Arrange the matrices in descending order of their ranks. Calculation of the rank of a matrix using the method of elementary transformations (Gauss algorithm)

To work with the concept of matrix rank, we will need information from the topic "Algebraic complements and minors. Types of minors and algebraic complements." First of all, this concerns the term “matrix minor”, since we will determine the rank of the matrix precisely through the minors.

Matrix rank is the maximum order of its minors, among which there is at least one that is not equal to zero.

Equivalent matrices- matrices whose ranks are equal to each other.

Let us explain in more detail. Suppose that among the second-order minors there is at least one that is different from zero. And all minors whose order is higher than two are equal to zero. Conclusion: the rank of the matrix is 2. Or, for example, among the minors of the tenth order there is at least one that is not equal to zero. And all minors whose order is higher than 10 are equal to zero. Conclusion: the rank of the matrix is 10.

The rank of the matrix $A$ is denoted as follows: $\rang A$ or $r(A)$. The rank of the zero matrix $O$ is assumed to be zero, $\rang O=0$. Let me remind you that to form a matrix minor you need to cross out rows and columns, but it is impossible to cross out more rows and columns than the matrix itself contains. For example, if the matrix $F$ has size $5\times 4$ (i.e. contains 5 rows and 4 columns), then the maximum order of its minors is four. It will no longer be possible to form minors of the fifth order, since they will require 5 columns (and we have only 4). This means that the rank of the matrix $F$ cannot be more than four, i.e. $\rang F≤4$.

In more general form, the above means that if a matrix contains $m$ rows and $n$ columns, then its rank cannot exceed the smallest of $m$ and $n$, i.e. $\rang A≤\min(m,n)$.

In principle, from the very definition of rank follows the method for finding it. The process of finding the rank of a matrix, by definition, can be schematically represented as follows:

Let me explain this diagram in more detail. Let's start reasoning from the very beginning, i.e. from the first order minors of some matrix $A$.

If all first-order minors (i.e., elements of the matrix $A$) are equal to zero, then $\rang A=0$. If among the first-order minors there is at least one that is not equal to zero, then $\rang A≥ 1$. Let's move on to checking second-order minors.
If all second-order minors are equal to zero, then $\rang A=1$. If among the second-order minors there is at least one that is not equal to zero, then $\rang A≥ 2$. Let's move on to checking third-order minors.
If all third-order minors are equal to zero, then $\rang A=2$. If among the third-order minors there is at least one that is not equal to zero, then $\rang A≥ 3$. Let's move on to checking fourth-order minors.
If all fourth-order minors are equal to zero, then $\rang A=3$. If among the fourth-order minors there is at least one that is not equal to zero, then $\rang A≥ 4$. We move on to checking fifth-order minors and so on.

What awaits us at the end of this procedure? It is possible that among the kth order minors there will be at least one that is different from zero, and all (k+1) order minors will be equal to zero. This means that k is the maximum order of minors, among which there is at least one that is not equal to zero, i.e. the rank will be equal to k. There may be a different situation: among the kth order minors there will be at least one that is not equal to zero, but it will no longer be possible to form (k+1) order minors. In this case, the rank of the matrix is also equal to k. In short, the order of the last composed non-zero minor will be equal to the rank of the matrix.

Let's move on to examples in which the process of finding the rank of a matrix, by definition, will be clearly illustrated. Let me emphasize once again that in the examples of this topic we will find the rank of matrices using only the definition of rank. Other methods (calculating the rank of a matrix using the method of bordering minors, calculating the rank of a matrix using the method of elementary transformations) are discussed in the following topics.

By the way, it is not at all necessary to start the procedure for finding the rank with minors of the smallest order, as was done in examples No. 1 and No. 2. You can immediately move on to minors of higher orders (see example No. 3).

Example No. 1

Find the rank of the matrix $A=\left(\begin(array)(ccccc) 5 & 0 & -3 & 0 & 2 \\ 7 & 0 & -4 & 0 & 3 \\ 2 & 0 & -1 & 0 & 1 \end(array) \right)$.

This matrix has size $3\times 5$, i.e. contains three rows and five columns. Of the numbers 3 and 5, the minimum is 3, therefore the rank of the matrix $A$ is no more than 3, i.e. $\rang A≤ 3$. And this inequality is obvious, since we will no longer be able to form fourth-order minors - they require 4 rows, and we have only 3. Let’s move on directly to the process of finding the rank of a given matrix.

Among the minors of the first order (i.e. among the elements of the matrix $A$) there are non-zero ones. For example, 5, -3, 2, 7. In general, we are not interested in the total number of non-zero elements. There is at least one non-zero element - and that's enough. Since among the first-order minors there is at least one non-zero, we conclude that $\rang A≥ 1$ and proceed to checking the second-order minors.

Let's start exploring second order minors. For example, at the intersection of rows No. 1, No. 2 and columns No. 1, No. 4 there are elements of the following minor: $\left|\begin(array)(cc) 5 & 0 \\ 7 & 0 \end(array) \right| $. For this determinant, all elements of the second column are equal to zero, therefore the determinant itself is equal to zero, i.e. $\left|\begin(array)(cc) 5 & 0 \\ 7 & 0 \end(array) \right|=0$ (see property No. 3 in the topic of properties of determinants). Or you can simply calculate this determinant using formula No. 1 from the section on calculating second- and third-order determinants:

$$ \left|\begin(array)(cc) 5 & 0 \\ 7 & 0 \end(array) \right|=5\cdot 0-0\cdot 7=0. $$

The first second-order minor we tested turned out to be equal to zero. What does this mean? About the need to further check second-order minors. Either they will all turn out to be zero (and then the rank will be equal to 1), or among them there will be at least one minor that is different from zero. Let's try to make a better choice by writing a second-order minor, the elements of which are located at the intersection of rows No. 1, No. 2 and columns No. 1 and No. 5: $\left|\begin(array)(cc) 5 & 2 \\ 7 & 3 \end(array) \right|$. Let's find the value of this second-order minor:

$$ \left|\begin(array)(cc) 5 & 2 \\ 7 & 3 \end(array) \right|=5\cdot 3-2\cdot 7=1. $$

This minor is not equal to zero. Conclusion: among the second-order minors there is at least one that is different from zero. Therefore $\rang A≥ 2$. We need to move on to the study of third-order minors.

If we choose column No. 2 or column No. 4 to form third-order minors, then such minors will be equal to zero (since they will contain a zero column). It remains to check only one third-order minor, the elements of which are located at the intersection of columns No. 1, No. 3, No. 5 and rows No. 1, No. 2, No. 3. Let's write down this minor and find its value:

$$ \left|\begin(array)(ccc) 5 & -3 & 2 \\ 7 & -4 & 3 \\ 2 & -1 & 1 \end(array) \right|=-20-18-14 +16+21+15=0. $$

So, all third order minors are equal to zero. The last non-zero minor we compiled was of second order. Conclusion: the maximum order of minors, among which there is at least one non-zero, is 2. Therefore, $\rang A=2$.

Answer: $\rang A=2$.

Example No. 2

Find the rank of the matrix $A=\left(\begin(array) (cccc) -1 & 3 & 2 & -3\\ 4 & -2 & 5 & 1\\ -5 & 0 & -4 & 0\\ 9 & 7 & 8 & -7 \end(array) \right)$.

We have a square matrix of the fourth order. Let us immediately note that the rank of this matrix does not exceed 4, i.e. $\rang A≤ 4$. Let's start finding the rank of the matrix.

Among the first-order minors (i.e., among the elements of the matrix $A$) there is at least one that is not equal to zero, therefore $\rang A≥ 1$. Let's move on to checking second-order minors. For example, at the intersection of rows No. 2, No. 3 and columns No. 1 and No. 2, we obtain the following second-order minor: $\left| \begin(array) (cc) 4 & -2 \\ -5 & 0 \end(array) \right|$. Let's calculate it:

$$\left| \begin(array) (cc) 4 & -2 \\ -5 & 0 \end(array) \right|=0-10=-10. $$

Among the second-order minors there is at least one that is not equal to zero, so $\rang A≥ 2$.

Let's move on to third-order minors. Let's find, for example, a minor whose elements are located at the intersection of rows No. 1, No. 3, No. 4 and columns No. 1, No. 2, No. 4:

$$\left | \begin(array) (cccc) -1 & 3 & -3\\ -5 & 0 & 0\\ 9 & 7 & -7 \end(array) \right|=105-105=0. $$

Since this third-order minor turned out to be equal to zero, it is necessary to investigate another third-order minor. Either all of them will be equal to zero (then the rank will be equal to 2), or among them there will be at least one that is not equal to zero (then we will begin to study fourth-order minors). Let's consider a third-order minor, the elements of which are located at the intersection of rows No. 2, No. 3, No. 4 and columns No. 2, No. 3, No. 4:

$$\left| \begin(array) (ccc) -2 & 5 & 1\\ 0 & -4 & 0\\ 7 & 8 & -7 \end(array) \right|=-28. $$

Among the third-order minors there is at least one non-zero, so $\rang A≥ 3$. Let's move on to checking fourth-order minors.

Any fourth-order minor is located at the intersection of four rows and four columns of the matrix $A$. In other words, the fourth order minor is the determinant of the matrix $A$, since this matrix contains 4 rows and 4 columns. The determinant of this matrix was calculated in example No. 2 of the topic “Reducing the order of the determinant. Decomposing the determinant in a row (column)”, so let’s just take the finished result:

$$\left| \begin(array) (cccc) -1 & 3 & 2 & -3\\ 4 & -2 & 5 & 1\\ -5 & 0 & -4 & 0\\ 9 & 7 & 8 & -7 \end (array)\right|=86. $$

So the fourth order minor is not equal to zero. We can no longer form minors of the fifth order. Conclusion: the highest order of minors, among which there is at least one non-zero, is 4. Result: $\rang A=4$.

Answer: $\rang A=4$.

Example No. 3

Find the rank of the matrix $A=\left(\begin(array) (cccc) -1 & 0 & 2 & -3\\ 4 & -2 & 5 & 1\\ 7 & -4 & 0 & -5 \end( array) \right)$.

Let's immediately note that this matrix contains 3 rows and 4 columns, so $\rang A≤ 3$. In the previous examples, we began the process of finding the rank by considering minors of the smallest (first) order. Here we will try to immediately check the minors of the highest possible order. For the matrix $A$ these are the third order minors. Let's consider a third-order minor, the elements of which lie at the intersection of rows No. 1, No. 2, No. 3 and columns No. 2, No. 3, No. 4:

$$\left| \begin(array) (ccc) 0 & 2 & -3\\ -2 & 5 & 1\\ -4 & 0 & -5 \end(array) \right|=-8-60-20=-88. $$

So, the highest order of minors, among which there is at least one that is not equal to zero, is 3. Therefore, the rank of the matrix is 3, i.e. $\rang A=3$.

Answer: $\rang A=3$.

In general, finding the rank of a matrix by definition is, in the general case, a rather labor-intensive task. For example, a relatively small matrix of size $5\times 4$ has 60 second-order minors. And even if 59 of them are equal to zero, then the 60th minor may turn out to be non-zero. Then you will have to study third-order minors, of which this matrix has 40 pieces. Usually they try to use less cumbersome methods, such as the method of bordering minors or the method of equivalent transformations.

Previously for a square matrix th order the concept of minor was introduced
element . Let us recall that this is the name given to the determinant of order
, obtained from the determinant
by crossing out th line and th column.

Let us now introduce the general concept of minor. Let's consider some not necessarily square matrix . Let's choose some line numbers
And column numbers
.

Definition. Minor order matrices (corresponding to the selected rows and columns) is called the order determinant , formed by the elements at the intersection of the selected rows and columns, i.e. number

Each matrix has as many minors of a given order , in how many ways can you select line numbers
and columns
.

Definition. In the matrix sizes
minor order called basic, if it is nonzero and all minors are of order
equal to zero or minor order
at the matrix not at all.

It is clear that a matrix can have several different basis minors, but all basis minors have the same order. Indeed, if all minors are of order
are equal to zero, then all minors of the order are equal to zero
, and, consequently, all higher orders.

Definition. Matrix rank The order of the basis minor is called, or, in other words, the largest order for which minors other than zero exist. If all elements of a matrix are equal to zero, then the rank of such a matrix, by definition, is considered zero.

Matrix rank we will denote by the symbol
. From the definition of rank it follows that for the matrix sizes
the ratio is correct.

Two ways to calculate the rank of a matrix

A) Bordering minor method

Let a minor be found in the matrix
-th order, different from zero. Let's consider only those minors
-th order, which contain (edge) a minor
: if they are all equal to zero, then the rank of the matrix is . Otherwise, among the bordering minors there is a non-zero minor
-th order, and the whole procedure is repeated.

Example 9 . Find the rank of a matrix by the method of bordering minors.

Let's choose a second-order minor
. There is only one third-order minor, bordering the selected minor
. Let's calculate it.

So it's minor
basic, and the rank of the matrix is equal to its order, i.e.

It is clear that iterating through the minors in this way in search of the basis is a task associated with large calculations, if the dimensions of the matrix are not very small. There is, however, a simpler way to find the rank of a matrix - using elementary transformations.

b) Elementary transformation method

Definition. Elementary matrix transformations The following transformations are called:

multiplying a string by a number other than zero;

adding another line to one line;

rearrangement of lines;

the same column transformations.

Transformations 1 and 2 are performed element by element.

By combining transformations of the first and second types, we can add a linear combination of the remaining strings to any string.

Theorem. Elementary transformations do not change the rank of the matrix.

(No proof)

The idea of a practical method for calculating the rank of a matrix

is that with the help of elementary transformations this matrix lead to the appearance

, (5)

in which the "diagonal" elements
are different from zero, and the elements located below the “diagonal” ones are equal to zero. Let's agree to call the matrix this type of triangular (otherwise, it is called diagonal, trapezoidal or ladder). After matrix reduction to the triangular form we can immediately write that
.

In fact,
(since elementary transformations do not change the rank). But the matrix there is a non-zero minor order :

and any minor of order
contains the null string and is therefore equal to zero.

Let us now formulate the practical rank calculation rule matrices using elementary transformations: to find the rank of the matrix it should be brought to a triangular form using elementary transformations . Then the rank of the matrix will be equal to the number of non-zero rows in the resulting matrix .

Example 10. Find the rank of a matrix by the method of elementary transformations

Solution.

Let's swap the first and second lines (since the first element of the second line is −1 and it will be convenient to perform transformations with it). As a result, we obtain a matrix equivalent to this one.

Let's denote -that row of the matrix – . We need to reduce the original matrix to triangular form. We will consider the first line to be the leading line; it will participate in all transformations, but itself remains unchanged.

At the first stage, we will perform transformations that allow us to get zeros in the first column, except for the first element. To do this, subtract the first line from the second line, multiplied by 2
, add the first to the third line
, and from the third we subtract the first, multiplied by 3
We obtain a matrix whose rank coincides with the rank of this matrix. Let's denote it with the same letter :

Since we need to reduce the matrix to form (5), we subtract the second from the fourth row. In this case we have:

A matrix of triangular form is obtained, and we can conclude that
, i.e. the number of non-zero lines. Briefly, the solution to the problem can be written as follows:

In each matrix, two ranks can be associated: a row rank (the rank of the row system) and a column rank (the rank of the column system).

Theorem

The row rank of a matrix is equal to its column rank.

Matrix rank

Definition

Matrix rank$A$ is the rank of its row or column system.

Denoted by $\operatorname(rang) A$

In practice, to find the rank of a matrix, the following statement is used: the rank of a matrix is equal to the number of non-zero rows after reducing the matrix to echelon form.

Elementary transformations over the rows (columns) of a matrix do not change its rank.

The rank of a step matrix is equal to the number of its non-zero rows.

Example

Exercise. Find the rank of the matrix $ A=\left(\begin(array)(cccc)(0) & (4) & (10) & (1) \\ (4) & (8) & (18) & (7) \ \ (10) & (18) & (40) & (17) \\ (1) & (7) & (17) & (3)\end(array)\right) $

Solution. Using elementary transformations on its rows, we reduce the matrix $A$ to echelon form. To do this, first subtract the second two from the third line:

$$ A \sim \left(\begin(array)(cccc)(0) & (4) & (10) & (1) \\ (4) & (8) & (18) & (7) \\ (2) & (2) & (4) & (3) \\ (1) & (7) & (17) & (3)\end(array)\right) $$

From the second line we subtract the fourth line, multiplied by 4; from the third - two fourths:

$$ A \sim \left(\begin(array)(rrrr)(0) & (4) & (10) & (1) \\ (0) & (-20) & (-50) & (-5 ) \\ (0) & (-12) & (-30) & (-3) \\ (1) & (7) & (17) & (3)\end(array)\right) $$

We add the first five to the second line, and the third three to the third:

$$ A \sim \left(\begin(array)(cccc)(0) & (4) & (10) & (1) \\ (0) & (0) & (0) & (0) \\ (0) & (0) & (0) & (0) \\ (1) & (7) & (17) & (3)\end(array)\right) $$

Swap the first and second lines:

$$ A \sim \left(\begin(array)(cccc)(0) & (0) & (0) & (0) \\ (0) & (4) & (10) & (1) \\ (0) & (0) & (0) & (0) \\ (1) & (7) & (17) & (3)\end(array)\right) $$

$$ A \sim \left(\begin(array)(cccc)(1) & (7) & (17) & (3) \\ (0) & (4) & (10) & (1) \\ (0) & (0) & (0) & (0) \\ (0) & (0) & (0) & (0)\end(array)\right) \Rightarrow \operatorname(rang) A=2 $$

Answer.$ \operatorname(rang) A=2 $

Method of bordering minors

Another method for finding the rank of a matrix is based on this theorem - minor edging method. The essence of this method is to find minors, starting from lower orders and moving to higher ones. If the minor of the $n$th order is not equal to zero, and all minors of the $n+1$th order are equal to zero, then the rank of the matrix will be equal to $n$ .

Example

Exercise. Find the rank of the matrix $ A=\left(\begin(array)(rrrr)(1) & (2) & (-1) & (-2) \\ (2) & (4) & (3) & (0 ) \\ (-1) & (-2) & (6) & (6)\end(array)\right) $ using the minor edging method.

Solution. Minors of minimal order are minors of first order, which are equal to the elements of the matrix $A$. Consider, for example, minor $ M_(1)=1 \neq 0 $ . located in the first row and first column. We border it with the help of the second row and the second column, we get the minor $ M_(2)^(1)=\left| \begin(array)(ll)(1) & (2) \\ (2) & (4)\end(array)\right|=0 $ ; Let's consider another minor of the second order, for this we border the minor $M_1$ with the help of the second row and the third column, then we have the minor $ M_(2)^(2)=\left| \begin(array)(rr)(1) & (-1) \\ (2) & (3)\end(array)\right|=5 \neq 0 $ , that is, the rank of the matrix is not less than two. Next, we consider third-order minors that border the minor $ M_(2)^(2) $ . There are two such minors: a combination of the third row with the second column or with the fourth column. Let's calculate these minors.

Any matrix A order m×n can be considered as a collection m string vectors or n column vectors.

Rank matrices A order m×n is the maximum number of linearly independent column vectors or row vectors.

If the matrix rank A equals r, then it is written:

Finding the rank of a matrix

Let A arbitrary order matrix m× n. To find the rank of a matrix A We apply the Gaussian elimination method to it.

Note that if at some stage of elimination the leading element is equal to zero, then we swap this line with the line in which the leading element is different from zero. If it turns out that there is no such line, then move on to the next column, etc.

After the forward Gaussian elimination process, we obtain a matrix whose elements under the main diagonal are equal to zero. In addition, there may be zero row vectors.

The number of non-zero row vectors will be the rank of the matrix A.

Let's look at all this using simple examples.

Example 1.

Multiplying the first line by 4 and adding to the second line and multiplying the first line by 2 and adding to the third line we have:

Multiply the second line by -1 and add it to the third line:

We received two non-zero rows and, therefore, the rank of the matrix is 2.

Example 2.

Let's find the rank of the following matrix:

Multiply the first line by -2 and add it to the second line. Similarly, we reset the elements of the third and fourth rows of the first column:

Let's reset the elements of the third and fourth rows of the second column by adding the corresponding rows to the second row multiplied by the number -1.

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, .

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 obtain 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.

Definition.

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. 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 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 exactly the same way, but only with the part of matrix A (2) marked in the figure.