How to solve the rank of a matrix. Find the rank of a matrix: methods and examples

This article will discuss such a concept as the rank of a matrix and the necessary additional concepts. We will give examples and proofs of finding the rank of a matrix, and also tell you what a matrix minor is and why it is so important.

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Minor matrix

To understand what the rank of a matrix is, it is necessary to understand such a concept as the minor of a matrix.

Definition 1

Minork-th order matrix is the determinant of a square matrix of order k × k, which is composed of the elements of the matrix A located in the preselected k-rows and k-columns, while maintaining the position of the elements of the matrix A.

Simply put, if we delete (p-k) rows and (n-k) columns in matrix A, and make up a matrix of those elements that remain, preserving the arrangement of the elements of matrix A, then the determinant of the resulting matrix is a minor of order k of matrix A.

From the example it follows that the first-order minors of the matrix A are the elements of the matrix themselves.

There are several examples of 2nd order minors. Let's select two rows and two columns. For example, 1st and 2nd row, 3rd and 4th column.

With this choice of elements, the second order minor will be - 1 3 0 2 = (- 1) × 2 - 3 × 0 = - 2

Another second-order minor of the matrix A is 0 0 1 1 = 0

Let us provide an illustration of the construction of the second-order minors of the matrix A:

The 3rd order minor is obtained by deleting the third column of the matrix A:

0 0 3 1 1 2 - 1 - 4 0 = 0 × 1 × 0 + 0 × 2 × (- 1) + 3 × 1 × (- 4) - 3 × 1 × (- 1) - 0 × 1 × 0 - 0 × 2 × (- 4) = - 9

An illustration of how the 3rd order minor of the matrix A is obtained:

For a given matrix, there are no minors higher than the 3rd order, because

k ≤ m i n (p, n) = m i n (3, 4) = 3

How many minors of order k are there for a matrix A of order p × n?

The number of minors is calculated using the following formula:

C p k × C n k, where e e C p k = p! k! (p - k)! and C n k = n! k! (n - k)! - the number of combinations from p to k, from n to k, respectively.

After we have decided what the minors of the matrix A are, we can proceed to determining the rank of the matrix A.

Matrix rank: methods of finding

Definition 2

Matrix rank - the highest order of the matrix other than zero.

Notation 1

Rank (A), Rg (A), Rang (A).

From the definition of the rank of the matrix and the minor of the matrix, it becomes clear that the rank of the zero matrix is zero, and the rank of the nonzero matrix is nonzero.

Finding the rank of a matrix by definition

Definition 3

Enumeration of Minors - a method based on determining the rank of a matrix.

Algorithm of actions by enumerating minors :

It is necessary to find the rank of the matrix A of order p× n... If there is at least one nonzero element, then the rank of the matrix is at least equal to one ( since is a minor of the 1st order, which is not equal to zero).

This is followed by an enumeration of the 2nd order minors. If all the 2nd order minors are equal to zero, then the rank is equal to one. If there is at least one nonzero minor of the 2nd order, it is necessary to go to an enumeration of the minors of the 3rd order, and the rank of the matrix, in this case, will be equal to at least two.

We will act in a similar way with the rank of the third order: if all the minors of the matrix are equal to zero, then the rank will be equal to two. If there is at least one nonzero third order minor, then the rank of the matrix is at least three. And so on, by analogy.

Example 2

Find the rank of a matrix:

A = - 1 1 - 1 - 2 0 2 2 6 0 - 4 4 3 11 1 - 7

Since the matrix is nonzero, its rank is at least equal to one.

The 2nd order minor - 1 1 2 2 = (- 1) × 2 - 1 × 2 = 4 is nonzero. Hence it follows that the rank of the matrix A is at least two.

We iterate over the minors of the 3rd order: С 3 3 × С 5 3 = 1 5! 3! (5 - 3)! = 10 pieces.

1 1 - 1 2 2 6 4 3 11 = (- 1) × 2 × 11 + 1 × 6 × 4 + (- 1) × 2 × 3 - (- 1) × 2 × 4 - 1 × 2 × 11 - (- 1) × 6 × 3 = 0

1 - 1 - 2 2 6 0 4 11 1 = (- 1) × 6 × 1 + (- 1) × 0 × 4 + (- 2) × 2 × 11 - (- 2) × 6 × 4 - (- 1) × 2 × 1 - (- 1) × 0 × 11 = 0

1 1 - 2 2 2 0 4 3 1 = (- 1) × 2 × 1 + 1 × 0 × 4 + (- 2) × 2 × 3 - (- 2) × 2 × 4 - 1 × 2 × 1 - (- 1) × 0 × 3 = 0

1 - 1 0 2 6 - 4 4 11 - 7 = (- 1) × 6 × (- 7) + (- 1) × (- 4) × 4 + 0 × 2 × 11 - 0 × 6 × 4 - ( - 1) × 2 × (- 7) - (- 1) × (- 4) × 11 = 0

1 - 1 0 2 6 - 4 3 11 - 7 = 1 × 6 × (- 7) + (- 1) × (- 4) × 3 + 0 × 2 × 11 - 0 × 6 × 3 - (- 1) × 2 × (- 7) - 1 × (- 4) × 11 = 0

1 - 2 0 2 0 - 4 3 1 - 7 = 1 × 0 × (- 7) + (- 2) × (- 4) × 3 + 0 × 2 × 1 - 0 × 0 × 3 - (- 2) × 2 × (- 7) - 1 × (- 4) × 1 = 0

1 - 2 0 6 0 - 4 11 1 - 7 = (- 1) × 0 × (- 7) + (- 2) × (- 4) × 11 + 0 × 6 × 1 - 0 × 0 × 11 - ( - 2) × 6 × (- 7) - (- 1) × (- 4) × 1 = 0

Minors of the 3rd order are equal to zero, therefore the rank of the matrix is equal to two.

Answer : Rank (A) = 2.

Finding the rank of a matrix by the bordering minors method

Definition 3

Border Minors Method - a method that allows you to get a result with less computational work.

Facing minor - the minor M o k (k + 1) -th order of the matrix A, which borders the minor M of order k of the matrix A, if the matrix that corresponds to the minor M o k "contains" the matrix that corresponds to the minor M.

Simply put, the matrix that corresponds to the bordered minor M is obtained from the matrix corresponding to the bordering minor M o k by deleting elements of one row and one column.

Example 3

Find the rank of a matrix:

A = 1 2 0 - 1 3 - 2 0 3 7 1 3 4 - 2 1 1 0 0 3 6 5

To find the rank, we take the 2nd order minor М = 2 - 1 4 1

We write down all the bordering minors:

1 2 - 1 - 2 0 7 3 4 1 , 2 0 - 1 0 3 7 4 - 2 1 , 2 - 1 3 0 7 1 4 1 1 , 1 2 - 1 3 4 1 0 0 6 , 2 0 - 1 4 - 2 1 0 3 6 , 2 - 1 3 4 1 1 0 6 5 .

To substantiate the method of bordering minors, we present a theorem, the formulation of which does not require a proof basis.

Theorem 1

If all the minors bordering the k-th order minor of the matrix A of order p by n are equal to zero, then all the minors of order (k + 1) of the matrix A are equal to zero.

Algorithm of actions :

To find the rank of a matrix, it is not necessary to iterate over all the minors; it is enough to look at the bordering ones.

If the bordering minors are equal to zero, then the rank of the matrix is zero. If there is at least one minor that is not equal to zero, then we consider the bordering minors.

If they are all zero, then Rank (A) is two. If there is at least one nonzero bordering minor, then we proceed to consider its bordering minors. And so on, in a similar way.

Example 4

Find the rank of a matrix by the bordering minors method

A = 2 1 0 - 1 3 4 2 1 0 - 1 2 1 1 1 - 4 0 0 2 4 - 14

How to solve?

Since the element a 11 of the matrix A is not equal to zero, then we take a minor of the 1st order. Let's start looking for a non-zero bordering minor:

2 1 4 2 = 2 × 2 - 1 × 4 = 0 2 0 4 1 = 2 × 1 - 0 × 4 = 2

We found a bordering 2nd order minor not equal to zero 2 0 4 1.

Let's iterate over the bordering minors - (there are (4 - 2) × (5 - 2) = 6 pieces).

2 1 0 4 2 1 2 1 1 = 0 ; 2 0 - 1 4 1 0 2 1 1 = 0 ; 2 0 3 4 1 - 1 2 1 - 4 = 0 ; 2 1 0 4 2 1 0 0 2 = 0 ; 2 0 - 1 4 1 0 0 2 4 = 0 ; 2 0 3 4 1 - 1 0 2 - 14 = 0

Answer : Rank (A) = 2.

Finding the rank of a matrix by the Gauss method (using elementary transformations)

Let's remember what elementary transformations are.

Elementary transformations:

by rearranging the rows (columns) of the matrix;
by multiplying all elements of any row (column) of the matrix by an arbitrary nonzero number k;

by adding to the elements of any row (column) elements that correspond to another row (column) of the matrix, which are multiplied by an arbitrary number k.

Definition 5

Finding the rank of a matrix by the Gauss method - a method based on the theory of equivalence of matrices: if matrix B is obtained from matrix A using a finite number of elementary transformations, then Rank (A) = Rank (B).

The validity of this statement follows from the definition of the matrix:

in the case of permutation of rows or columns of a matrix, its determinant changes sign. If it is equal to zero, then it remains equal to zero when rearranging rows or columns;
in the case of multiplying all elements of any row (column) of the matrix by an arbitrary number k, which is not equal to zero, the determinant of the resulting matrix is equal to the determinant of the original matrix, which is multiplied by k;

in the case of adding to the elements of a certain row or column of the matrix the corresponding elements of another row or column, which are multiplied by the number k, does not change its determinant.

The essence of the method of elementary transformations : reduce the matrix whose rank is to be found to a trapezoidal one using elementary transformations.

For what?

The rank of matrices of this kind is quite easy to find. It is equal to the number of rows that contain at least one nonzero element. And since the rank does not change during elementary transformations, this will be the rank of the matrix.

Let's illustrate this process:

for rectangular matrices A of order p by n, the number of rows of which is greater than the number of columns:

A ~ 1 b 12 b 13 ⋯ b 1 n - 1 b 1 n 0 1 b 23 ⋯ b 2 n - 2 b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 bn - 1 n 0 0 0 ⋯ 0 1 0 0 0 ⋯ 0 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 0 0, R ank (A) = n

А ~ 1 b 12 b 13 ⋯ b 1 kb 1 k + 1 ⋯ b 1 n 0 1 b 23 ⋯ b 2 kb 2 k + 1 ⋯ b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 bkk + 1 ⋯ bkn 0 0 0 ⋯ 0 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 0 0 ⋯ 0, R ank (A) = k

for rectangular matrices A of order p by n, the number of rows of which is less than the number of columns:

A ~ 1 b 12 b 13 ⋯ b 1 pb 1 p + 1 ⋯ b 1 n 0 1 b 23 ⋯ b 2 pb 2 p + 1 ⋯ b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 bpp + 1 ⋯ bpn, R ank (A) = p

А ~ 1 b 12 b 13 ⋯ b 1 kb 1 k + 1 ⋯ b 1 n 0 1 b 23 ⋯ b 2 kb 2 k + 1 ⋯ b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 bkk + 1 ⋯ bkn 0 0 0 ⋯ 0 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 0 0 ⋯ 0

for square matrices A of order n by n:

A ~ 1 b 12 b 13 ⋯ b 1 n - 1 b 1 n 0 1 b 23 ⋯ b 2 n - 1 b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 bn - 1 n 0 0 0 ⋯ 0 1 , R ank (A) = n

A ~ 1 b 12 b 13 ⋯ b 1 kb 1 k + 1 ⋯ b 1 n 0 1 b 23 ⋯ b 2 kb 2 k + 1 ⋯ b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 bkk + 1 ⋯ bkn 0 0 0 ⋯ 0 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 0 0 ⋯ 0, R ank (A) = k, k< n

Example 5

Find the rank of matrix A using elementary transformations:

A = 2 1 - 2 6 3 0 0 - 1 1 - 1 2 - 7 5 - 2 4 - 15 7 2 - 4 11

How to solve?

Since the element a 11 is nonzero, it is necessary to multiply the elements of the first row of the matrix A by 1 a 11 = 1 2:

A = 2 1 - 2 6 3 0 0 - 1 1 - 1 2 - 7 5 - 2 4 - 15 7 2 - 4 11 ~

Add to the elements of the 2nd row the corresponding elements of the 1st row, which are multiplied by (-3). To the elements of the 3rd line, add the elements of the 1st line, which are multiplied by (-1):

~ A (1) = 1 1 2 - 1 3 3 0 0 - 1 1 - 1 2 - 7 5 - 2 4 - 15 7 2 - 4 11 ~ A (2) = = 1 1 2 - 1 3 3 + 1 (- 3) 0 + 1 2 (- 3) 0 + (- 1) (- 3) - 1 + 3 (- 3) 1 + 1 (- 3) - 1 + 1 2 (- 3) 2 + (- 1) (- 1) - 7 + 3 (- 1) 5 + 1 (- 5) - 2 + 1 2 (- 5) 4 + (- 1) (- 5) - 15 + 3 (- 5) 7 + 1 (- 7) 2 + 1 2 (- 7) - 4 + (- 1) (- 7) 11 + 3 (- 7) =

1 1 2 - 1 3 0 - 3 2 3 - 10 0 - 3 2 3 - 10 0 - 9 2 9 - 30 0 - 3 2 3 - 10

The element a 22 (2) is nonzero, so we multiply the elements of the 2nd row of the matrix A by A (2) by a 1 a 22 (2) = - 2 3:

A (3) = 1 1 2 - 1 3 0 1 - 2 20 3 0 - 3 2 3 - 10 0 - 9 2 9 - 30 0 - 3 2 3 - 10 ~ A (4) = 1 1 2 - 1 3 0 1 - 2 20 3 0 - 3 2 + 1 3 2 3 + (- 2) 3 2 - 10 + 20 3 × 3 2 0 - 9 2 + 1 9 2 9 + (- 2) 9 2 - 30 + 20 3 × 9 2 0 - 3 2 + 1 3 2 3 + (- 2) 3 2 - 10 + 20 3 × 3 2 = = 1 1 2 - 1 3 0 1 - 2 20 3 0 0 0 0 0 0 0 0 0 0 0 0

To the elements of the 3rd row of the resulting matrix, add the corresponding elements of the 2nd row, which are multiplied by 3 2;
to the elements of the 4th row - the elements of the 2nd row, which are multiplied by 9 2;
to the elements of the 5th row - the elements of the 2nd row, which are multiplied by 3 2.

All row elements are zero. Thus, with the help of elementary transformations, we brought the matrix to a trapezoidal form, from which it is seen that R a n k (A (4)) = 2. Hence it follows that the rank of the original matrix is also equal to two.

Comment

If you carry out elementary transformations, then approximate values are not allowed!

If you notice an error in the text, please select it and press Ctrl + Enter

Theorem (on the correctness of the definition of ranks). Let all the minors of the matrix A m × n (\ displaystyle A_ (m \ times n)) order k (\ displaystyle k) equal to zero ( M k = 0 (\ displaystyle M_ (k) = 0)). Then ∀ M k + 1 = 0 (\ displaystyle \ forall M_ (k + 1) = 0) if they exist. Pattern: / frame

Related definitions

Properties

Theorem (about basic minor): Let r = rang ⁡ A, M r (\ displaystyle r = \ operatorname (rang) A, M_ (r))- base minor of the matrix A (\ displaystyle A), then:
Consequences:
Theorem (on rank invariance under elementary transformations): Let us introduce a notation for matrices obtained from each other by elementary transformations. Then the following statement is true: If A ∼ B (\ displaystyle A \ sim B), then their ranks are equal.
The Kronecker - Capelli theorem: A system of linear algebraic equations is consistent if and only if the rank of its main matrix is equal to the rank of its extended matrix. In particular:
- The number of main variables of the system is equal to the rank of the system.
- A joint system will be determined (its solution is unique) if the rank of the system is equal to the number of all its variables.
Sylvester's inequality: If A and B size matrices m x n and n x k, then

r a n k A B ≥ r a n k A + r a n k B - n (\ displaystyle rankAB \ geq rankA + rankB-n)

This is a special case of the following inequality.

Frobenius inequality: If AB, BC, ABC are well defined, then

r a n k A B C ≥ r a n k A B + r a n k B C - r a n k B (\ displaystyle rankABC \ geq rankAB + rankBC-rankB)

Linear transformation and rank of a matrix

Let A (\ displaystyle A)- size matrix m × n (\ displaystyle m \ times n) over the field C (\ displaystyle C)(or R (\ displaystyle R)). Let T (\ displaystyle T)- linear transformation corresponding A (\ displaystyle A) in a standard basis; it means that T (x) = A x (\ displaystyle T (x) = Ax). Matrix rank A (\ displaystyle A) is the dimension of the range of values of the transformation T (\ displaystyle T).

Methods

There are several methods for finding the rank of a matrix:

Elementary transformation method

The rank of the matrix is equal to the number of nonzero rows in the matrix after reducing it to a stepped form using elementary transformations over the rows of the matrix.

Border Minors Method

Let in the matrix A (\ displaystyle A) nonzero minor found k (\ displaystyle k)-th order M (\ displaystyle M)... Consider all the minors (k + 1) (\ displaystyle (k + 1))-th order, including (bordering) minor M (\ displaystyle M); if they are all equal to zero, then the rank of the matrix is k (\ displaystyle k)... Otherwise, there is a nonzero one among the bordering minors, and the whole procedure is repeated.

Definition. By the rank of the matrix is the maximum number of linearly independent lines considered as vectors.

Theorem 1 on the rank of a matrix. By the rank of the matrix is the maximum order of a nonzero minor of the matrix.

We have already analyzed the concept of a minor in the lesson using determinants, and now we will generalize it. Let us take in the matrix some rows and some columns, and this “some” should be less than the number of rows and columns of the matrix, and for rows and columns this “some” should be the same number. Then at the intersection of some rows and how many columns there will be a matrix of lower order than our original matrix. The determinant of this matrix will be the k-th order minor if the mentioned "some" (the number of rows and columns) is denoted by k.

Definition. Minor ( r+1) th order, within which the selected minor lies r-th order is called bordering for a given minor.

The two most commonly used are finding the rank of the matrix... This bordering minors way and method of elementary transformations(by the Gauss method).

The following theorem is used for the bordering minors method.

Theorem 2 on the rank of a matrix. If from the elements of the matrix it is possible to compose a minor r th order, not equal to zero, then the rank of the matrix is r.

In the method of elementary transformations, the following property is used:

If, by elementary transformations, a trapezoidal matrix is obtained that is equivalent to the original one, then the rank of this matrix is the number of lines in it, except for lines consisting entirely of zeros.

Finding the rank of a matrix by the bordering minors method

A bordering minor is a minor of a higher order in relation to a given one, if this minor of a higher order contains a given minor.

For example, given the matrix

Let's take a minor

bordering will be the following minors:

Algorithm for finding the rank of a matrix next.

1. Find non-zero minors of the second order. If all the second-order minors are equal to zero, then the rank of the matrix will be equal to one ( r =1 ).

2. If there is at least one second-order minor that is not equal to zero, then compose the bordering third-order minors. If all the bordering minors of the third order are equal to zero, then the rank of the matrix is equal to two ( r =2 ).

3. If at least one of the bordering minors of the third order is not equal to zero, then we compose the bordering minors. If all the bordering minors of the fourth order are equal to zero, then the rank of the matrix is three ( r =2 ).

4. Continue as long as the size of the matrix allows.

Example 1. Find the rank of a matrix

Solution. Minor of the second order .

We frame it. There will be four bordering minors:

Thus, all bordering minors of the third order are equal to zero, therefore, the rank of this matrix is equal to two ( r =2 ).

Example 2. Find the rank of a matrix

Solution. The rank of this matrix is 1, since all the second-order minors of this matrix are equal to zero (in this, as in the cases of bordering minors in the next two examples, dear students are invited to check for themselves, possibly using the rules for calculating determinants), and among the first-order minors , that is, among the elements of the matrix, there are not equal to zero.

Example 3. Find the rank of a matrix

Solution. Minor of the second order of this matrix, in all the minors of the third order of this matrix are equal to zero. Therefore, the rank of this matrix is two.

Example 4. Find the rank of a matrix

Solution. The rank of this matrix is 3, since the only third-order minor of this matrix is 3.

Finding the rank of a matrix by the method of elementary transformations (Gauss's method)

Already in Example 1, it can be seen that the problem of determining the rank of a matrix by the method of bordering minors requires calculating a large number of determinants. There is, however, a way to keep the amount of computation to a minimum. This method is based on the use of elementary matrix transformations and is also called the Gauss method.

Elementary matrix transformations are understood as the following operations:

1) multiplying any row or any column of the matrix by a number other than zero;

2) adding to the elements of any row or any column of the matrix the corresponding elements of another row or column, multiplied by the same number;

3) interchanging two rows or columns of the matrix;

4) removal of "zero" lines, that is, those, all elements of which are equal to zero;

5) deletion of all proportional lines, except for one.

Theorem. An elementary transformation does not change the rank of the matrix. In other words, if we use elementary transformations from the matrix A went to the matrix B, then .

The rank of a matrix is an important numerical characteristic. The most typical problem requiring finding the rank of a matrix is checking the consistency of a system of linear algebraic equations. In this article, we will give the concept of the rank of a matrix and consider methods for finding it. For a better assimilation of the material, we will analyze in detail the solutions of several examples.

Page navigation.

Determination of the rank of a matrix and the necessary additional concepts.

Before announcing the definition of the rank of a matrix, one should understand well the concept of a minor, and finding the minors of a matrix implies the ability to calculate the determinant. So we recommend, if necessary, to recall the theory of the article, the methods of finding the determinant of the matrix, the properties of the determinant.

Take a matrix A of order. Let k be some natural number not exceeding the smallest of the numbers m and n, that is, .

Definition.

Minor of the kth order of the matrix A is called the determinant of the square matrix of the order, composed of the elements of the matrix A, which are in the pre-selected k rows and k columns, and the arrangement of the elements of the matrix A is preserved.

In other words, if we delete (p – k) rows and (n – k) columns in matrix A, and form a matrix from the remaining elements, preserving the arrangement of the elements of matrix A, then the determinant of the resulting matrix is a minor of order k of matrix A.

Let's look at the definition of a matrix minor using an example.

Consider the matrix .

Let us write several first-order minors of this matrix. For example, if we choose the third row and the second column of the matrix A, then our choice corresponds to the first-order minor ... In other words, to obtain this minor, we crossed out the first and second rows, as well as the first, third and fourth columns from the matrix A, and made up the determinant from the remaining element. If we select the first row and third column of the matrix A, then we get a minor .

Let us illustrate the procedure for obtaining the considered first-order minors
and .

Thus, the first-order minors of the matrix are the elements of the matrix themselves.

We show several second-order minors. Select two rows and two columns. For example, let's take the first and second rows and the third and fourth columns. With this choice, we have a minor of the second order ... This minor could also be formed by deleting the third row, first and second columns from matrix A.

Another second-order minor of the matrix A is.

Let us illustrate the construction of these second-order minors
and .

The minors of the third order of the matrix A can be found similarly. Since there are only three rows in matrix A, we select all of them. If we choose the first three columns for these rows, then we get a third-order minor

It can also be constructed by deleting the last column of the matrix A.

Another third-order minor is

obtained by deleting the third column of the matrix A.

Here is a drawing showing the construction of these third-order minors.
and .

For a given matrix A, minors of order higher than the third do not exist, since.

How many minors of the kth order of the matrix A of order exist?

The number of minors of order k can be calculated as, where and - the number of combinations from p to k and from n to k, respectively.

How to construct all the minors of order k of the matrix A of order p by n?

We need many matrix row numbers and many column numbers. We write down everything combinations of p elements by k(they will correspond to the selected rows of matrix A when constructing a minor of order k). To each combination of line numbers, we successively add all combinations of n elements with k column numbers. These sets of combinations of row numbers and column numbers of matrix A will help to compose all the minors of order k.

Let's take an example.

Example.

Find all the second order minors of the matrix.

Solution.

Since the order of the original matrix is 3 by 3, then the total minors of the second order will be .

Let's write down all combinations of 3 by 2 numbers of rows of matrix A: 1, 2; 1, 3 and 2, 3. All combinations of 3 by 2 column numbers are 1, 2; 1, 3 and 2, 3.

Take the first and second rows of matrix A. Choosing to these rows the first and second columns, the first and third columns, the second and third columns, we get the minors, respectively

For the first and third rows, with a similar choice of columns, we have

It remains to add the first and second, first and third, second and third columns to the second and third rows:

So, all nine second-order minors of the matrix A are found.

Now you can move on to determining the rank of the matrix.

Definition.

Matrix rank Is the highest order of a nonzero minor in a matrix.

The rank of the matrix A is referred to as Rank (A). You can also find the designations Rg (A) or Rang (A).

From the definitions of the rank of a matrix and a minor of a matrix, we can conclude that the rank of a zero matrix is zero, and the rank of a nonzero matrix is at least one.

Finding the rank of a matrix by definition.

So, the first method for finding the rank of a matrix is brute force method... This method is based on determining the rank of the matrix.

Suppose we need to find the rank of a matrix A of order.

Let's briefly describe algorithm solving this problem by enumerating the minors.

If there is at least one element of the matrix that is different from zero, then the rank of the matrix is at least equal to one (since there is a first-order minor that is not equal to zero).

Next, we iterate over the second-order minors. If all the second order minors are equal to zero, then the rank of the matrix is equal to one. If there is at least one nonzero second-order minor, then we pass to the enumeration of the third-order minors, and the rank of the matrix is at least two.

Similarly, if all third-order minors are zero, then the rank of the matrix is two. If there is at least one third-order minor other than zero, then the rank of the matrix is at least three, and we go over the fourth-order minors.

Note that the rank of the matrix cannot exceed the smallest of the numbers p and n.

Example.

Find the rank of the matrix .

Solution.

Since the matrix is nonzero, its rank is at least one.

Minor of the second order is nonzero, therefore, the rank of the matrix A is at least two. We pass to the enumeration of the minors of the third order. All of them things.

All third-order minors are zero. Therefore, the rank of the matrix is two.

Answer:

Rank (A) = 2.

Finding the rank of a matrix by the bordering minors method.

There are other methods for finding the rank of a matrix that allow you to get the result with less computational work.

One such method is bordering minor method.

Let's deal with bordering minor.

It is said that the minor M ok of the (k + 1) th order of the matrix A borders the minor M of order k of the matrix A, if the matrix corresponding to the minor M ok "contains" the matrix corresponding to the minor M.

In other words, the matrix corresponding to the bordered minor M is obtained from the matrix corresponding to the bordering minor M ok by deleting the elements of one row and one column.

For example, consider the matrix and take a minor of the second order. Let's write down all the bordering minors:

The method of bordering minors is substantiated by the following theorem (we present its formulation without proof).

Theorem.

If all the minors bordering the k-th order minor of the matrix A of order p by n are equal to zero, then all the minors of order (k + 1) of the matrix A are equal to zero.

Thus, to find the rank of a matrix, it is not necessary to iterate over all the minors that are sufficiently bordering. The number of minors bordering the k-th order minor of the order matrix A is found by the formula ... Note that the minors bordering the k-th order minor of the matrix A are not more than the (k + 1) -th order minors of the matrix A. Therefore, in most cases, the use of the bordering minors method is more profitable than a simple enumeration of all the minors.

Let us proceed to finding the rank of the matrix by the bordering minors method. Let's briefly describe algorithm this method.

If the matrix A is nonzero, then as a first-order minor we take any element of the matrix A other than zero. Consider its bordering minors. If they are all equal to zero, then the rank of the matrix is equal to one. If there is at least one nonzero bordering minor (its order is two), then we proceed to consider its bordering minors. If they are all zero, then Rank (A) = 2. If at least one bordering minor is nonzero (its order is three), then we consider its bordering minors. Etc. As a result, Rank (A) = k, if all bordering minors of the (k + 1) th order of matrix A are equal to zero, or Rank (A) = min (p, n) if there is a nonzero minor bordering the minor of order (min ( p, n) - 1).

Let us analyze the bordering minors method for finding the rank of a matrix using an example.

Example.

Find the rank of the matrix by the method of bordering minors.

Solution.

Since the element a 1 1 of the matrix A is nonzero, we take it as a first-order minor. Let's start looking for a non-zero bordering minor:

Found a bordering minor of the second order, other than zero. Let's sort out its bordering minors (their things):

All the minors bordering the second-order minor are equal to zero, therefore, the rank of the matrix A is equal to two.

Answer:

Rank (A) = 2.

Example.

Find the rank of the matrix using bordering minors.

Solution.

As a nonzero minor of the first order, we take the element a 1 1 = 1 of the matrix A. The flanking minor of the second order is not zero. This minor is bordered by a third-order minor.
... Since it is not equal to zero and there is not a single bordering minor for it, the rank of the matrix A is equal to three.

Answer:

Rank (A) = 3.

Finding the rank using elementary matrix transformations (Gauss method).

Consider another way to find the rank of a matrix.

The following matrix transformations are called elementary:

permutation of rows (or columns) of the matrix;
multiplication of all elements of any row (column) of the matrix by an arbitrary number k other than zero;
adding to the elements of any row (column) the corresponding elements of another row (column) of the matrix, multiplied by an arbitrary number k.

Matrix B is called equivalent to matrix A if B is obtained from A using a finite number of elementary transformations. Equivalence of matrices is denoted by the symbol "~", that is, written A ~ B.

Finding the rank of a matrix using elementary matrix transformations is based on the statement: if matrix B is obtained from matrix A using a finite number of elementary transformations, then Rank (A) = Rank (B).

The validity of this statement follows from the properties of the determinant of the matrix:

When the rows (or columns) of a matrix are rearranged, its determinant changes sign. If it is equal to zero, then upon permutation of rows (columns) it remains equal to zero.
When all the elements of any row (column) of the matrix are multiplied by an arbitrary number k other than zero, the determinant of the resulting matrix is equal to the determinant of the original matrix multiplied by k. If the determinant of the original matrix is equal to zero, then after multiplying all the elements of any row or column by the number k, the determinant of the resulting matrix will also be equal to zero.
Adding to the elements of some row (column) of the matrix the corresponding elements of another row (column) of the matrix, multiplied by some number k, does not change its determinant.

The essence of the method of elementary transformations consists in reducing the matrix, the rank of which we need to find, to a trapezoidal (in a particular case, to the upper triangular) using elementary transformations.

Why is this done? The rank of matrices of this kind is very easy to find. It is equal to the number of lines containing at least one nonzero element. And since the rank of the matrix does not change during elementary transformations, the resulting value will be the rank of the original matrix.

Here are some illustrations of matrices, one of which should be obtained after transformations. Their form depends on the order of the matrix.

These illustrations are templates to which we will transform the matrix A.

Let's describe method algorithm.

Suppose we need to find the rank of a nonzero matrix A of order (p can be equal to n).

So, . Let's multiply all the elements of the first row of the matrix A by. In this case, we obtain an equivalent matrix, denote it by A (1):

To the elements of the second row of the resulting matrix A (1), add the corresponding elements of the first row, multiplied by. To the elements of the third row, add the corresponding elements of the first row, multiplied by. And so on up to the p-th line. We obtain an equivalent matrix, denote it by A (2):

If all the elements of the resulting matrix, located in rows from the second to the p-th, are equal to zero, then the rank of this matrix is equal to one, and, consequently, the rank of the original matrix is equal to one.

If there is at least one nonzero element in rows from the second to the pth, then we continue to carry out the transformations. Moreover, we act in absolutely the same way, but only with the part of the matrix A marked in the figure (2)

If, then we rearrange the rows and (or) columns of the matrix A (2) so that the “new” element becomes nonzero.

The number r is called the rank of the matrix A if:
1) the matrix A contains a minor of order r, different from zero;
2) all minors of order (r + 1) and higher, if they exist, are equal to zero.
Otherwise, the rank of the matrix is the highest nonzero minor order.
Designations: rangA, r A, or r.
It follows from the definition that r is a positive integer. For a null matrix, the rank is considered to be zero.

Service purpose... The online calculator is designed to find rank of the matrix... In this case, the solution is saved in Word and Excel format. see solution example.

Instruction. Select the dimension of the matrix, click Next.

Definition . Let a matrix of rank r be given. Any minor of a matrix other than zero and having order r is called basic, and the rows and columns of its components are called basic rows and columns.
According to this definition, the matrix A can have several basic minors.

The rank of the identity matrix E is n (the number of rows).

Example 1. Two matrices are given, and their minors , ... Which one can be taken as the baseline?
Solution... Minor M 1 = 0, so it cannot be basic for any of the matrices. Minor M 2 = -9 ≠ 0 and has order 2, so it can be taken as the basis matrices A or / and B, provided that they have ranks equal to 2. Since detB = 0 (as a determinant with two proportional columns), then rangB = 2 and M 2 can be taken as the base minor of the matrix B. The rank of the matrix A is 3, since detA = -27 ≠ 0 and, therefore, the order the basic minor of this matrix must be equal to 3, that is, M 2 is not basic for the matrix A. Note that the matrix A has a single basic minor, which is equal to the determinant of the matrix A.

Theorem (on basic minor). Any row (column) of a matrix is a linear combination of its base rows (columns).
Corollaries from the theorem.

Any (r + 1) columns (rows) of a matrix of rank r are linearly dependent.
If the rank of a matrix is less than the number of its rows (columns), then its rows (columns) are linearly dependent. If rangA is equal to the number of its rows (columns), then the rows (columns) are linearly independent.
The determinant of the matrix A is equal to zero if and only if its rows (columns) are linearly dependent.
If to a row (column) of the matrix we add another row (column) multiplied by any number other than zero, then the rank of the matrix will not change.
If a row (column) in the matrix is crossed out, which is a linear combination of other rows (columns), then the rank of the matrix will not change.
The rank of a matrix is equal to the maximum number of its linearly independent rows (columns).
The maximum number of linearly independent rows is the same as the maximum number of linearly independent columns.

Example 2. Find the rank of a matrix .
Solution. Based on the definition of the rank of a matrix, we will look for a minor of the highest order, other than zero. First, we transform the matrix to a simpler form. To do this, multiply the first row of the matrix by (-2) and add to the second, then multiply it by (-1) and add to the third.