Matrix determinant minor. Algebraic complements and minors. Types of minors and algebraic complements

Let the matrix be highlighted
any k rows and k columns, k and k. The elements located at the intersection of these rows and columns form a square matrix A¢ of order k ( submatrix matrix A).
Its determinant is called the kth order minor of a given matrix A. Obviously, in the general case there can be several such minors of the matrix A. In this case, the maximum order of the minors is equal to the minimum of the numbers m and n, i.e. . From all possible minors of matrix A, we select those that are different from zero. In turn, among these minors one can find at least one minor of the highest order.

Definition. The highest order of a non-zero minor is called the rank of the matrix.

Definition. A nonzero minor of a matrix whose order is equal to the rank of the matrix is called the basis minor of this matrix.

Rows and columns at the intersection of which there is a basis minor are called basic.

In general, a matrix can have several basis minors.

The following main theorem, which we present without proof, plays an important role.

Theorem 3.6.(about the basic minor). The basis rows (basis columns) of the matrix are linearly independent. Any row (any column) of matrix A is a linear combination of basic rows (basic columns).

Thus, if the rank of matrix A is r, then this matrix must have a minor r th order, different from zero, and all minors whose order is greater r, are equal to zero.

Previously, the rank of a matrix was defined as the largest number of linearly independent vector rows (columns). In an algebra course it is proven that these two definitions are equivalent. This makes it possible to calculate the rank of the matrix, and therefore the rank of the system of vectors.

Example. Find all basis minors of a matrix

A= .

○ Any minor of the third-order matrix A is equal to zero, since it contains a zero row. We will find second-order minors other than zero.

, , , , .

Among the minors of the second order there are non-zero ones, which means that the rank of matrix A is 2 and the basis minors are . ●

Theorem 3.7. In order for the nth order determinant to be equal to zero, it is necessary and sufficient that its rows (columns) be linearly dependent.

□ 1) Let the determinant of a square matrix A of order n equal to zero. Then the maximum order of non-zero minors must be less than n; therefore, the rank of matrix A is less n. This means that the system of all rows of the matrix is linearly dependent.

2) If the lines A 1, A 2,…, A m of the determinant are linearly dependent,
then by the property of 6° linear dependence one line A i is a linear combination of the remaining rows of the determinant, i.e.

Adding to line A i This linear combination, multiplied by (–1), will result in one line consisting entirely of zeros, and based on the 7° property of the determinant, the value of the determinant will not change. But then, by the 2° property, the determinant is equal to zero. ■

Example. Prove that vectors a 1 =(2;–1;3), a 2 =(–1;1;0), a 3 =(1;1;6) are coplanar.

○ Three non-zero three-dimensional vectors are coplanar if they are linearly dependent. Let's compose a determinant from the coordinates of these vectors

Since the determinant is equal to zero, its rows are linearly dependent, which means the vectors are linearly dependent a 1 =(2;–1;3), a 2 =(–1;1;0), a 3 =(1;1;6), therefore, they are coplanar. ●

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

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

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

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

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

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

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

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

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

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

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

For example, given the matrix

Let's take a minor

The bordering minors will be:

Algorithm for finding the rank of a matrix next.

1. Find minors of the second order that are not equal to zero. If all second-order minors are equal to zero, then the rank of the matrix will be equal to one ( r =1 ).

2. If there is at least one minor of the second order that is not equal to zero, then we compose the bordering minors of the third order. If all bordering minors of the third order are equal to zero, then the rank of the matrix is equal to two ( r =2 ).

3. If at least one of the bordering minors of the third order is not equal to zero, then we compose the bordering minors. If all the bordering minors of the fourth order are equal to zero, then the rank of the matrix is equal to three ( r =2 ).

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

Example 1. Find the rank of a matrix

Solution. Minor of the second order .

Let's border it. There will be four bordering minors:

Thus, all bordering minors of the third order are equal to zero, therefore, the rank of this matrix is equal to two ( r =2 ).

Example 2. Find the rank of a matrix

Solution. The rank of this matrix is equal to 1, since all the second-order minors of this matrix are equal to zero (in this, as in the cases of bordering minors in the two following examples, dear students are invited to verify for themselves, perhaps using the rules for calculating determinants), and among the first-order minors , that is, among the elements of the matrix, there are non-zero ones.

Example 3. Find the rank of a matrix

Solution. The second order minor of this matrix is , and all third order minors of this matrix are equal to zero. Therefore, the rank of this matrix is two.

Example 4. Find the rank of a matrix

Solution. The rank of this matrix is 3, since the only third-order minor of this matrix is 3.

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

Already in example 1 it is clear that the task of determining the rank of a matrix using the method of bordering minors requires the calculation of a large number of determinants. There is, however, a way to reduce the amount of computation to a minimum. This method is based on the use of elementary matrix transformations and is also called the Gauss method.

The following operations are understood as elementary matrix transformations:

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

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

3) swapping two rows or columns of the matrix;

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

5) deleting all proportional lines except one.

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

Matrix determinants are often used in calculus, linear algebra, and analytic geometry. Outside the academic world, matrix determinants are constantly needed by engineers and programmers, especially those who work with computer graphics. If you already know how to find the determinant of a 2x2 matrix, then the only tools you need to find the determinant of a 3x3 matrix are addition, subtraction, and multiplication.

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. For 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. Draw a horizontal 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 sign diagram 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 a brief description of 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 is the 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 row with the most 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. In this case we are talking about triangular matrices with dimensions 3x3. Triangular matrices can be of the following types, depending on the 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 be equal to the sum of all elements of this line by 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) is not 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 of 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:

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 this topic contains a lot of terms related to minors and algebraic complements, I will add a brief 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 the i-th row and j-th column (i.e., the row and column at the intersection of which the element is located $a_(ij)$).

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 obtain the minor of the element $a_(ij)$, you need to cross out the i-th row and j-th column from the original determinant. 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 finding algebraic complements, the minor is not calculated 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 pink in the figure:

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 a second-order principal minor:

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. In general, there can be several basic minors, and usually the process of searching for 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 $M^*$:

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