Matrix rank: definition, methods of finding, examples, solutions. Finding the rank of a matrix What is the rank of a matrix 1123 5814

And also consider an important practical application of the topic: study of a system of linear equations for compatibility.

What is the rank of a matrix?

The humorous epigraph of the article contains a great deal of truth. The word "rank" itself is usually associated with some kind of hierarchy, most often with the 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 overall degree of "toughness".

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

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

Zero matrix rankany size is zero.

Note : the zero matrix is denoted by the Greek letter "theta"

In order to better understand the rank of the matrix, hereinafter I will draw on the materials analytical geometry. Consider zero vector of our three-dimensional space, which does not set a certain direction and is useless for building affine basis. From an algebraic point of view, the coordinates of a given vector are written in matrix"one by three" and logical (in the specified 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-null element, and that's something!

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

And generally speaking - if in matrix arbitrary sizes has at least one non-zero element, then its rank not less units.

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

Theoretical background : in linear algebra, a vector is an element of a vector space (defined through 8 axioms), which, in particular, can be an ordered row (or column) of real numbers with the operations of addition and multiplication by a real number defined for them. For more information about vectors, see 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 in building three-dimensional basis, being redundant in this sense. Thus, the rank of this matrix is also equal to one.

We rewrite the coordinates of the vectors in 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 in full agreement with our geometric sense of rank.

An important statement follows from the above example:

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

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

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

Did he help us in building 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 \u200b\u200bone. You can take as many collinear vectors as you like, say 100, put their coordinates into a 100 by 3 matrix, and the rank of such a skyscraper will still remain one.

Let's get acquainted with the matrix whose rows 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 do not seem to be proportional ... so, in theory, three. However, the rank of this matrix is also equal to two. I added the first two lines and wrote down the result at the bottom, i.e. linearly expressed third line through the first two. Geometrically, the rows of the matrix correspond to the coordinates of three coplanar vectors, and among this triple there is a pair of non-collinear comrades.

As you can see linear dependence in the considered matrix is not obvious, and today we will just learn how to bring it “to clean water”.

I think many people 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 any number of rows is added to the matrix, then its rank will still be three.

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

Definition : matrix rank is the maximum number of linearly independent rows. Or: the rank of a matrix is the maximum number of linearly independent columns. Yes, they always match.

An important practical guideline follows from the above: the rank of a 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 will certainly not exceed 4.

Notation: in world theory and practice there is no generally accepted standard for designating the rank of the matrix, the most common one can be found: - as they say, an Englishman writes one thing, a German another. Therefore, based on the well-known anecdote about the American and Russian hell, let's designate the rank of the matrix with a native word. For example: . And if the matrix is \u200b\u200b"nameless", of which there are a lot, then you can simply write .

How to find the rank of a matrix using minors?

If our grandmother had a fifth column in the matrix, then another 4th order minor (“blue”, “raspberry” + 5th column) should have been calculated.

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

Perhaps not everyone fully comprehended this phrase: the 4th order minor is equal to zero, but among the 3rd order minors there was a non-zero one - therefore, the maximum order non-zero minor and equal to 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, then the task will be rejected with a high probability, since it usually implies a standard “bottom-up” solution. And in the considered example, the zero determinant of the 4th order even allows us to assert that the rank of the matrix is only less than four.

I must admit that I came up with the analyzed problem 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 by the method of fringing minors

Solution and answer at the end of the lesson.

When is the algorithm running fastest? Let's go back 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 - .

Thinking is not at all hypothetical - there are many examples where the whole thing 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 Gauss method?

This section is intended for readers who are already familiar with Gauss method and little by little got their hands on it.

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

1) using elementary transformations, we bring the matrix to a step form;

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

It is quite clear that using the Gauss method does not change the rank of the matrix, and the essence here is extremely simple: according to the algorithm, in the course of elementary transformations, all unnecessary proportional (linearly dependent) lines are identified and removed, as a result of which a “dry residue” remains - the maximum number of linearly independent lines.

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

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

(2) Zero lines are removed.

So 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 finding out the rank! By the way, let's dwell on the question again, why not? Source Matrix carries information that is fundamentally different from matrix and row information. In some mathematical models (without exaggeration), the difference in one number can be a matter of life and death. ... I remembered the school mathematics teachers of primary and secondary classes, who ruthlessly cut off the grade by 1-2 points for the slightest inaccuracy or deviation from the algorithm. And it was terribly disappointing when, instead of the seemingly guaranteed "five", it turned out "good" or even worse. Understanding came much later - how else to entrust a person with satellites, nuclear warheads and power plants? 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: given a four-by-five matrix, which means that its rank is certainly no more than 4.

In the first column, there is no 1 or -1, therefore, additional steps are needed to obtain at least one unit. Over the entire existence of the site, I have been repeatedly asked the question: “Is it possible to rearrange the columns during elementary transformations?”. Here - rearranged the first or second column, and everything is fine! In most tasks where Gauss method, columns can really be rearranged. BUT DON'T. And the point is not even a possible confusion with variables, the point is that in the classical course of teaching higher mathematics this action is traditionally not considered, therefore, such a curtsy will be looked at VERY crookedly (or even forced to redo everything).

The second point concerns numbers. In the course of the decision, it is useful to be guided by the following rule of thumb: elementary transformations should, if possible, reduce the numbers of the matrix. Indeed, it is much easier to work with a one-two-three than, for example, with 23, 45 and 97. And the first action is aimed not only at obtaining a unit in the first column, but also at eliminating the numbers 7 and 11.

First the full solution, then the comments:

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

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

(3) The first row was added to the second row, multiplied by -3.

The matrix reduced to a stepped 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 Gauss method does not imply unambiguous rigidity, and your solution will most likely be different from my solution. A brief sample of the task at the end of the lesson.

What method to use to find the rank of a matrix?

In practice, it is often not said at all which method should be used to find the rank. In such a situation, one should analyze the condition - for some matrices it is more rational to carry out the solution 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 way 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 / matching columns, then it is still worth amputating:

(1) The fifth column is zero, we remove it from the matrix. Thus, the rank of the matrix is at most four. The first row is multiplied by -1. This is another signature feature of the Gaussian method, which makes the following action a pleasant walk:

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

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

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

(5) The last two lines are proportional, we delete the fifth one.

The result is 4 rows.

Answer:

Standard five-story building for self-exploration:

Example 6

Find the rank of a matrix

Short solution and answer at the end of the lesson.

It should be noted that the phrase "matrix rank" is not so common in practice, and in most problems you can do without it. But there is one task where the concept under consideration is the main character, and in conclusion of the article we will consider this practical application:

How to investigate the system of linear equations for compatibility?

Often, in addition to solving 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 this verification is played by Kronecker-Capelli theorem, which I will formulate in the required form:

If rank system matrices equal to rank augmented matrix system, then the system is consistent, and if the 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 - augmented matrix system(i.e. matrix with coefficients at variables + column of free terms).

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

Matrix minor

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

Definition 1

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

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

It follows from the example that the first-order minors of the matrix A are the matrix elements themselves.

We can give several examples of minors of the 2nd order. Let's choose two rows and two columns. For example, 1st and 2nd row, 3rd and 4th column.

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

Another 2nd order minor of matrix A is 0 0 1 1 = 0

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

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

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

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

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

k ≤ m i n (p , n) = m i n (3 , 4) = 3

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

The number of minors is calculated using the following formula:

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

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

Matrix rank: methods of finding

Definition 2

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

Designation 1

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

From the definition of the rank of a matrix and the minor of a matrix, it becomes clear that the rank of a zero matrix is equal to zero, and the rank of a non-zero matrix is different from zero.

Finding the rank of a matrix by definition

Definition 3

Minor enumeration method - a method based on determining the rank of a matrix.

Algorithm of actions by enumeration of minors :

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

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

Let's do the same with the rank of the 3rd order: if all the minors of the matrix are equal to zero, then the rank will be equal to two. If there is at least one non-zero third-order minor, then the rank of the matrix is at least three. And so on, by analogy.

Example 2

Find the rank of a matrix:

A \u003d - 1 1 - 1 - 2 0 2 2 6 0 - 4 4 3 11 1 - 7

Since the matrix is non-zero, its rank is at least equal to one.

The 2nd order minor - 1 1 2 2 = (- 1) × 2 - 1 × 2 = 4 is non-zero. This implies that the rank of the matrix A is at least two.

We sort through the minors of the 3rd order: C 3 3 × C 5 3 \u003d 1 5 ! 3! (5 - 3) ! = 10 pieces.

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

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

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

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

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

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

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

The 3rd order minors are zero, so the rank of the matrix is two.

Answer : Rank (A) = 2.

Finding the rank of a matrix by the method of fringing minors

Definition 3

Fringing Minor Method - a method that allows you to get a result with less computational work.

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

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

Example 3

Find the rank of a matrix:

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

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

We write down all bordering minors:

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

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

Theorem 1

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

Action algorithm :

To find the rank of a matrix, it is not necessary to go through all the minors, just look at the borders.

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

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

Example 4

Find the rank of a matrix by the method of fringing minors

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

How to decide?

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

2 1 4 2 = 2 x 2 - 1 x 4 = 0 2 0 4 1 = 2 x 1 - 0 x 4 = 2

We have found a bordering minor of the 2nd order which is not equal to zero 2 0 4 1 .

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

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

Answer : Rank(A) = 2.

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

Recall what elementary transformations are.

Elementary transformations:

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

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

Definition 5

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

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

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

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

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

For what?

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

Let's illustrate this process:

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

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

A ~ 1 b 12 b 13 ⋯ b 1 k b 1 k + 1 ⋯ b 1 n 0 1 b 23 ⋯ b 2 k b 2 k + 1 ⋯ b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 b k k + 1 ⋯ b k n 0 0 0 ⋯ 0 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 0 0 ⋯ 0 , R a n k (A) = k

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

A ~ 1 b 12 b 13 ⋯ b 1 p b 1 p + 1 ⋯ b 1 n 0 1 b 23 ⋯ b 2 p b 2 p + 1 ⋯ b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 b p p + 1 ⋯ b p n , R a n k (A) = p

A ~ 1 b 12 b 13 ⋯ b 1 k b 1 k + 1 ⋯ b 1 n 0 1 b 23 ⋯ b 2 k b 2 k + 1 ⋯ b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 b k k + 1 ⋯ b k n 0 0 0 ⋯ 0 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 0 0 ⋯ 0

for square matrices A of order n by n:

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

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

Example 5

Find the rank of matrix A using elementary transformations:

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

How to decide?

Since the element a 11 is non-zero, it is necessary to multiply the elements of the first row of the matrix A by 1 a 11 \u003d 1 2:

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

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

~ A (1) \u003d 1 1 2 - 1 3 3 0 0 - 1 1 - 1 2 - 7 5 - 2 4 - 15 7 2 - 4 11 ~ A (2) \u003d \u003d 1 1 2 - 1 3 3 + 1 (- 3) 0 + 1 2 (- 3) 0 + (- 1) (- 3) - 1 + 3 (- 3) 1 + 1 (- 3) - 1 + 1 2 (- 3) 2 + (- 1) (- 1) - 7 + 3 (- 1) 5 + 1 (- 5) - 2 + 1 2 (- 5) 4 + (- 1) (- 5) - 15 + 3 (- 5) 7 + 1 (- 7) 2 + 1 2 (- 7) - 4 + (- 1) (- 7) 11 + 3 (- 7) =

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

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

A (3) \u003d 1 1 2 - 1 3 0 1 - 2 20 3 0 - 3 2 3 - 10 0 - 9 2 9 - 30 0 - 3 2 3 - 10 ~ A (4) \u003d 1 1 2 - 1 3 0 1 - 2 20 3 0 - 3 2 + 1 3 2 3 + (- 2) 3 2 - 10 + 20 3 × 3 2 0 - 9 2 + 1 9 2 9 + (- 2) 9 2 - 30 + 20 3 × 9 2 0 - 3 2 + 1 3 2 3 + (- 2) 3 2 - 10 + 20 3 × 3 2 = = 1 1 2 - 1 3 0 1 - 2 20 3 0 0 0 0 0 0 0 0 0 0 0 0

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

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

Comment

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

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

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 system of rows or columns.

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 the matrix has been reduced to a stepped form.

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

The rank of a step matrix is equal to the number of its nonzero rows.

Example

Exercise. Find the rank of a 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 over its rows, we reduce the matrix $A$ to a step 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 three thirds 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(rank) A=2 $

Minor bordering method

Another method for finding the rank of a matrix is based on this theorem - minor bordering method. The essence of this method is to find minors, starting from lower orders and moving to higher ones. If the $n$-th order minor is non-zero, and all $n+1$-th minors are equal to zero, then the rank of the matrix will be equal to $n$ .

Example

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

Solution. The minors of the minimum order are the first order minors, which are equal to the elements of the matrix $A$ . Consider, for example, the minor $ M_(1)=1 \neq 0 $ . located in the first row and first column. Bordering it with the second row and second column, we get the minor $ M_(2)^(1)=\left| \begin(array)(ll)(1) & (2) \\ (2) & (4)\end(array)\right|=0 $ ; 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 at least two. Next, we consider third-order minors that surround 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. We calculate these minors.

§3. Matrix rank

Determining the rank of a matrix

Linear dependent rows

Elementary Matrix Transformations

Equivalent matrices

Algorithm for finding the rank of a matrix using elementary transformations

§four. Determinants of the first, second and third order

First order determinant

Second order determinant

Third order determinant

Sarrus rule

§5. Calculation of determinants of large orders

Algebraic addition

Laplace's theorem

Triangular matrix determinant

Application. The concept of a determinant P th order in general.

§ 3. Matrix rank

Each matrix is characterized by a certain number that is important in solving systems of linear equations. This number is called matrix rank.

Matrix rank is equal to the number of its linearly independent rows (columns), through which all other rows (columns) of it are linearly expressed.

Rows (columns) of a matrix are called linearly dependent if their corresponding elements are proportional.

In other words, the elements of one of the linearly dependent rows are equal to the elements of the other, multiplied by the same number. For example, rows 1 and 2 of the matrix BUT are linearly dependent if , where (λ is some number).

Example. Find the rank of a matrix

Solution.

The second row is obtained from the first if its elements are multiplied by -3, the third is obtained from the first if its elements are multiplied by 0, and the fourth row cannot be expressed in terms of the first. It turns out that the matrix has two linearly independent rows, because the first and fourth rows are not proportional, hence the rank of the matrix is 2.

Matrix rank BUT denoted rank A or r(A).

From the definition of the rank of a matrix, it follows:

1. The rank of a matrix does not exceed the smallest of its dimensions, i.e. for matrix A m × n .

2. The rank of a matrix is zero only if it is a zero matrix.

In the general case, determining the rank of a matrix is quite laborious. To facilitate this task, transformations are used that preserve the rank of the matrix, which are called elementary transformations:

1) discarding a zero row (column);

2) multiplication of all elements of a row (column) by a number other than zero;

3) changing the order of rows (columns);

4) adding to the elements of one row (column) the corresponding elements of another row (column), multiplied by any number;

5) matrix transposition.

The two matrices are called equivalent if one is obtained from the other by a finite number of elementary transformations.

The equivalence of matrices is indicated by the sign "~" (equivalent).

With the help of elementary transformations, any matrix can be reduced to a triangular form, then the calculation of its rank is not difficult.

The process of calculating the rank of a matrix using elementary transformations let's look at an example.

Example. Find the rank of a matrix

A =

Solution.

Our task is to bring the matrix to a triangular form, i.e. using elementary transformations, ensure that only zeros are below the main diagonal in the matrix.

1. Consider the first line. If element a 11 = 0, then when permuting rows or columns, we achieve that a 11 ¹ 0. In our example, let's swap, for example, the first and second rows of the matrix:

A =

Now element a 11 ¹ 0. Multiplying the first row by suitable numbers and adding with other rows, we will ensure that all elements of the first column (except a 11) were equal to zero.

2. Consider now the second line. If element a 22 = 0, then when permuting rows or columns, we achieve that a 22 ¹ 0. If the element a 22 ¹ 0 (and we have a 22 = –1 ¹ 0), then by multiplying the second row by suitable numbers and adding to other rows, we will ensure that all elements of the second column (except a 22) were equal to zero.

3. If in the process of transformations rows (columns) are obtained that consist entirely of zeros, then we discard them. In our example, we will discard lines 3 and 4:

The last matrix has a stepped form and contains two rows. They are linearly independent, hence the rank of the matrix is 2.

§ four. Determinants of the first, second and third order

Among the variety of matrices, square matrices are singled out separately. This type of matrix is good because:

1. The identity matrices are square.

2. You can multiply and add any square matrices of the same order, and you get a matrix of the same order.

3. Square matrices can be raised to a power.

Also, only square matrices can have a determinant.

Matrix determinant is a special number calculated according to some rule. Matrix determinant BUT denoted:

Or with straight brackets: ,

Or capital Greek letter "delta": Δ( A),

Or the symbol "determinant": det ( A).

Determinant of a first order matrix BUT= (a 11) or first order determinant, is a number equal to the matrix element:

∆1 = =a 11

Second order matrix determinant or second order determinant

Example:

The determinant of a matrix of the third order or third order determinant, is a number that is calculated by the formula:

The third order determinant can be calculated using Sarrus rule .

Sarrus rule. The first two columns are signed to the third-order determinant on the right and with a plus sign (+) they take the sum of the products of the three elements located on the main diagonal of the determinant and on the "straight lines" parallel to the main diagonal, with a minus sign (-) they take the sum of the products of the elements located on the second diagonal and on the "straight lines" parallel to it.

Example:

It is easy to see that the number of terms in the determinant increases with its order. In general, in the determinant P th order, the number of terms is 1 2 3 . . . P = P!.

Let's check: for Δ 1 the number of terms is equal to 1! = 1,

for Δ 2 the number of terms is 2! = 1 2 = 2,

for Δ 3 the number of terms is 3! = 1 2 3 = 6.

It follows that for the 4th order determinant, the number of terms is 4! = 1 2 3 4 = 24, which means that the calculation of such a determinant is quite laborious, not to mention higher-order determinants. Considering this, they try to reduce the calculation of determinants of large orders to the calculation of determinants of the second or third orders.

§ 5. Calculation of determinants of large orders

Let us introduce a number of concepts.

Let a square matrix be given A n-th order:

Minor M ij element a ij is called the determinant ( P– 1)th order obtained from the matrix BUT strikeout i-th line and j-th column.

For example, the element's minor a 12 third order matrices will be:

Algebraic addition BUT ij element a ij is its minor, taken with the sign (−1) i + j:

BUT ij = (−1) i + jM ij

In other words, BUT ij = M ij if i+j even number,

BUT ij = − M ij if i+j odd number.

Example. Find algebraic complements of elements of the second row of a matrix

Solution.

With the help of algebraic complements, one can calculate determinants of large orders, based on Laplace's theorem.

Laplace's theorem. The determinant of a square matrix is equal to the sum of the products of the elements of any of its rows (columns) and their algebraic complements:

– decomposition on the i-th line;

( is the expansion in the jth column).

Example. Calculate matrix determinant decomposition on the first line.

Solution.

Thus, a determinant of any order can be reduced to the calculation of several determinants of a smaller order. It is obvious that for the expansion it is convenient to choose a row or column containing as many zeros as possible.

Let's consider one more example.

Example. Calculate the determinant of a triangular matrix

Solution.

Got that the determinant of a triangular matrix is equal to the product of the elements of its main diagonal .

This important conclusion makes it easy to calculate the determinant of any triangular matrix. This is all the more useful because, if necessary, any determinant can be reduced to a triangular form. In this case, some properties of determinants are used.

Application

The concept of a determinant P th order in general.

In general, one can give a rigorous definition for the matrix determinant P th order, but for this it is necessary to introduce a number of concepts.

permutation numbers 1, 2, ..., n any arrangement of these numbers in a certain order is called. In elementary algebra, it is proved that the number of all permutations that can be formed from n numbers is 12...n = n!. For example, three numbers 1, 2, 3 can form 3! = 6 permutations: 123, 132, 312, 321, 231, 213.

They say that in a given permutation of the number i and j constitute inversion(disorder) if i> j, but i stands in this permutation before j, that is, if the larger number is to the left of the smaller one.

The permutation is called even(or odd) if the total number of inversions is even (odd), respectively.

An operation by means of which one passes from one permutation to another, composed of the same n numbers is called substitution n th degree.

A substitution that transforms one permutation into another is written in two lines in common brackets, and the numbers that occupy the same places in the permutations under consideration are called corresponding and are written one under the other. For example, the symbol

denotes a permutation in which 3 goes to 4, 1 to 2, 2 to 1, 4 to 3. A permutation is called even (or odd) if the total number of inversions in both rows of the substitution is even (odd). Any substitution n th degree can be written as

those. with the natural arrangement of numbers in the top line.

Let us be given a square matrix of order n

Consider all possible products n elements of this matrix, taken one and only one from each row and each column, i.e. works of the form:

where indices q 1 , q 2 ,..., q n make up some permutation of numbers
1, 2,..., n. The number of such products is equal to the number of different permutations from n characters, i.e. equals n!. Work sign , equals (–1) q, where q is the number of inversions in the permutation of the second indices of elements.

determinant n-th order is called the algebraic sum of all possible products over n matrix elements, taken one and only one from each row and each column, i.e. works of the form: . At the same time, the sign of the work equals (–1) q, where q is the number of inversions in the permutation of the second indices of elements.

Linear algebra

Consider a matrix A of size .

A=
Select k rows and k columns in it (
).

Definition 26:Minor The k-th order of the matrix A is the determinant of the square matrix, which is obtained from the given one by selection in it.

k rows and k columns.

Definition 27:rank matrix is called the largest of the non-zero orders of its minors, r(A).

Definition 28: A minor whose order is the same as its rank is called basic minor.

Statement:

1. The rank is expressed as an integer.(
)

2.r=0,
when A is zero.

Elementary transformations of matrices.

The elementary transformations of matrices include the following:

1) multiplication of all elements of any row (column) of the matrix by the same number.

2) addition to the elements of any row (column) of the matrix of the corresponding elements of another row (column) multiplied by the same number;

3) permutation of the rows (columns) of the matrix;

4) discarding the zero row (column);

5) replacement of matrix rows with corresponding columns.

Definition 29: Matrices obtained from one another, under elementary transformations, are called equivalent matrices, denoted by “~“

The main property of equivalent matrices: The ranks of equivalent matrices are equal.

Example 18: Calculate r(A),

Solution: Multiply the first line step by step by (-4)(-2)

(-7) and then add to the second, third and fourth rows respectively.

swap the second and fourth lines
multiply the second row by (-2) and add to the fourth row; add the second and third rows.

add the third and fourth rows.

~
discard the null line

~
r(A)=3
rank of original matrix

equals three.

Definition 30: We call a matrix A a step matrix if all elements of the main diagonal 0, and the elements under the main diagonal are zero.

Sentence:

1) the rank of a step matrix is equal to the number of its rows;

2) any matrix can be reduced to a step form with the help of elementary transformations.

Example 19: At what values of  matrix
has rank equal to one?

Solution: The rank is equal to one if the second-order determinant is equal to zero, i.e.

§6. Systems of linear equations of general form.

view system
---(9) is called a system of general form.

Definition 31: Two systems are said to be equivalent (equivalent) if every solution of the first system is a solution of the second and vice versa.

In system (1) the matrix A=
will be called the main matrix of the system, and =
expanded matrix system

Theorem. Kronecker-Cappelli

For system (9) to be consistent, it is necessary and sufficient that the rank of the main matrix of the system is equal to the rank of the extended matrix, i.e., r(A)=r( )

Theorem 1. If the rank of the matrix of a consistent system is equal to the number of unknowns, then the system has a unique solution.

Theorem 2. If the rank of the matrix of a joint system is less than the number of unknowns, then the system has an infinite number of solutions.

The rule for solving an arbitrary system of linear equations:

1) find the ranks of the main and extended matrices of the system. If a
, then the system is inconsistent.

2) If
=r, then the system is compatible. Find some basic minor of order r. We will call the basic minor, on the basis of which the rank of the matrix was determined.

The unknowns whose coefficients are included in the basic minor are called the main (basic) and left on the left, while the remaining unknowns are called free and transferred to the right side of the equation.

3) Find the expressions of the main unknowns in terms of the free ones. The general solution of the system is obtained.

Example 20: Investigate the system and, in case of its compatibility, find either a unique or general solution