How to determine the rank of a matrix example solution. Matrix rank and matrix basis minor

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
is the largest order of the non-zero minor of this matrix.

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 matrix representation when in the bordering minor of the highest order non-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.










.

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, let's transform the matrix to a simpler form. 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.

Elementary The following matrix transformations are called:

1) permutation of any two rows (or columns),

2) multiplying a row (or column) by a non-zero number,

3) adding to one row (or column) another row (or column), multiplied by a certain number.

The two matrices are called equivalent, if one of them is obtained from the other using a finite set of elementary transformations.

Equivalent matrices are not, generally speaking, equal, but their ranks are equal. If matrices A and B are equivalent, then it is written as follows: A ~ B.

Canonical A matrix is ​​a matrix in which at the beginning of the main diagonal there are several ones in a row (the number of which can be zero), and all other elements are equal to zero, for example,

Using elementary transformations of rows and columns, any matrix can be reduced to canonical. The rank of a canonical matrix is ​​equal to the number of ones on its main diagonal.

Example 2 Find the rank of a matrix

A=

and bring it to canonical form.

Solution. From the second line, subtract the first and rearrange these lines:

.

Now from the second and third lines we subtract the first, multiplied by 2 and 5, respectively:

;

subtract the first from the third line; we get a matrix

B = ,

which is equivalent to matrix A, since it is obtained from it using a finite set of elementary transformations. Obviously, the rank of matrix B is 2, and therefore r(A)=2. Matrix B can easily be reduced to canonical. By subtracting the first column, multiplied by suitable numbers, from all subsequent ones, we turn to zero all the elements of the first row, except the first, and the elements of the remaining rows do not change.

.

Then, subtracting the second column, multiplied by suitable numbers, from all subsequent ones, we turn to zero all elements of the second row, except the second, and obtain the canonical matrix: Kronecker - Capelli theorem

- compatibility criterion for a system of linear algebraic equations:

In order for a linear system to be consistent, it is necessary and sufficient that the rank of the extended matrix of this system be equal to the rank of its main matrix.

Proof (system compatibility conditions)

Necessity Let system

joint

Let . Let's take some basic minor in the matrix. Since, then it will also be the basis minor of the matrix. Then, according to the basis theorem minor

, the last column of the matrix will be a linear combination of the basis columns, that is, the columns of the matrix.

Therefore, the column of free terms of the system is a linear combination of the columns of the matrix. Consequences Number of main variables

systems Let equal to the rank of the system.

Joint

will be defined (its solution is unique) if the rank of the system is equal to the number of all its variables.15 . 2 Homogeneous system of equations

Offer

Homogeneous system of equations is always joint.

Proof

will be defined (its solution is unique) if the rank of the system is equal to the number of all its variables.15 . 3 . For this system, the set of numbers , , , is a solution.

Homogeneous system of equations In this section we will use the matrix notation of the system: .

The sum of solutions to a homogeneous system of linear equations is a solution to this system. A solution multiplied by a number is also a solution.

.

The sum of solutions to a homogeneous system of linear equations is a solution to this system. A solution multiplied by a number is also a solution.

Let them serve as solutions to the system.15 . 1 Then and.

Let . Then

Since, then - the solution.15 . 5 Let be an arbitrary number, . Then Consequence If a homogeneous system of linear equations has a nonzero solution, then it has infinitely many different solutions. Indeed, multiplying a non-zero solution by various numbers, we will obtain different solutions.

Definition We will say that the solutions

systems form fundamental system of solutions

, if columns

form a linearly independent system and any solution to the system is a linear combination of these columns. 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 the maximum number of linearly independent rows considered as vectors. 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 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 is the maximum number of linearly independent rows considered as vectors. th order, not equal to zero, then the rank of the matrix is ​​equal to is the maximum number of linearly independent rows considered as vectors..

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 ( is the maximum number of linearly independent rows considered as vectors. =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 ( is the maximum number of linearly independent rows considered as vectors. =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 ( is the maximum number of linearly independent rows considered as vectors. =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 ( is the maximum number of linearly independent rows considered as vectors. =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 the calculation of a large number of 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 .

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 consistency 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.

Page navigation.

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, .

form a linearly independent system and any solution to the system is a linear combination of these columns.

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, the total of second order minors 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.

form a linearly independent system and any solution to the system is a linear combination of these columns.

Matrix rank is the highest order of the non-zero minor of the matrix.

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. We 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 k-th 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 a finite number of 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 absolutely 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.

So, . We multiply each element of the second row of matrix A (2) by . We obtain the equivalent matrix A (3):

To the elements of the third row of the resulting matrix A (3) we add the corresponding elements of the second row, multiplied by . To the elements of the fourth line we add the corresponding elements of the second line, multiplied by . And so on until the p-th line. Let's get an equivalent matrix, denote it A (4):

If all elements of the resulting matrix located in rows from the third to the p-th are equal to zero, then the rank of this matrix is ​​equal to two, and, therefore, Rank(A) = 2.

If the lines from the third to the p-th contain 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 the matrix marked in the figure

The element is non-zero, so we can multiply the elements of the second row of matrix A (2) by:

To the elements of the third row of the resulting matrix we add the corresponding elements of the second row, multiplied by ; to the elements of the fourth line – the elements of the second line multiplied by ; to the elements of the fifth line – the elements of the second line, multiplied by:

All elements of the third, fourth and fifth rows of the resulting matrix are equal to zero. So, using elementary transformations, we brought matrix A to trapezoidal form, from which it can be seen that Rank(A (4)) = 2. Therefore, the rank of the original matrix is ​​also two.

So the first column is converted to the desired form.

The element in the resulting matrix is ​​different from zero. Multiply the elements of the second line by:

The second column of the resulting matrix has the desired form, since the element is already equal to zero.

Since , a , then swap the third and fourth columns:

Let's multiply the third row of the resulting matrix by:

This completes the transformation. We get Rank(A (5))=3, therefore, Rank(A)=3.

Answer:

The rank of the original matrix is ​​three.

Summarize.

We examined the concept of matrix rank and looked at three ways to find it:

• by definition by enumerating all minors;
• the method of bordering minors;
• by the method of elementary transformations.

It is advisable to always use the method of elementary transformations when finding the rank of a matrix, since it leads to a result with less computation compared to the method of bordering minors, and even more so in comparison with the method of enumerating all minors of a matrix.