Examples of finding the rank of a matrix with a detailed solution. Find the rank of a matrix: methods and examples. How to investigate a system of linear equations for consistency

The rank of a matrix is ​​an important numerical characteristic. The most typical problem that requires finding the rank of a matrix is ​​checking the compatibility of a system of linear algebraic equations. In this article we will give the concept of matrix rank and consider methods for finding it. To better understand the material, we will analyze in detail the solutions to several examples.

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

Before voicing the definition of the rank of a matrix, you should have a good understanding of the concept of a minor, and finding the minors of a matrix implies the ability to calculate the determinant. So, if necessary, we recommend that you recall the theory of the article, methods for finding the determinant of a matrix, and properties of the determinant.

Let's 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 kth order matrix A is the determinant of a square matrix of order, composed of elements of matrix A, which are located in pre-selected k rows and k columns, and the arrangement of elements of matrix A is preserved.

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

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

Consider the matrix .

Let's write down several first-order minors of this matrix. For example, if we choose the third row and second column of matrix A, then our choice corresponds to a 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 a determinant from the remaining element. If we choose the first row and third column of 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 a matrix are the matrix elements themselves.

Let's show several second-order minors. Select two rows and two columns. For example, take the first and second rows and the third and fourth columns. With this choice we have a second-order minor . This minor could also be created 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 .

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

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

Another third order minor is

obtained by deleting the third column of matrix A.

Here is a picture showing the construction of these third order minors
And .

For a given matrix A there are no minors of order higher than third, since .

How many minors of the kth order are there of a matrix A of order ?

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 minors of order k of matrix A of order p by n?

We will need many matrix row numbers and many column numbers. We write everything down 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 row numbers we sequentially add all combinations of n elements of k column numbers. These sets of combinations of row numbers and column numbers of matrix A will help to compose all minors of order k.

Let's look at it with an example.

Example.

Find all 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 to 2 row numbers of matrix A: 1, 2; 1, 3 and 2, 3. All combinations of 3 to 2 column numbers are 1, 2; 1, 3 and 2, 3.

Let's take the first and second rows of matrix A. By selecting the first and second columns, the first and third columns, the second and third columns for these rows, we obtain 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 matrix A have been found.

Now we can proceed to determining the rank of the matrix.

Definition.

Matrix rank- This highest order matrix minor, different from zero.

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

From the definitions of matrix rank and matrix minor, we can conclude that the rank of a zero matrix is ​​equal to zero, and the rank of a nonzero matrix is ​​not less than one.

Finding the rank of a matrix by definition.

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

Let us need to find the rank of a matrix A of order .

Let's briefly describe algorithm solving this problem by enumerating 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 look at the second order minors. If all second-order minors are equal to zero, then the rank of the matrix is ​​equal to one. If there is at least one non-zero minor of the second order, then we proceed to enumerate the minors of the third order, and the rank of the matrix is ​​at least equal to 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 move on to enumerating 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 ​​non-zero, its rank is not less than one.

Minor of the second order is different from zero, therefore, the rank of matrix A is at least two. Let's move on to enumerating third-order minors. Total of them things.

All third order minors are equal to zero. Therefore, the rank of the matrix is ​​two.

Answer:

Rank(A) = 2 .

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

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

One such method is edge minor method.

Let's deal with concept of edge minor.

It is said that a minor M ok of the (k+1)th order of the matrix A borders a 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 bordering 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 second order minor. Let's write down all the bordering minors:

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

Theorem.

If all minors bordering the kth order minor of a matrix A of order p by n are equal to zero, then all 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 go through all the minors that are sufficiently bordering. The number of minors bordering the minor of the kth order of a matrix A of order , is found by the formula . Note that there are no more minors bordering the kth order minor of the matrix A than there are (k + 1) order minors of the matrix A. Therefore, in most cases, using the method of bordering minors is more profitable than simply enumerating all the minors.

Let's move on to finding the rank of the matrix using the method of bordering minors. 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 that is different from zero. Let's look at 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 non-zero 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 non-zero (its order is three), then we consider its bordering minors. And so on. As a result, Rank(A) = k if all bordering minors of the (k + 1)th order of the matrix A are equal to zero, or Rank(A) = min(p, n) if there is a non-zero minor bordering a minor of order (min( p, n) – 1) .

Let's look at the method of bordering minors to find the rank of a matrix using an example.

Example.

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

Solution.

Since element a 1 1 of matrix A is nonzero, we take it as a first-order minor. Let's start searching for a bordering minor that is different from zero:

An edge minor of the second order, different from zero, is found. Let's look at its bordering minors (their things):

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

Answer:

Rank(A) = 2 .

Example.

Find the rank of the matrix using bordering minors.

Solution.

As a non-zero minor of the first order, we take the element a 1 1 = 1 of the matrix A. The surrounding minor of the second order not equal to 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 matrix A is equal to three.

Answer:

Rank(A) = 3 .

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

Let's consider another way to find the rank of a matrix.

The following matrix transformations are called elementary:

• rearranging rows (or columns) of a matrix;
• multiplying all elements of any row (column) of a matrix by an arbitrary number k, different from zero;
• adding to the elements of a 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 finite number elementary transformations. The 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 rearranging the rows (or columns) of a matrix, its determinant changes sign. If it is equal to zero, then when the rows (columns) are rearranged, it remains equal to zero.
• When multiplying all elements of any row (column) of a matrix 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 a certain row (column) of a matrix the corresponding elements of another row (column) of the matrix, multiplied by a certain number k, does not change its determinant.

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

Why is this being done? The rank of matrices of this type is very easy to find. It is equal to the number of lines containing at least one non-zero element. And since the rank of the matrix does not change when carrying out elementary transformations, the resulting value will be the rank of the original matrix.

We give illustrations of matrices, one of which should be obtained after transformations. Their appearance depends on the order of the matrix.

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

Let's describe method algorithm.

Let us need to find the rank of a non-zero matrix A of order (p can be equal to n).

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

To the elements of the second row of the resulting matrix A (1) we add the corresponding elements of the first row, multiplied by . To the elements of the third line we add the corresponding elements of the first line, multiplied by . And so on until the p-th line. Let's get an equivalent matrix, denote it A (2):

If all 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 in the lines from the second to the p-th there is at least one non-zero element, then we continue to carry out transformations. Moreover, we act in exactly the same way, but only with the part of matrix A (2) marked in the figure.

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

A number r is called the rank of matrix A if:
1) in the matrix A there is 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 a matrix is ​​the highest minor order other than zero.
Designations: rangA, r A or r.
From the definition it follows that r is a positive integer. For a null matrix, the rank is considered to be zero.

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

Instructions. Select the matrix dimension, click Next.

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

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

Example 1. Given two matrices, and their minors , . Which of them can be taken as the basic one?
Solution. Minor M 1 =0, so it cannot be a basis for any of the matrices. Minor M 2 =-9≠0 and has order 2, which means it can be taken as the basis of 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 basis minor of matrix B. The rank of matrix A is 3, due to the fact that detA=-27≠0 and, therefore, the order the basis minor of this matrix must be equal to 3, that is, M 2 is not a basis for the matrix A. Note that the matrix A has a single basis minor, equal to the determinant of the matrix A.

Theorem (about the basis minor). Any row (column) of a matrix is ​​a linear combination of its basis rows (columns).
Corollaries from the theorem.

1. Every (r+1) column (row) matrix of rank r is linearly dependent.
2. 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.
3. The determinant of a matrix A is equal to zero if and only if its rows (columns) are linearly dependent.
4. If you add another row (column) to a row (column) of a matrix, multiplied by any number other than zero, then the rank of the matrix will not change.
5. If you cross out a row (column) in a matrix, which is a linear combination of other rows (columns), then the rank of the matrix will not change.
6. The rank of a matrix is ​​equal to the maximum number of its linearly independent rows (columns).
7. 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 matrix rank, we will look for a minor of the highest order, different from zero. First we transform the matrix to more simple view. To do this, multiply the first row of the matrix by (-2) and add it to the second, then multiply it by (-1) and add it to the third.

Definition. Matrix rank is the maximum number of linearly independent rows considered as vectors.

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

We already discussed the concept of a minor in the lesson on determinants, but now we will generalize it. Let's take a certain number of rows and a certain number of columns in the matrix, and this “how much” should be less than the number of rows and columns of the matrix, and for rows and columns this “how many” should be the same number. Then at the intersection of how many rows and how many columns there will be a matrix of lower order than our original matrix. The determinant is a matrix and will be a minor of the kth order if the mentioned “some” (the number of rows and columns) is denoted by k.

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

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

When using the bordering minors method, the following theorem is used.

Theorem 2 on the rank of the matrix. If a minor can be composed from matrix elements r th order, not equal to zero, then the rank of the matrix is ​​equal to r.

When using the elementary transformation method, the following property is used:

If, through elementary transformations, a trapezoidal matrix is ​​obtained that is equivalent to the original one, then rank of this matrix is the number of lines in it other than lines consisting entirely of zeros.

Finding the rank of a matrix using the method of bordering minors

An enclosing minor is a minor of a higher order relative to the given one if this minor of a higher order contains the given minor.

For example, given the matrix

Let's take a minor

The bordering minors will be:

Algorithm for finding the rank of a matrix next.

1. Find minors of the second order that are not equal to zero. If all 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 minor of the second order that is not equal to zero, then we compose the bordering minors of the third order. If all 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 ​​equal to three ( r =2 ).

4. Continue this way as long as the matrix size allows.

Example 1. Find the rank of a matrix

.

Solution. Minor of the second order .

Let's border 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 ​​equal to 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 two following examples, dear students are invited to verify for themselves, perhaps using the rules for calculating determinants), and among the first-order minors , that is, among the elements of the matrix, there are non-zero ones.

Example 3. Find the rank of a matrix

Solution. The second order minor of this matrix is ​​, and all third order minors 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 using the method of elementary transformations (Gauss method)

Already in example 1 it is clear that the task of determining the rank of a matrix using the method of bordering minors requires calculating large number determinants. There is, however, a way to reduce 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.

The following operations are understood as elementary matrix transformations:

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

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

3) swapping two rows or columns of the matrix;

4) removing “null” rows, that is, those whose elements are all equal to zero;

5) deleting all proportional lines except one.

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

Let some matrix be given:

.

Let us select in this matrix arbitrary strings and arbitrary columns
. Then the determinant th order, composed of matrix elements
, located at the intersection of selected rows and columns, is called a minor th order matrix
.

Definition 1.13. Matrix rank
called highest order minor of this matrix, different from zero.

To calculate the rank of a matrix, one should consider all its minors of the lowest order and, if at least one of them is different from zero, proceed to considering the minors of the highest order. This approach to determining the rank of a matrix is ​​called the bordering method (or the method of bordering minors).

Problem 1.4. Using the method of bordering minors, determine the rank of the matrix
.

.

Consider first-order edging, for example,
. Then we move on to consider some second-order edging.

For example,
.

Finally, let's analyze the third-order bordering.

.

So the highest order of a non-zero minor is 2, hence
.

When solving Problem 1.4, you can notice that a number of second-order bordering minors are nonzero. In this regard, the following concept applies.

Definition 1.14. A basis minor of a matrix is ​​any non-zero minor whose order is equal to the rank of the matrix.

Theorem 1.2.(Basis minor theorem). The basis rows (basis columns) are linearly independent.

Note that the rows (columns) of a matrix are linearly dependent if and only if at least one of them can be represented as a linear combination of the others.

Theorem 1.3. The number of linearly independent matrix rows is equal to the number of linearly independent matrix columns and is equal to the rank of the matrix.

Theorem 1.4.(Necessary and sufficient condition for the determinant to be equal to zero). In order for the determinant -th order was equal to zero, it is necessary and sufficient that its rows (columns) be linearly dependent.

Calculating the rank of a matrix based on its definition is too cumbersome. This becomes especially important for matrices of high orders. In this regard, in practice, the rank of a matrix is ​​calculated based on the application of Theorems 10.2 - 10.4, as well as the use of the concepts of matrix equivalence and elementary transformations.

Definition 1.15. Two matrices
And are called equivalent if their ranks are equal, i.e.
.

If matrices
And are equivalent, then note
 .

Theorem 1.5. The rank of the matrix does not change due to elementary transformations.

We will call elementary matrix transformations
any of the following operations on a matrix:

Replacing rows with columns and columns with corresponding rows;

Rearranging matrix rows;

Crossing out a line whose elements are all zero;

Multiplying a string by a number other than zero;

Adding to the elements of one line the corresponding elements of another line multiplied by the same number
.

Corollary of Theorem 1.5. If matrix
obtained from matrix using a finite number of elementary transformations, then the matrix
And are equivalent.

When calculating the rank of a matrix, it should be reduced to a trapezoidal form using a finite number of elementary transformations.

Definition 1.16. We will call trapezoidal a form of representation of a matrix when, in the bordering minor of the highest order other than zero, all elements below the diagonal ones vanish. For example:

.

Here
, matrix elements
go to zero. Then the form of representation of such a matrix will be trapezoidal.

As a rule, matrices are reduced to a trapezoidal shape using the Gaussian algorithm. The idea of ​​the Gauss algorithm is that, by multiplying the elements of the first row of the matrix by the corresponding factors, it is achieved that all elements of the first column located below the element
, would turn to zero. Then, multiplying the elements of the second column by the corresponding factors, we ensure that all elements of the second column located below the element
, would turn to zero. Then proceed in the same way.

Problem 1.5. Determine the rank of a matrix by reducing it to a trapezoidal shape.

.

To make it easier to use the Gaussian algorithm, you can swap the first and third lines.








.

It's obvious that here
. However, to bring the result to a more elegant form, you can further continue transforming the columns.










.

We will also consider an important practical application of the topic: system research linear equations for jointness.

What is the rank of a matrix?

The humorous epigraph of the article contains a large amount of truth. We usually associate the word “rank” with some kind of hierarchy, most often with a career ladder. The more knowledge, experience, abilities, connections, etc. a person has. – the higher his position and range of opportunities. In youth terms, rank refers to the general degree of “steepness.”

And our mathematical brothers live by the same principles. Let's take a few random ones for a walk zero matrices:

Let's think about it, if in the matrix all zeros, then what rank can we talk about? Everyone is familiar with the informal expression “total zero”. In the society of matrices everything is exactly the same:

Rank of the zero matrixany size equals zero.

Note : The zero matrix is ​​denoted Greek letter"theta"

In order to better understand the rank of the matrix, hereinafter I will use materials to help analytical geometry. Consider zero vector our three-dimensional space, which does not set a specific direction and is useless for building affine basis. From an algebraic point of view, the coordinates given vector recorded in matrix“one by three” and logical (in the indicated geometric sense) assume that the rank of this matrix is ​​zero.

Now let's look at a few non-zero column vectors And row vectors:


Each instance has at least one non-zero element, and that's something!

The rank of any non-zero row vector (column vector) is equal to one

And in general - if in the matrix arbitrary sizes there is at least one non-zero element, then its rank no less units.

Algebraic row vectors and column vectors are to a certain extent abstract, so let's turn again to the geometric association. Non-zero vector sets a very definite direction in space and is suitable for constructing basis, therefore the rank of the matrix will be considered equal to one.

Theoretical information : V linear algebra a vector is an element vector space(defined through 8 axioms), which, in particular, can be an ordered row (or column) of real numbers with operations of addition and multiplication defined for them by real number. With more detailed information about vectors can be found in the article Linear transformations.

linearly dependent(expressed through each other). From a geometric point of view, the second line contains the coordinates of the collinear vector , which did not advance the matter at all in building three-dimensional basis, being in this sense superfluous. Thus, the rank of this matrix is ​​also equal to one.

Let's rewrite the coordinates of the vectors into columns ( transpose the matrix):

What has changed in terms of rank? Nothing. The columns are proportional, which means the rank is equal to one. By the way, note that all three lines are also proportional. They can be identified with the coordinates three collinear vectors of the plane, of which only one useful for constructing a "flat" basis. And this is completely consistent with our geometric sense rank.

An important statement follows from the above example:

The rank of the matrix in rows is equal to the rank of the matrix in columns. I already mentioned this a little in the lesson about effective methods for calculating the determinant.

Note : from the linear dependence of the rows it follows linear dependence columns (and vice versa). But in order to save time, and out of habit, I will almost always talk about linear dependence of strings.

Let's continue training our beloved pet. Let's add the coordinates of another collinear vector to the matrix in the third row :

Did he help us in constructing a three-dimensional basis? Of course not. All three vectors walk back and forth along the same path, and the rank of the matrix is ​​equal to one. You can take as many collinear vectors as you like, say, 100, put their coordinates into a “one hundred by three” matrix, and the rank of such a skyscraper will still remain one.

Let's get acquainted with the matrix, the rows of which linearly independent. A pair of non-collinear vectors is suitable for constructing a three-dimensional basis. The rank of this matrix is ​​two.

What is the rank of the matrix? The lines don’t seem to be proportional... so, in theory, they are three. However, the rank of this matrix is ​​also two. I added the first two lines and wrote the result at the bottom, i.e. linearly expressed the third line through the first two. Geometrically, the rows of the matrix correspond to the coordinates of three coplanar vectors, and among this three there are a pair of non-collinear comrades.

As you can see, linear dependence in the considered matrix is ​​not obvious, and today we will learn how to bring it out into the open.

I think many people can guess what the rank of a matrix is!

Consider a matrix whose rows linearly independent. Vectors form affine basis, and the rank of this matrix is ​​three.

As you know, any fourth, fifth, tenth vector of three-dimensional space will be linearly expressed in terms of basis vectors. Therefore, if you add any number of rows to a matrix, then its rank will still be equal to three.

Similar reasoning can be carried out for matrices of larger sizes (of course, without any geometric meaning).

Definition : The rank of a matrix is ​​the maximum number of linearly independent rows. Or: The rank of a matrix is ​​the maximum number of linearly independent columns. Yes, their number is always the same.

An important practical guideline also follows from the above: the rank of the matrix does not exceed its minimum dimension. For example, in the matrix four rows and five columns. The minimum dimension is four, therefore, the rank of this matrix certainly will not exceed 4.

Designations: in world theory and practice there is no generally accepted standard for designating the rank of a matrix; most often you can find: - as they say, an Englishman writes one thing, a German another. Therefore, based on the famous joke about American and Russian hell, let’s denote the rank of the matrix with a native word. For example: . And if the matrix is ​​“unnamed”, of which there are many, then you can simply write .

How to find the rank of a matrix using minors?

If my grandmother had a fifth column in her matrix, then she would have to calculate another minor of the 4th order (“blue”, “raspberry” + 5th column).

Conclusion: the maximum order of a non-zero minor is three, which means .

Perhaps not everyone has fully comprehended this phrase: a minor of the 4th order is equal to zero, but among the minors of the 3rd order there was a non-zero one - therefore the maximum order non-zero minor and equals three.

The question arises, why not immediately calculate the determinant? Well, firstly, in most tasks the matrix is ​​not square, and secondly, even if you get a non-zero value, the task will most likely be rejected, since it usually involves a standard “bottom-up” solution. And in the example considered, the zero determinant of the 4th order allows us to state that the rank of the matrix is ​​only less than four.

I must admit, I came up with the problem I analyzed myself in order to better explain the method of bordering minors. In real practice, everything is simpler:

Example 2

Find the rank of a matrix using the edge minors method

The solution and answer are at the end of the lesson.

When does the algorithm work fastest? Let's return to the same four-by-four matrix. . Obviously, the solution will be the shortest in the case of “good” corner minors:

And, if , then , otherwise – .

The thinking is not at all hypothetical - there are many examples where the whole matter is limited only to angular minors.

However, in some cases another method is more effective and preferable:

How to find the rank of a matrix using the Gaussian method?

The paragraph is intended for readers who are already familiar with Gaussian method and more or less got their hands on it.

From a technical point of view, the method is not novel:

1) using elementary transformations, we reduce the matrix to a stepwise form;

2) the rank of the matrix is ​​equal to the number of rows.

It is absolutely clear that using the Gaussian method does not change the rank of the matrix, and the essence here is extremely simple: according to the algorithm, during elementary transformations, all unnecessary proportional (linearly dependent) rows are identified and removed, resulting in a “dry residue” - the maximum number of linearly independent rows.

Let's transform the old familiar matrix with the coordinates of three collinear vectors:

(1) The first line was added to the second line, multiplied by –2. The first line was added to the third line.

(2) Zero lines are removed.

Thus, there is one line left, hence . Needless to say, this is much faster than calculating nine zero minors of the 2nd order and only then drawing a conclusion.

I remind you that in itself algebraic matrix nothing can be changed, and transformations are performed only for the purpose of determining the rank! By the way, let’s dwell once again on the question, why not? Source matrix carries information that is fundamentally different from the information of the matrix and row. In some mathematical models(no exaggeration) the difference in one number can be a matter of life and death. ...Remembered school teachers mathematicians of primary and secondary classes who mercilessly cut the grade by 1-2 points for the slightest inaccuracy or deviation from the algorithm. And it was terribly disappointing when, instead of a seemingly guaranteed “A”, it turned out “good” or even worse. Understanding came much later - how else to entrust satellites, nuclear warheads and power plants to a person? But don't worry, I don't work in these areas =)

Let's move on to more meaningful tasks, where, among other things, we will get acquainted with important computational techniques Gauss method:

Example 3

Find the rank of a matrix using elementary transformations

Solution: a “four by five” matrix is ​​given, which means that its rank is certainly no more than 4.

In the first column, there is no 1 or –1, therefore, additional actions are required to obtain at least one unit. Throughout the existence of the site, I have been repeatedly asked the question: “Is it possible to rearrange columns during elementary transformations?” Here, we rearranged the first and second columns, and everything is fine! In most tasks where it is used Gaussian method, the columns can indeed be rearranged. BUT NOT NEEDED. And the point is not even in possible confusion with variables, the point is that in the classical course of higher mathematics this action is traditionally not considered, so such a nod will be looked at VERY crookedly (or even forced to redo everything).

The second point concerns numbers. As you make your decision, it is helpful to use the following rule of thumb: elementary transformations if possible, reduce the matrix numbers. After all, it is much easier to work with one, two, three than, for example, with 23, 45 and 97. And the first action is aimed not only at obtaining a one in the first column, but also at eliminating the numbers 7 and 11.

At first complete solution, then comments:

(1) The first line was added to the second line, multiplied by –2. The first line was added to the third line, multiplied by –3. And to the heap: the 1st line was added to the 4th line, multiplied by -1.

(2) The last three lines are proportional. The 3rd and 4th lines were removed, the second line was moved to the first place.

(3) The first line was added to the second line, multiplied by –3.

The matrix reduced to echelon form has two rows.

Answer:

Now it's your turn to torture the four-by-four matrix:

Example 4

Find the rank of a matrix using the Gaussian method

I remind you that Gaussian method does not imply unambiguous rigidity, and your decision will most likely differ from my decision. A brief example of a task at the end of the lesson.

Which method should I use to find the rank of a matrix?

In practice, it is often not stated at all which method should be used to find the rank. In such a situation, the condition should be analyzed - for some matrices it is more rational to solve through minors, while for others it is much more profitable to apply elementary transformations:

Example 5

Find the rank of a matrix

Solution: the first method somehow immediately disappears =)

A little higher, I advised not to touch the columns of the matrix, but when there is a zero column, or proportional/coinciding columns, then it is still worth amputating:

(1) The fifth column is zero, remove it from the matrix. Thus, the rank of the matrix is ​​no more than four. The first line was multiplied by –1. This is another signature feature of the Gauss method, which turns the following action into a pleasant walk:

(2) To all lines, starting from the second, the first line was added.

(3) The first line was multiplied by –1, the third line was divided by 2, the fourth line was divided by 3. The second line was added to the fifth line, multiplied by –1.

(4) The third line was added to the fifth line, multiplied by –2.

(5) The last two lines are proportional, the fifth is deleted.

The result is 4 lines.

Answer:

Standard five-story building for independent study:

Example 6

Find the rank of a matrix

A short solution and answer at the end of the lesson.

It should be noted that the phrase “matrix rank” is not so often seen in practice, and in most problems you can do without it altogether. But there is one task where the concept in question is the main one actor, and to conclude the article we will look at this practical application:

How to study a system of linear equations for consistency?

Often, in addition to the solution systems of linear equations according to the condition, it is first required to examine it for compatibility, that is, to prove that any solution exists at all. A key role in such verification is played by Kronecker-Capelli theorem, which I will formulate in the necessary form:

If rank system matrices equal to rank extended matrix system, then the system is consistent, and if given number coincides with the number of unknowns, then the solution is unique.

Thus, to study the system for compatibility it is necessary to check the equality , Where - system matrix(remember the terminology from the lesson Gauss method), A - extended system matrix(i.e. a matrix with coefficients of variables + a column of free terms).