 1 - System of linear equations
IE University
22 / 22