Ranks of matrices and the Rouché-Capelli Theorem
Marco Tolotti
Introduction
Consider the linear system
Ax = b,
0
where A = (aij )m×n , x = (x1 , ..., xn ) and b = (b1 , ..., bm )0 .
To solve the system means to identify the set S ∈ Rn defined by
S = {x ∈ Rn |Ax = b}.
There are different methods to solve it (Gaussian elimination, Substitution, ...)
We say that a linear system is consistent if it is solvable, i.e., when S 6= ∅.
A linear system may be consistent or not; in particular, if consistent, it can admit one
unique solution or infinite solutions, depending on the matrix A of the coefficients.
No other possibilities apply.
Is it possible to determine the number of solutions without necessarily solve it?
The answer is affirmative: by means of the Theorem of Rouché-Capelli.
We need some prerequisites in order to state the theorem.
The rank of a matrix
We recall the definition of minor of a matrix.
Definition 1 Given the matrix A = (aij )m×n , we call minor of order k the determinant of any square submatrix that can be constructed by A cutting (a certain number)
of rows and/or columns.
From this definition it is clear that the order of the minors that can be extracted by
A cannot be larger than the minimum between m and n. Suppose indeed that m ≤ n
(i.e. there are less rows than columns), it is not possible to obtain a square matrix
from A cutting rows or columns, whose dimension is larger than m.
Example 2 Consider
2 1 0
A=
1 6 4
Find all the minors of A of order 2 and at least one minor of order 1 that is non-zero.
Is it possible to obtain a minor of order k > 2?
Answer:
The minors of order two are obtained by cutting one of the three columns of A. Hence
we can define three minors of order 2. These are
2 1 2 0 1 1 1 6 ;
1 4 ;
6 4 .
M.Tolotti - Mathematics (Part One) - September 2011
Recall that the notation |A| states for the
2 1 1 6 = 2 · 6 − 1 · 1 = 12 − 1 = 11 ;
2
determinant of the matrix A. Hence
2 0 1 0 = ... = 8 ;
1 4 6 4 = ... = 4.
A minor of order one is the determinant of any entry of the matrix. For instance
|a11 | = |2| is a minor of A of order 1. The determinant of a scalar is the scalar itself
so |2| = 2 which is different from zero. This is one example of non-zero minor of A of
order 1.
Concerning the existence of minors of order k > 2. Take for instance k = 3. It is not
possible to find such a minor. In fact, cutting rows or columns of A, we can not obtain
a 3 × 3 matrix. Obviously it is not possible to find an even larger one. 2
Now we state an important result that can be useful when computing the rank of a
matrix.
Definition 3 We define ρ(A) ( rank of A) the maximum number of linearly independent rows or columns of A.
Proposition 4 The rank of the matrix A = (aij )m×n coincides with the order of the
largest non-zero minor that can be extracted by A.
Put differently we say that a matrix A has rank k if and only if there exists at least
one minor different from zero of order k whereas all the minors of order larger than k
are indeed zero.
If there were a minor of order k + 1 different from zero, then the rank of A would be
at least k + 1.
Example 5 Look at the matrix
A=
2 1 0
1 6 4
defined in Example 2. Find its rank.
Answer:
2 1 ,
Since there exists at least one minor of A different from zero, for instance 1 6 the rank of A is at least 2. Moreover we have saw that it is not possible to extract a
minor of order k > 2. Thus the rank of A is 2. 2
Rules for the calculus of the rank.
Given the matrix A = (aij )m×n .
i) rank(A) is an integer number.
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ii) rank(A) ≥ 0, in particular rank(A) = 0 if and only if A = 0, where 0 denotes
the zero matrix.
iii) rank(A) ≤ min(m; n): the rank of A is at most equal to the minimum between
the number of rows and columns.
iv) As a consequence the following relationship holds:
0 ≤ rank(A) ≤ min(m; n).
Example 6 Find the rank of


2 3
1
M =  1 −1 3  .
0 4 −4
Answer:
We start from the smallest minors:
|2| =
6 0 ⇒ rank(M ) ≥ 1;
2 3 1 −1 6= 0 ⇒ rank(M ) ≥ 2;
2 3
1
1 −1 3 = 0 ⇒ rank(A) < 3.
0 4 −4 What about the last implication? The rank of M is equal to three only if we can
find a minor of order 3 different from zero. The only minor of order 3 is the determinant of the matrix itself! (it is a 3 × 3 matrix). Thus the fact that the determinant
of M is equal to zero, implies that there are no minors of order 3 different from zero,
hence rank(M ) can not be 3 (it is smaller). Since rank(M ) < 3 and rank(M ) ≥ 2 it
is equal to 2. 2
Remark 7 As a consequence of the previous example, we conclude (and it can be
proved formally) that the rank of a square matrix of order n is equal to n if
and only if its determinant is different from zero.
Number of solutions of a linear system
Consider the linear system Ax = b where A = (aij )m×n , x = (x1 , ..., xn )0 and b =
(b1 , ..., bm )0 . We say that s ∈ Rn is a solution of the linear system if As = b, i.e. if
s ∈ S.
As we have already discussed, a linear system may have zero, one or infinite solutions
depending on the matrix of the coefficients and on the vector b.
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The following theorem tells us how to choose among these three possibilities just
looking at the rank of the matrix A and the rank of the so called augmented matrix
[A|b] obtained adding a column to A equal to the vector b. It is explicitly defined as


a11 ... a1n b1

..
..  .
[A|b] =  ... ...
.
. 
am1 ... amn bm
The last concept we need is the concept of degree of freedom. When solving a system
of equations, a degree of freedom is one of the variables that can be freely chosen,
while the remaining variables are then determined once the free variables has been
assigned.
As a consequence, if there are zero degrees of freedom, the solution (if it exists) is
necessarily unique. If there are degrees of freedom, the solutions are infinite.
Consider for example the following linear system
x1 + x2 + x3 = 0
x2
= 2
It is easy to see that either x1 or x3 can be chosen freely. Suppose to choose x3 = α
where α ∈ R. x3 is the free parameter of this system. Then the first equation
x1 + x2 + x3 = 0 can be written as x1 = −x2 − x3 . The second equation tells us that
x2 = 2 is fixed and x3 = α, so that x1 = −2 − α. The solutions of the system are thus


−2 − α

 ;
2
α ∈ R.
α
Notice that we have one degree of freedom. So that the solutions of the system are
infinite.
Theorem 8 (Rouché-Capelli) The system Ax = b admits solutions (it is consistent) if and only if
rank(A) = rank(A|b).
Moreover if the system is consistent, the number of degrees of freedom is equal to
n − rank(A),
where n is the number of unknowns of the system.
The first part of the theorem tells us wether there are solutions or not. The second
part tells us that the solution is unique only if rank(A) = n. In this case in fact the
number of degrees of freedom is zero. Otherwise there is a positive number of degrees
of freedom and thus there are infinite solutions.
Example 9 Consider the system

 3x1 + 5x2 = 9
2x1 − 13 x2 = −5

−x1
= 2
Is this system consistent? How many solutions does it admit?
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Answer:
We rely on the Theorem of Rouché-Capelli. We need to compute the rank of A and
of [A|b], where




3
5
9
3
5
[A|b] =  2 − 13 −5  .
A =  2 − 13  ;
−1 0
−1 0
2
3 5 6= 0, the rank of A is at least 2. On the other hand min(m; n) = 2 so
Since 2 − 31 that its rank is at most 2. Hence rank(A) = 2. Concerning [A|b], its rank must be at
least 2 since necessarily1
rank(A) ≤ rank(A|b).
[A|b] is a 3 × 3 matrix, its rank can be at most 3. We have to check if it is exactly 3.
To do so we compute its determinant
3
5
9
1
2 − −5 = ... = 0.
3
−1 0
2 As a consequence rank(A|b) = 2. We now apply the theorem. First of all we see that
rank(A) = rank(A|b), so that the system is consistent.
Moreover n − rank(A) = 2 − 2 = 0 this means that the number of degrees of freedom
is zero. Hence the solution is unique. 2
We stress the fact that the Theorem of Rouché-Capelli does not help you in finding
the solutions of the system! It gives only information about the number of solutions.
Nevertheless, it is possible to rely on this theorem to reduce any linear system to a
square linear system (same number of equations and unknowns) that can be solved
by Cramer’s method.
Example 10 Consider the system

 x1 + 2x2 + x3 = 1
x1 − x2 − x3 = 2

2x1 + x2
= 3
Is this system consistent? How many solutions does it admit? Find all the solutions
of the system.
Answer:
We rely on the Theorem of Rouché-Capelli. We need to compute the rank of A and
of [A|b], where




1 2
1
1 2
1 1
A =  1 −1 −1  ;
[A|b] =  1 −1 −1 2  .
2 1
0
2 1
0 3
1
Why? Because A is a submatrix of [A|b]. Then, looking at the definition of rank, the rank of a
matrix is at least equal (or larger) than the rank of any submatrix.
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1 2 6= 0, the rank of A is at least 2. We compute the determinant of A to
Since 1 −1 check whether the rank is 3 or less.
1 2
1
1 −1 −1 = ... = 0 ⇒ rank(A) < 3.
2 1
0 So we conclude that rank(A) = 2.
Concerning [A|b], its rank must be at least 2 since necessarily rank(A) ≤ rank(A|b).
[A|b] is a 3 × 4 matrix, its rank can be at most 3. We have to check if it is exactly
3. To do so we compute its minor of order 3. Notice that (arguing similarly as in
Example 2) we need to compute 4 different minors! In fact if at least one is non-zero
then the rank is equal to three. The minors are
1 2
1 2 1 1 1 1 2
1 1 1 1 −1 −1 ; 1 −1 2 ; 1 −1 2 ; −1 −1 2 .
2 1
2 1 3 2 0 3 1
0 0 3 It can be verified that all these minors are worth zero. As a consequence rank(A|b) =
2. We now apply the theorem. First of all we see that rank(A) = rank(A|b), so that
the system is consistent.
Moreover n − rank(A) = 3 − 2 = 1 this means that the number of degrees of freedom
is 1. The solutions are thus infinite.
We have now to find the solutions. One possible way is to reduce the system to
an auxiliary square system that can be used to determine easily the solutions of the
system. Here we explain this procedure.
• Since the rank of A is 2, identify a 2 × 2 submatrix à of A whose determinant
1 2
is non-zero. E.g. Ã =
.
1 −1
• Delete all the rows of A out of Ã. In our example: (2, 1, 0). We then have
1 2
1
.
1 −1 −1
1
• Look at the columns that are out of Ã. In our example
. Substitute b
−1
1
with b̃ = b − α
, where α is a constant.2
−1
• Cut the columns that are out of Ã. We obtain the 2 × 2 system
1 2
x1
1−α
=
1 −1
x2
2+α
2
In the case in which there were more than one column out of Ã, we do the same operation for
each column, multiplying
For example, consider the case
it by different
constants.
in which
we have
1
0
1
0
two columns out of Ã,
and
. In this case b̃ = b − α
−β
, where
−1
1
−1
1
α, β ∈ R.
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Notice that in doing this we are considering x3 as a free parameter (a degree of
freedom) for the system, indeed x3 = α.
• Solve the reduced linear system in (x1 , x2 ). It admits one unique solution that
can be found, for instance, using Cramer’s method. In this case we find
α+5 x1
3
=
.
−1−2α
x2
3
The final solution of the initial system will be

  α+5 
x1
3
 x2  =  −1−2α  ; α ∈ R.
3
x3
α
2
Example 11 Consider the parametric system

+ αx3 = 1
 x1
− x2 + 2x3 = β

−x1 + x2
= 0
where α, β ∈ R.
(a) Find the rank of the matrix of the coefficients for all α and determine the number
of solutions of the system for any α and β.
(b) Solve the system in the case of α = −2 and β = −1.
Answer:


1
0 α
(a) A =  0 −1 2 . It can be verified that |A| = −2 − α.
−1 1 0
As a consequence, having a 3 × 3 matrix, rank(A)
= 3 if and only if α 6= −2.
1 0 6= 0).
When α = −2, rank(A) = 2 (notice in fact that 0 −1 Concerning the number of solutions of the system, we study the rank of [A|b].
If α 6= −2, rank(A|b) = 3 since it is not smaller than rank(A) but it is not
bigger because of its dimension. Then in this case the Rouché-Capelli Theorem
says that there is one and only one solution.
It remains to study the case α = −2. In this case we need to study the minors
of [A|b] in order to determine its rank. The 3 × 3 minors of [A|b] are:
1
1
1 1 −2 1 0 −2 0
−2
0
1
0 −1 2 ; 0 −1 β ; 0 β 2 ; β −1 2 .
−1 1
−1 1 0 −1 0 0 0 1
0 0 M.Tolotti - Mathematics (Part One) - September 2011
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Notice that the system admits solutions only in the case rank(A|b) = rank(A) =
2. The condition to have solution is thus that the minors above are all zero (in
this case the rank cannot be 3).
It can be seen that the minors are all zero if and only if β = −1. Thus when
α = −2 the system is consistent only if β = −1; in this case it admits infinite
solutions (one degree of freedom).
(b) α = −2, β = −1 is the case where the system has one degree of freedom. We can
solve the system implementing the method of the reduction to a smaller system
as seen in Example 10.
Firstly, since A has rank 2, we
choose one 2 × 2 matrix whose determinant is
1 0
non-zero. E.g. Ã =
.
0 −1
Then we cut the third
row and the third column, changing the vector b to the
−2
new one b̃ = b − c
. We obtain the system
2
1 0
x1
1 + 2c
=
; x3 = c ∈ R.
0 −1
x2
−1 − 2c
We can solve it using Cramer’s method:
1 + 2c
0 −1 − 2c −1 ;
x1 =
−1
1 1 + 2c 0 −1 − 2c x2 =
.
−1


1 + 2c
We conclude that the solutions of the starting system are  1 + 2c  ; c ∈ R.
c
2