Main minor. Find the rank of a matrix: methods and examples. Finding the rank of a matrix using the bordering minors method

Matrices whose elements are in a given rectangular matrix of order k(which is also called the order of this minor) at the intersection of rows with numbers and columns with numbers .

If the numbers of the marked rows coincide with the numbers of the marked columns, then the minor is called main, and if the first ones are marked k lines and first k columns - corner or leading chief.

The additional minor of an nth-order matrix element is the order determinant (n-1), corresponding to the matrix that is obtained from the matrix by deleting the i-th row and j-th column.

Basic A matrix minor is any non-zero minor of a matrix. maximum order. In order for a minor to be basic, it is necessary and sufficient that all minors bordering it (that is, containing minors of one higher order) are equal to zero. The system of rows (columns) of a matrix associated with a basis minor is a maximal linearly independent subsystem of the system of all rows (columns) of the matrix.

Example

For example, there is a matrix:

Suppose we need to find an additional minor M 23. This minor is the determinant of the matrix obtained by crossing out row 2 and column 3:

We get M 23 = 13

Steps

Finding a determinant

Write down a 3 x 3 matrix. Let us write down a matrix of dimension 3 x 3, which we denote as M, and find its determinant |M|. The following is the general matrix notation we will use and the matrix for our example:

M = (a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33) = (1 5 3 2 4 7 4 6 2) (\displaystyle M=(\begin(pmatrix)a_(11)&a_ (12)&a_(13)\\a_(21)&a_(22)&a_(23)\\a_(31)&a_(32)&a_(33)\end(pmatrix))=(\begin(pmatrix)1&5&3\ \2&4&7\\4&6&2\end(pmatrix)))

Select a row or column of the matrix. This row (or column) will be the reference. The result will be the same no matter which row or column you select. IN in this example let's take the first line. You'll find some tips later on how to select a row or column to make calculations easier.
- Let's select the first row of the matrix M in our example. Circle the numbers 1 5 3. In general form, circle a 11 a 12 a 13 .
Cross out the row or column with the first item. Refer to the reference row (or reference column) and select the first element. Swipe horizontally and vertical line through this element, thus crossing out the column and row with this element. There should be four numbers left. We will consider these elements to be a new 2 x 2 matrix.
- In our example, the reference row would be 1 5 3. The first element is at the intersection of the first column and the first row. Cross out the row and column with this element, that is, the first line and the first column. Write the remaining elements as a 2 x 2 matrix:
- 1 5 3
- 2 4 7
- 4 6 2
Find the determinant of a 2 x 2 matrix. Remember that the determinant of a matrix (a b c d) (\displaystyle (\begin(pmatrix)a&b\\c&d\end(pmatrix))) calculated as ad-bc. From this, you can calculate the determinant of the resulting 2 x 2 matrix, which you can denote as X if you wish. Multiply the two numbers of the matrix X, connected diagonally from left to right (that is, like this: \). Then subtract the result of multiplying the other two numbers diagonally from right to left (that is, like this: /). Use this formula to calculate the determinant of the matrix you just obtained.

Multiply the resulting answer by the selected matrix element M. Remember which element from the reference row (or column) we used when we crossed out other elements in the row and column to get a new matrix. Multiply this element by the resulting minor (the determinant of the 2x2 matrix, which we denoted as X).
- In our example, we chose the element a 11, which was equal to 1. Multiply it by -34 (the determinant of a 2x2 matrix), and we get 1*-34 = -34 .
Determine the sign of the result obtained. Next, you will need to multiply the result by 1, or by -1 to get algebraic complement (cofactor) selected element. The sign of the cofactor will depend on where the element is located in the 3x3 matrix. Remember this simple diagram signs to know the sign of the cofactor:
Repeat all the above steps with the second element of the reference row (or column). Return to the original 3x3 matrix and the row we circled at the very beginning of the calculation. Repeat all actions with this element:
- Cross out the row and column with this element. In our example, we must select element a 12 (equal to 5). Cross out the first row (1 5 3) and the second column (5 4 6) (\displaystyle (\begin(pmatrix)5\\4\\6\end(pmatrix))) matrices.
- Write the remaining elements as a 2x2 matrix. In our example, the matrix will look like (2 7 4 2) (\displaystyle (\begin(pmatrix)2&7\\4&2\end(pmatrix)))
- Find the determinant of this new 2x2 matrix. Use the ad - bc formula above. (2*2 - 7*4 = -24)
- Multiply the resulting determinant by the selected element of the 3x3 matrix. -24 * 5 = -120
- Check to see if you need to multiply the result by -1. Let's use the formula (-1) ij to determine the sign of the algebraic complement. For the element a 12 we selected, the table shows a “-” sign; the formula gives a similar result. That is, we must change the sign: (-1)*(-120) = 120 .
Repeat with the third element. Next you will need to find one more algebraic complement. Calculate it for the last element of the reference row or reference column. The following is short description how the algebraic complement of a 13 is calculated in our example:
- Cross out the first row and third column to get a matrix (2 4 4 6) (\displaystyle (\begin(pmatrix)2&4\\4&6\end(pmatrix)))
- Its determinant is 2*6 - 4*4 = -4.
- Multiply the result by element a 13: -4 * 3 = -12.
- Element a 13 has a + sign in the table above, so the answer will be -12 .
Add up the results. This last step. You need to add the resulting algebraic complements of the elements of the reference row (or reference column). Add them together and you get the value of the determinant of a 3x3 matrix.
- In our example, the determinant is equal to -34 + 120 + -12 = 74 .
How to simplify the task
1. Choose as a reference row (or column) the one that has more zeros. Remember that you can choose as a reference any row or column. The choice of reference row or column does not affect the result. If you select the line with the largest number zeros, you will have to do fewer calculations because you will only need to calculate the algebraic complements for the non-zero elements. That's why:
  - Let's say you selected row 2 with the elements a 21 , a 22 , and a 23 . To find the determinant, you will need to find the determinants of three different 2x2 matrices. Let's call them A 21, A 22, and A 23.
  - That is, the determinant of a 3x3 matrix is equal to a 21 |A 21 | - a 22 |A 22 | + a 23 |A 23 |.
  - If both a 22 and a 23 are 0, then our formula becomes much shorter a 21 |A 21 | - 0*|A 22 | + 0*|A 23 | = a 21 |A 21 | - 0 + 0 = a 21 |A 21 |. That is, it is necessary to calculate only the algebraic complement of one element.
2. Use row addition to simplify a matrix. If you take one row and add another to it, the determinant of the matrix will not change. The same is true for columns. You can do this multiple times, or you can multiply the string values by a constant (before adding) to get as many zeros as possible. Doing this can save a lot of time.
  - For example, we have a matrix of three rows: (9 − 1 2 3 1 0 7 5 − 2) (\displaystyle (\begin(pmatrix)9&-1&2\\3&1&0\\7&5&-2\end(pmatrix)))
  - To get rid of the 9 in place of element a 11 , we can multiply the second line by -3 and add the result to the first. The new first line will be + [-9 -3 0] = .
  - That is, we get a new matrix (0 − 4 2 3 1 0 7 5 − 2) (\displaystyle (\begin(pmatrix)0&-4&2\\3&1&0\\7&5&-2\end(pmatrix))) Try doing the same with the columns to get a zero in place of element a 12.
3. Remember that calculating the determinant of triangular matrices is much easier. The determinant of triangular matrices is calculated as the product of the elements on the main diagonal, from a 11 in the upper left corner to a 33 in the lower right corner. Speech in in this case is about triangular matrices with dimensions 3x3. Triangular matrices can be the following types, depending on location non-zero values:
  - Upper triangular matrix: All non-zero elements are on and above the main diagonal. All elements below the main diagonal are zero.
  - Lower triangular matrix: All non-zero elements are below and on the main diagonal.
  - Diagonal matrix: All non-zero elements are on the main diagonal. It is a special case of the matrices described above.
  - The described method applies to square matrices of any rank. For example, if you use it for a 4x4 matrix, then after the “crossing out” there will be 3x3 matrices left, for which the determinant will be calculated in the above way. Be prepared for the fact that calculating the determinant for matrices of such dimensions manually is a very labor-intensive task!
  - If all elements of a row or column are 0, then the determinant of the matrix is also 0.

Matrix minors

Let given a square matrix A, nth order. Minor some element a ij , determinant of the matrix nth order is called determinant(n - 1)th order, obtained from the original one by crossing out the row and column at the intersection of which the selected element a ij is located. Denoted by M ij.

Let's look at an example determinant of the matrix 3 - its order:

Then according to the definition minor, minor M 12, corresponding to element a 12, will be determinant:

At the same time, with the help minors can make the calculation task easier determinant of the matrix. We need to spread it out matrix determinant along some line and then determinant will equal to the sum all elements of this line to their minors. Decomposition determinant of the matrix 3 - its order will look like this:

The sign in front of the product is (-1) n, where n = i + j.

Algebraic additions:

Algebraic complement element a ij is called its minor, taken with a "+" sign if the sum (i + j) is an even number, and with a "-" sign if this sum is an odd number. Denoted by A ij. A ij = (-1) i+j × M ij.

Then we can reformulate the property stated above. Matrix determinant equal to the sum of the product of the elements of a certain row (row or column) matrices to their corresponding algebraic additions. Example:

4. Inverse matrix and its calculation.

Let A be square matrix nth order.

Square matrix A is called non-degenerate if matrix determinant(Δ = det A) not equal to zero(Δ = det A ≠ 0). Otherwise (Δ = 0) matrix A is called degenerate.

Matrix, allied to matrix Ah, it's called matrix

Where A ij - algebraic complement element a ij given matrices(it is defined in the same way as algebraic complement element determinant of the matrix).

Matrix A -1 is called inverse matrix A, if the condition is met: A × A -1 = A -1 × A = E, where E is unit matrix same order as matrix A. Matrix A -1 has the same dimensions as matrix A.

inverse matrix

If there are square matrices X and A, satisfying the condition: X × A = A × X = E, where E is the unit matrix of the same order, then matrix X is called inverse matrix to the matrix A and is denoted by A -1. Any non-degenerate matrix It has inverse matrix and, moreover, only one, i.e., in order for it to be square matrix A had inverse matrix, it is necessary and sufficient for it determinant was different from zero.

For getting inverse matrix use the formula:

Where M ji is additional minor element a ji matrices A.

5. Matrix rank. Calculating rank using elementary transformations.

Consider a rectangular matrix mxn. Let us select some k rows and k columns in this matrix, 1 £ k £ min (m, n) . From the elements located at the intersection of the selected rows and columns, we compose a k-th order determinant. All such determinants are called matrix minors. For example, for a matrix you can compose second-order minors and first order minors 1, 0, -1, 2, 4, 3.

Definition. The rank of a matrix is the highest order of the non-zero minor of this matrix. Denote the rank of the matrix r(A).

In the example given, the rank of the matrix is two, since, for example, minor

It is convenient to calculate the rank of a matrix using the method elementary transformations. Elementary transformations include the following:

1) rearrangement of rows (columns);

2) multiplying a row (column) by a number other than zero;

3) adding to the elements of a row (column) the corresponding elements of another row (column), previously multiplied by a certain number.

These transformations do not change the rank of the matrix, since it is known that 1) when the rows are rearranged, the determinant changes sign and, if it was not equal to zero, then it will no longer be; 2) when multiplying a string of a determinant by a number that is not equal to zero, the determinant is multiplied by this number; 3) the third elementary transformation does not change the determinant at all. Thus, by performing elementary transformations on a matrix, one can obtain a matrix for which it is easy to calculate the rank of it and, consequently, of the original matrix.

Definition. A matrix obtained from a matrix using elementary transformations is called equivalent and is denoted A IN.

Theorem. The rank of the matrix does not change during elementary matrix transformations.

Using elementary transformations, you can reduce the matrix to the so-called step form, when calculating its rank is not difficult.

Matrix is called stepwise if it has the form:

Obviously, the rank of the echelon matrix is equal to the number of non-zero rows , because there is a minor of order not equal to zero:

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

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

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

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

Definition.

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

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

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

Consider the matrix .

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

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

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

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

Another second-order minor of the matrix A is .

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

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

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

Another third order minor is

obtained by deleting the third column of matrix A.

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

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

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

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

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

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

Let's look at it with an example.

Example.

Find all second order minors of the matrix.

Solution.

Since the order of the original matrix is 3 by 3, then the total 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- This highest order matrix minor, different from zero.

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

From the definitions of matrix rank and matrix minor we can conclude that the rank zero matrix is equal to zero, and the rank non-zero matrix no 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 matrix element 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.

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

In this topic we will consider the concepts of algebraic complement and minor. The presentation of the material is based on the terms explained in the topic "Matrixes. Types of matrices. Basic terms". We will also need some formulas for calculating determinants. Since there are a lot of terms in this topic related to minors and algebraic complements, I will add summary to make it easier to navigate the material.

Minor $M_(ij)$ of element $a_(ij)$

$M_(ij)$ element$a_(ij)$ matrices $A_(n\times n)$ name the determinant of the matrix obtained from matrix $A$ by deleting i-th line and the j-th column (i.e. the row and column at the intersection of which the element $a_(ij)$ is located).

For example, consider a fourth-order square matrix: $A=\left(\begin(array) (ccc) 1 & 0 & -3 & 9\\ 2 & -7 & 11 & 5 \\ -9 & 4 & 25 & 84 \\ 3 & 12 & -5 & 58 \end(array) \right)$. Let's find the minor of the element $a_(32)$, i.e. let's find $M_(32)$. First, let's write down the minor $M_(32)$ and then calculate its value. In order to compose $M_(32)$, we delete the third row and second column from the matrix $A$ (it is at the intersection of the third row and the second column that the element $a_(32)$ is located). We will obtain a new matrix, the determinant of which is the required minor $M_(32)$:

This minor is easy to calculate using formula No. 2 from the calculation topic:

$$ M_(32)=\left| \begin(array) (ccc) 1 & -3 & 9\\ 2 & 11 & 5 \\ 3 & -5 & 58 \end(array) \right|= 1\cdot 11\cdot 58+(-3) \cdot 5\cdot 3+2\cdot (-5)\cdot 9-9\cdot 11\cdot 3-(-3)\cdot 2\cdot 58-5\cdot (-5)\cdot 1=579. $$

So, the minor of the element $a_(32)$ is 579, i.e. $M_(32)=579$.

Often, instead of the phrase “matrix element minor” in the literature, “determinant element minor” is found. The essence remains the same: to get the minor of the element $a_(ij)$ you need to cross out from the original determinant of the i-th line and jth column. The remaining elements are written into a new determinant, which is the minor of the element $a_(ij)$. For example, let's find the minor of the element $a_(12)$ of the determinant $\left| \begin(array) (ccc) -1 & 3 & 2\\ 9 & 0 & -5 \\ 4 & -3 & 7 \end(array) \right|$. To write down the required minor $M_(12)$ we need to delete the first row and second column from the given determinant:

To find the value of this minor, we use formula No. 1 from the topic of calculating determinants of the second and third orders:

$$ M_(12)=\left| \begin(array) (ccc) 9 & -5\\ 4 & 7 \end(array) \right|=9\cdot 7-(-5)\cdot 4=83. $$

So, the minor of the element $a_(12)$ is 83, i.e. $M_(12)=83$.

Algebraic complement $A_(ij)$ of element $a_(ij)$

Let a square matrix $A_(n\times n)$ be given (i.e., a square matrix of nth order).

Algebraic complement$A_(ij)$ element$a_(ij)$ of matrix $A_(n\times n)$ is found by the following formula: $$ A_(ij)=(-1)^(i+j)\cdot M_(ij), $$

where $M_(ij)$ is the minor of element $a_(ij)$.

Let us find the algebraic complement of element $a_(32)$ of the matrix $A=\left(\begin(array) (ccc) 1 & 0 & -3 & 9\\ 2 & -7 & 11 & 5 \\ -9 & 4 & 25 & 84\\ 3 & 12 & -5 & 58 \end(array) \right)$, i.e. let's find $A_(32)$. We previously found the minor $M_(32)=579$, so we use the result obtained:

Usually when you find algebraic additions They don’t calculate the minor separately, and only then the complement itself. The minor note is omitted. For example, let's find $A_(12)$ if $A=\left(\begin(array) (ccc) -5 & 10 & 2\\ 6 & 9 & -4 \\ 4 & -3 & 1 \end( array) \right)$. According to the formula $A_(12)=(-1)^(1+2)\cdot M_(12)=-M_(12)$. However, to get $M_(12)$ it is enough to cross out the first row and second column of the matrix $A$, so why introduce an extra notation for the minor? Let us immediately write down the expression for the algebraic complement $A_(12)$:

Minor of the kth order of the matrix $A_(m\times n)$

If in the previous two paragraphs we talked only about square matrices, then here we will also talk about rectangular matrices, in which the number of rows does not necessarily equal the number of columns. So, let the matrix $A_(m\times n)$ be given, i.e. a matrix containing m rows and n columns.

Minor kth order matrix $A_(m\times n)$ is a determinant whose elements are located at the intersection of k rows and k columns of matrix $A$ (it is assumed that $k≤ m$ and $k≤ n$).

For example, consider the matrix $A=\left(\begin(array) (ccc) -1 & 0 & -3 & 9\\ 2 & 7 & 14 & 6 \\ 15 & -27 & 18 & 31\\ 0 & 1 & 19 & 8\\ 0 & -12 & 20 & 14\\ 5 & 3 & -21 & 9\\ 23 & -10 & -5 & 58 \end(array) \right)$ and write down what -or third order minor. To write a third-order minor, we need to select any three rows and three columns of this matrix. For example, take rows numbered 2, 4, 6 and columns numbered 1, 2, 4. At the intersection of these rows and columns the elements of the required minor will be located. In the figure, the minor elements are shown in blue:

First order minors are found at the intersection of one row and one column, i.e. first order minors are equal to the elements of a given matrix.

The kth order minor of the matrix $A_(m\times n)=(a_(ij))$ is called main, if on the main diagonal of a given minor there are only the main diagonal elements of the matrix $A$.

Let me remind you that the main diagonal elements are those elements of the matrix whose indices are equal: $a_(11)$, $a_(22)$, $a_(33)$ and so on. For example, for the matrix $A$ considered above, such elements will be $a_(11)=-1$, $a_(22)=7$, $a_(33)=18$, $a_(44)=8$. They are highlighted in the figure pink:

For example, if in the matrix $A$ we cross out the rows and columns numbered 1 and 3, then at their intersection there will be elements of a minor of the second order, on the main diagonal of which there will be only diagonal elements of the matrix $A$ (elements $a_(11) =-1$ and $a_(33)=18$ of matrix $A$). Therefore we get major minor second order:

Naturally, we could take other rows and columns, for example, with numbers 2 and 4, thereby obtaining a different principal minor of the second order.

Let some minor $M$ of the kth order of the matrix $A_(m\times n)$ be not equal to zero, i.e. $M\neq 0$. In this case, all minors whose order is higher than k are equal to zero. Then the minor $M$ is called basic, and the rows and columns on which the elements of the basic minor are located are called base strings And base columns.

For example, consider the matrix $A=\left(\begin(array) (ccc) -1 & 0 & 3 & 0 & 0 \\ 2 & 0 & 4 & 1 & 0\\ 1 & 0 & -2 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 \end(array) \right)$. Let us write the minor of this matrix, the elements of which are located at the intersection of rows numbered 1, 2, 3 and columns numbered 1, 3, 4. We get a third-order minor:

Let's find the value of this minor using formula No. 2 from the topic of calculating determinants of the second and third orders:

$$ M=\left| \begin(array) (ccc) -1 & 3 & 0\\ 2 & 4 & 1 \\ 1 & -2 & -1 \end(array) \right|=4+3+6-2=11. $$

So, $M=11\neq 0$. Now let's try to compose any minor whose order is higher than three. To make a fourth-order minor, we have to use the fourth row, but all the elements of this row are zero. Therefore, any fourth-order minor will have a zero row, which means that all fourth-order minors are equal to zero. We cannot create minors of the fifth and higher orders, since the matrix $A$ has only 4 rows.

We have found a third order minor that is not equal to zero. In this case, all minors of higher orders are equal to zero, therefore, the minor we considered is basic. The rows of the matrix $A$ on which the elements of this minor are located (the first, second and third) are the basic rows, and the first, third and fourth columns of the matrix $A$ are the basic columns.

This example, of course, is trivial, since its purpose is to clearly show the essence of the basic minor. At all, basis minors there may be several, and usually the process of finding such a minor is much more complex and extensive.

Let's introduce another concept - bordering minor.

Let some kth order minor $M$ of the matrix $A_(m\times n)$ be located at the intersection of k rows and k columns. Let's add another row and column to the set of these rows and columns. The resulting minor of (k+1)th order is called edge minor for minor $M$.

For example, let's look at the matrix $A=\left(\begin(array) (ccc) -1 & 2 & 0 & -2 & -14\\ 3 & -17 & -3 & 19 & 29\\ 5 & -6 & 8 & -9 & 41\\ -5 & 11 & 19 & -20 & -98\\ 6 & 12 & 20 & 21 & 54\\ -7 & 10 & 14 & -36 & 79 \end(array) \right)$. Let's write a second-order minor, the elements of which are located at the intersection of rows No. 2 and No. 5, as well as columns No. 2 and No. 4.

Let's add another row No. 1 to the set of rows on which the elements of the minor $M$ lie, and column No. 5 to the set of columns. We obtain a new minor $M"$ (already of the third order), the elements of which are located at the intersection of rows No. 1, No. 2, No. 5 and columns No. 2, No. 4, No. 5. The elements of the minor $M$ in the figure are highlighted in pink, and The elements we add to the minor $M$ are green:

The minor $M"$ is the bordering minor for the minor $M$. Similarly, adding row No. 4 to the set of rows on which the elements of the minor $M$ lie, and column No. 3 to the set of columns, we obtain the minor $M""$ (third order minor):

The minor $M""$ is also a bordering minor for the minor $M$.

Minor of the kth order of the matrix $A_(n\times n)$. Additional minor. Algebraic complement to the minor of a square matrix.

Let's return to square matrices again. Let us introduce the concept of an additional minor.

Let a certain minor $M$ of the kth order of the matrix $A_(n\times n)$ be given. A determinant of (n-k)th order, the elements of which are obtained from the matrix $A$ after deleting the rows and columns containing the minor $M$, is called a minor, complementary to minor$M$.

For example, consider a fifth-order square matrix: $A=\left(\begin(array) (ccc) -1 & 2 & 0 & -2 & -14\\ 3 & -17 & -3 & 19 & 29\\ 5 & -6 & 8 & -9 & 41\\ -5 & 11 & 16 & -20 & -98\\ -7 & 10 & 14 & -36 & 79 \end(array) \right)$. Let's select rows No. 1 and No. 3, as well as columns No. 2 and No. 5. At the intersection of these rows and columns there will be elements of the minor $M$ of the second order:

Now let’s remove from the matrix $A$ rows No. 1 and No. 3 and columns No. 2 and No. 5, at the intersection of which there are elements of the minor $M$ (the removed rows and columns are shown in red in the figure below). The remaining elements form the minor $M"$:

The minor $M"$, whose order is $5-2=3$, is the complementary minor to the minor $M$.

Algebraic complement to a minor$M$ of a square matrix $A_(n\times n)$ is called the expression $(-1)^(\alpha)\cdot M"$, where $\alpha$ is the sum of the row and column numbers of the matrix $A$, on which the elements of the minor $M$ are located, and $M"$ is the minor complementary to the minor $M$.

The phrase "algebraic complement to the minor $M$" is often replaced by the phrase "algebraic complement to the minor $M$".

For example, consider the matrix $A$, for which we found the second-order minor $ M=\left| \begin(array) (ccc) 2 & -14 \\ -6 & 41 \end(array) \right| $ and its additional third-order minor: $M"=\left| \begin(array) (ccc) 3 & -3 & 19\\ -5 & 16 & -20 \\ -7 & 14 & -36 \end (array) \right|$. Let us denote the algebraic complement of the minor $M$ as $M^*$. Then, according to the definition:

$$ M^*=(-1)^\alpha\cdot M". $$

The $\alpha$ parameter is equal to the sum of the row and column numbers on which the minor $M$ is located. This minor is located at the intersection of rows No. 1, No. 3 and columns No. 2, No. 5. Therefore, $\alpha=1+3+2+5=11$. So:

$$ M^*=(-1)^(11)\cdot M"=-\left| \begin(array) (ccc) 3 & -3 & 19\\ -5 & 16 & -20 \\ -7 & 14 & -36 \end(array) \right|.

In principle, using formula No. 2 from the topic of calculating determinants of the second and third orders, you can complete the calculations, obtaining the value $M^*$:

$$ M^*=-\left| \begin(array) (ccc) 3 & -3 & 19\\ -5 & 16 & -20 \\ -7 & 14 & -36 \end(array) \right|=-30. $$