How to compute the algebraic complements to the elements of a matrix. Algebraic Complements and Minors

Minors of the matrix

Given a square matrix A, n - th order. Minor of some element аij, the determinant of the nth order matrix is called determinant(n - 1) - th order, obtained from the initial one by deleting the row and column at the intersection of which the selected element аij is located. It is designated Мij.

Let's consider an example determinant of matrix 3 - its order:
Minors and algebraic complements, determinant of matrix 3 - its order, then according to the definition minor, minor M12, corresponding to element a12, will be determinant:Moreover, using minors it is possible to facilitate the task of calculating determinant of matrix... Gotta decompose determinant of a matrix on some line and then determinant will be equal to the sum of all elements of this line by their minors. Decomposition determinant of matrix 3 - its order will look like this:

, the sign before the product is (-1) n, where n = i + j.

Algebraic additions:

Algebraic complement element аij 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. Aij is designated.
Аij = (-1) i + j × Мij.

Then we can reformulate the above property. Determinant of a matrix is equal to the sum of the product of elements of some row (row or column) matrices to their respective algebraic complements... Example.

Without matrix transformation, the determinant is easy to calculate only for 2 × 2 and 3 × 3 matrices. This is done using the formulas:

For matrix

the determinant is:

For matrix

the determinant is:

a11 * (a22 * a33-a23 * a32) -a12 * (a21 * a33-a23 * a31) + a13 * (a21 * a32-a22 * a31)

Calculations for matrices of 4 × 4 and larger are difficult, so they need to be transformed in accordance with the properties of the determinant. It is necessary to strive to obtain a matrix in which all values except one any column or any row are equal to zero. An example of such a matrix:

For it, the determinant is:

A12 * (a21 * (a33 * a44-a34 * a43) -a23 * (a31 * a44-a34 * a41) + a24 * (a31 * a43-a33 * a41))

note that

a21 * (a33 * a44-a34 * a43) -a23 * (a31 * a44-a34 * a41) + a24 * (a31 * a43-a33 * a41)

this is the calculation of the determinant of the matrix obtained by subtracting the row and column, at the intersection of which there is the only non-zero row / column number, according to which we decompose the matrix:

And we multiply the resulting value by the same number from the "zero" column / row, while the number can be multiplied by -1 (all the details below).

If the matrix is reduced to a triangular form, then its determinant is calculated as the product of the digits along the diagonal. For example, for the matrix

The determinant is:

The same should be done with matrices 5 × 5, 6 × 6 and other large dimensions.

Matrix transformations must be performed in accordance with the properties of the determinant. But before moving on to the practice of calculating the determinant for 4 × 4 matrices, let's go back to 3 × 3 matrices and take a closer look at how the determinant is calculated for them.

Minor

The determinant of a matrix is not very easy to understand, since there is recursion in its concept: the determinant of a matrix consists of several elements, including the determinant of (other) matrices.

In order not to get stuck on this, let's assume right now (temporarily) that the determinant of the matrix

is calculated like this:

Let's also understand the conventions and concepts such as minor and algebraic complement.

With the letter i we denote the ordinal number of the drain, with the letter j - the ordinal number of the column.

a ij means the element of the matrix (digit) at the intersection of row i and column j.

Imagine a matrix that is obtained from the original by deleting row i and column j. The determinant of the new matrix, which is obtained from the original by deleting row i and column j, is called the minor M ij of the element a ij.

Let us illustrate what has been said. Suppose given the matrix

Then, to determine the minor M 11 of the element a 11, we need to compose a new matrix, which is obtained from the original by removing the first row and the first column:

And calculate the determinant for it: 2 * 1 - (-4) * 0 = 2

To determine the minor M 22 of the element a 22, we need to compose a new matrix, which is obtained from the original by removing the second row and the second column:

And calculate the determinant for it: 1 * 1 -3 * 3 = -8

Algebraic complement

The algebraic complement A ij for an element a ij is the minor M ij of this element, taken with a "+" sign, if the sum of the row and column indices (i + j), at the intersection of which this element is located, is even and with a "-" sign, if the sum of the indices is odd.

Thus,

For the matrix from the previous example

A 11 = (-1) (1 + 1) * (2 * 1 - (-4) * 0) = 2

A 22 = (-1) (2 + 2) * (1 * 1 -3 * 3) = -8

Calculating the determinant for matrices

The determinant of order n corresponding to matrix A is a number denoted by det A and calculated by the formula:

Everything in this formula is already familiar to us, let's now calculate the determinant of the matrix for

Whatever the number of the row i = 1,2, ..., n or the column j = 1, 2, ..., n, the determinant of the nth order is equal to the sum of the products of the elements of this row or this column by their algebraic complements, i.e.

Those. the determinant can be calculated for any column or for any row.

To verify this, we calculate the determinant for the matrix from the last example for the second column

As you can see, the result is identical and for this matrix the determinant will always be -52, regardless of which row or which column we will count it.

Matrix determinant properties

The rows and columns of the determinant are equal, that is, the value of the determinant will not change if you swap its rows and columns while maintaining their order. This operation is called qualifier transposition. In accordance with the formulated property det A = det AT.
When swapping two rows (or two columns), the determinant retains its absolute value, but changes its sign to the opposite.
A determinant with two identical rows (or columns) is zero.
Multiplication of all elements of some row (or some column) of the determinant by the number λ is equivalent to multiplying the determinant by the number λ.
If all elements of any row (or any column) of the determinant are equal to zero, then the determinant itself is equal to zero.
If the elements of two rows (or two columns) of the determinant are proportional, then the determinant is zero.
If to the elements of some row (or some column) of the determinant we add the corresponding elements of another row (other column), multiplied by an arbitrary factor λ, then the value of the determinant will not change.
The sum of the products of elements of any row (any column) of the determinant by the corresponding algebraic complements of elements of any other row (any other column) is equal to zero.
If all elements of the i-th row of the determinant are presented as the sum of two terms a ij = bj + cj, then the determinant is equal to the sum of two determinants, in which all rows, except for the i-th, are the same as in the given determinant, i-th row in one of the terms it consists of elements bj, and in the other, it consists of elements cj. A similar property is true for the determinant columns.
The determinant of the product of two square matrices is equal to the product of their determinants: det (A * B) = det A * det B.

To calculate a determinant of any order, the method of successive decreasing the order of the determinant can be used. To do this, use the rule of decomposition of the determinant by the elements of a row or column. Another way of calculating determinants is to use elementary transformations with rows (or columns), primarily in accordance with properties 4 and 7 of determinants, to bring the determinant to the form when under the main diagonal of the determinant (defined in the same way as for square matrices) all elements are equal to zero. Then the determinant is equal to the product of the elements located on the main diagonal.

When calculating the determinant by successive decreasing the order to reduce the amount of computational work, it is advisable to use property 7 of determinants to zero out some of the elements of any row or any column of the determinant, which will reduce the number of computed algebraic complements.

Reduction of a matrix to a triangular form, transformation of a matrix that facilitates the calculation of the determinant

The methods shown below are impractical to use for 3 × 3 matrices, but I propose to consider the essence of the methods with a simple example. Let's use the matrix for which we have already calculated the determinant - it will be easier for us to check the correctness of the calculations:

Using the 7th property of the determinant, subtract from the second row the third, multiplied by 2:

from the third line, we subtract the corresponding elements of the first line of the determinant, multiplied by 3:

Since the elements of the determinant located under its main diagonal are equal to 0, then, therefore, determine is equal to the product of the elements located on the main diagonal:

1*2*(-26) = -52.

As you can see, the answer coincided with those received earlier.

Let's remember the formula for the determinant of a matrix:

The determinant is the sum of the algebraic complements multiplied by the members of one of the rows or one of the columns.

If, as a result of the transformations, we make one of the rows (or column) consist entirely of zeros except for one position, then we will not need to count all the algebraic complements, since they will certainly be equal to zero. Like the previous method, it is advisable to use this method for large matrices.

Let's show an example on the same matrix:

Note that the second column of the determinant already contains one zero element. Add to the elements of the second row the elements of the first row multiplied by -1. We get:

Let's calculate the determinant for the second column. We only need to calculate one algebraic complement, since the rest are certainly reduced to zero:

Calculation of the determinant for matrices 4 × 4, 5 × 5 and large dimensions

To avoid too large calculations for large matrices, you should do the transformations described above. Here are a couple of examples.

Calculate Define Matrix

Solution. Using the 7th property of the determinant, we subtract the third from the second row, and from the fourth row the corresponding elements of the first row of the determinant, multiplied by 3, 4, 5, respectively. These actions will be abbreviated as follows: (2) - (13; (3) - (1) * 4; (4) - (1) * 5. We get:

Let's perform actions

Minors of the matrix

Given a square matrix A, n - th order. Minor of some element a ij, determinant of matrix n - th order is called determinant(n - 1) - th order, obtained from the original by deleting the row and column, at the intersection of which the selected element a ij is located. It is denoted by М ij.

Let's consider an example determinant of matrix 3 - its order:

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

Moreover, using minors it is possible to facilitate the task of calculating determinant of matrix... Gotta decompose determinant of a matrix on some line and then determinant will be equal to the sum of all elements of this line by their minors. Decomposition determinant of 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. It is denoted by А ij. And ij = (-1) i + j × M ij.

Then we can reformulate the above property. Determinant of a matrix is equal to the sum of the product of elements of a certain row (row or column) matrices to their respective algebraic complements... Example:

4. The inverse matrix and its calculation.

Let A be a square matrix n - th order.

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

Matrix, allied to matrix Ah, called matrix

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

Matrix A -1 is called inverse matrix A, if the condition is met: A × A -1 = A -1 × A = E, where E is a unit matrix of the 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, that is, in order for a square matrix A had inverse matrix, it is necessary and sufficient that it determinant was nonzero.

To receive inverse matrix use the formula:

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

5. The rank of the matrix. Calculation of rank using elementary transformations.

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

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

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

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

1) permutations 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 some 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 not become; 2) when multiplying the string of the 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, performing elementary transformations on the matrix, one can obtain a matrix for which it is easy to calculate its rank and, therefore, the original matrix.

Definition. The matrix obtained from the matrix using elementary transformations is called equivalent and denoted A V.

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

With the help of elementary transformations, the matrix can be reduced to the so-called stepped form, when the calculation of its rank is not difficult.

Matrix is called stepped if it has the form:

Obviously, the rank of the stepped matrix is equal to the number of nonzero rows since there is a minor of the th order, not equal to zero:

Example. Determine the rank of the matrix using elementary transformations.

The rank of the matrix is equal to the number of nonzero rows, i.e. ...

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 "Matrices. Types of matrices. Basic terms". We also need some formulas to calculate the determinants. Since there are a lot of terms related to minors and algebraic additions in this topic, 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 the matrix $ A $ by deleting the i-th row and j-th column (i.e., the row and column at the intersection of which there is an element $ a_ (ij) $).

For example, consider a fourth-order square matrix: $ A = \ left (\ begin (array) (cccc) 1 & 0 & -3 & 9 \\ 2 & -7 & 11 & 5 \\ -9 & 4 & 25 & 84 \\ 3 & 12 & -5 & 58 \ end (array) \ right) $. Find the minor element $ a_ (32) $, i.e. find $ M_ (32) $. First, we write down the minor $ M_ (32) $, and then we calculate its value. In order to compose $ M_ (32) $, delete the third row and the 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 get a new matrix, the determinant of which is the required minor $ M_ (32) $:

This minor is easy to calculate using formula # 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 $ a_ (32) $ is 579, i.e. $ M_ (32) = 579 $.

Often, instead of the phrase "minor of the matrix element" in the literature, there is "minor of the element of the determinant". The essence remains unchanged: to get 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 the new determinant, which is the minor of the element $ a_ (ij) $. For example, let us find the minor element $ a_ (12) $ of the determinant $ \ left | \ begin (array) (ccc) -1 & 3 & 2 \\ 9 & 0 & -5 \\ 4 & -3 & 7 \ end (array) \ right | $. To write the required minor $ M_ (12) $, we need to cross out the first row and the second column from the given determinant:

To find the value of this minor, we use formula # 1 from the topic of calculating determinants of the second and third orders:

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

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

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

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

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

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

Usually, when finding algebraic complements, the minor is not calculated separately, and only then the complement itself. The minor entry is omitted. For example, 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 the second column of the matrix $ A $, so why introduce an unnecessary notation for the minor? Let's write down the expression for the algebraic complement $ A_ (12) $ right away:

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

For example, consider a matrix like this:

$$ A = \ left (\ begin (array) (cccc) -1 & 0 & -3 & 9 \\ 2 & 7 & 14 & 6 \\ 15 & -27 & 18 & 31 \\ 0 & 1 & 19 & 8 \\ 0 & -12 & 20 & 14 \\ 5 & 3 & -21 & 9 \\ 23 & -10 & -5 & 58 \ end (array) \ right) $$

Let's write down some third-order minor for it. To write a third-order minor, we need to select any three rows and three columns of this matrix. For example, let's take rows # 2, # 4, # 6 and columns # 1, # 2, # 4. The elements of the required minor will be located at the intersection of these rows and columns. In the figure, the minor elements are shown in blue:

$$ \ left (\ begin (array) (cccc) -1 & 0 & -3 & 9 \\ \ boldblue (2) & \ boldblue (7) & 14 & \ boldblue (6) \\ 15 & -27 & 18 & 31 \\ \ boldblue (0) & \ boldblue (1) & 19 & \ boldblue (8) \\ 0 & -12 & 20 & 14 \\ \ boldblue (5) & \ boldblue (3) & -21 & \ boldblue (9) \\ 23 & -10 & -5 & 58 \ end (array) \ right); \; M = \ left | \ begin (array) (ccc) 2 & 7 & 6 \\ 0 & 1 & 8 \\ 5 & 3 & 9 \ end (array) \ right |. $$

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

The kth order minor of the matrix $ A_ (m \ times n) = (a_ (ij)) $ is called the main if the main diagonal of this minor contains 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 green in the figure:

$$ \ left (\ begin (array) (cccc) \ boldgreen (-1) & 0 & -3 & 9 \\ 2 & \ boldgreen (7) & 14 & 6 \\ 15 & -27 & \ boldgreen (18 ) & 31 \\ 0 & 1 & 19 & \ boldgreen (8) \\ 0 & -12 & 20 & 14 \\ 5 & 3 & -21 & 9 \\ 23 & -10 & -5 & 58 \ end ( array) \ right) $$

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

$$ M = \ left | \ begin (array) (cc) \ boldgreen (-1) & -3 \\ 15 & \ boldgreen (18) \ end (array) \ right | $$

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

Suppose that some minor $ M $ of the kth order of the matrix $ A_ (m \ times n) $ is 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 lines and base columns.

For example, consider the following 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's write down the minor of this matrix, the elements of which are located at the intersection of rows No. 1, No. 2, No. 3 and columns No. 1, No. 3, No. 4. We will get a third-order minor (its elements are highlighted in the $ A $ matrix in purple):

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

Let's find the value of this minor, using formula # 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, the order of which is higher than three. To make a fourth order minor, we have to use the fourth line, but all elements of this line are equal to zero. Therefore, any fourth-order minor will have a zero line, which means that all fourth-order minors are equal to zero. We cannot compose minors of the fifth and higher orders, since the matrix $ A $ has only 4 rows.

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

This example is, of course, 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 finding such a minor is much more complicated and voluminous.

Let's introduce one more concept - the bordering minor.

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

For example, consider the following matrix:

$$ A = \ left (\ begin (array) (ccccc) -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 down the minor of the second order, the elements of which are located at the intersection of lines №2 and №5, as well as columns №2 and №4. These elements are highlighted in red in the matrix:

$$ \ left (\ begin (array) (ccccc) -1 & 2 & 0 & -2 & -14 \\ 3 & \ boldred (-17) & -3 & \ boldred (19) & 29 \\ 5 & -6 & 8 & -9 & 41 \\ -5 & 11 & 19 & -20 & -98 \\ 6 & \ boldred (12) & 20 & \ boldred (21) & 54 \\ -7 & 10 & 14 & -36 & 79 \ end (array) \ right); \; M = \ left | \ begin (array) (ccc) -17 & 19 \\ 12 & 21 \ end (array) \ right |. $$

Let's add to the set of rows on which the minor elements $ M $ lie, another row # 1, and to the set of columns - column # 5. We will get 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 red, and the elements we add to the $ M $ minor are blue:

$$ \ left (\ begin (array) (ccccc) -1 & \ boldblue (2) & 0 & \ boldblue (-2) & \ boldblue (-14) \\ 3 & \ boldred (-17) & -3 & \ boldred (19) & \ boldblue (29) \\ 5 & -6 & 8 & -9 & 41 \\ -5 & 11 & 19 & -20 & -98 \\ 6 & \ boldred (12) & 20 & \ boldred (21) & \ boldblue (54) \\ -7 & 10 & 14 & -36 & 79 \ end (array) \ right); \; M "= \ left | \ begin (array) (ccc) 2 & -2 & -14 \\ -17 & 19 & 29 \\ 12 & 21 & 54 \ end (array) \ right |. $$

Minor $ M "$ is a bordering minor for minor $ M $. Similarly, adding to the set of rows on which the elements of the minor $ M $ lie, row # 4, and to the set of columns - column # 3, we get minor $ M" "$ (third-order minor):

$$ \ left (\ begin (array) (ccccc) -1 & 2 & 0 & -2 & -14 \\ 3 & \ boldred (-17) & \ boldblue (-3) & \ boldred (19) & 29 \\ 5 & -6 & 8 & -9 & 41 \\ -5 & \ boldblue (11) & \ boldblue (19) & \ boldblue (-20) & -98 \\ 6 & \ boldred (12) & \ boldblue (20) & \ boldred (21) & 54 \\ -7 & 10 & 14 & -36 & 79 \ end (array) \ right); \; M "" = \ left | \ begin (array) (ccc) -17 & -3 & 19 \\ 11 & 19 & -20 \\ 12 & 20 & 21 \ end (array) \ right |. $$

Minor $ M "" $ is also a bordering minor for minor $ M $.

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

Let's go back to square matrices again. Let's introduce the concept of an additional minor.

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

For example, consider a square matrix of the fifth order:

$$ A = \ left (\ begin (array) (ccccc) -1 & 2 & 0 & -2 & -14 \\ 3 & -17 & -3 & 19 & 29 \\ 5 & -6 & 8 & - 9 & 41 \\ -5 & 11 & 16 & -20 & -98 \\ -7 & 10 & 14 & -36 & 79 \ end (array) \ right) $$

Let's select rows No. 1 and No. 3 in it, as well as columns No. 2 and No. 5. At the intersection of these rows and columns there will be second-order minor elements $ M $. These elements are highlighted in green in the $ A $ matrix:

$$ \ left (\ begin (array) (ccccc) -1 & \ boldgreen (2) & 0 & -2 & \ boldgreen (-14) \\ 3 & -17 & -3 & 19 & 29 \\ 5 & \ boldgreen (-6) & 8 & -9 & \ boldgreen (41) \\ -5 & 11 & 16 & -20 & -98 \\ -7 & 10 & 14 & -36 & 79 \ end (array) \ right); \; M = \ left | \ begin (array) (cc) 2 & -14 \\ -6 & 41 \ end (array) \ right |. $$

Now let's remove from the matrix $ A $ rows # 1 and # 3 and columns # 2 and # 5, at the intersection of which there are elements of the minor $ M $ (the elements of the rows and columns to be removed are shown in red in the figure below). The remaining elements form a minor $ M "$:

$$ \ left (\ begin (array) (ccccc) \ boldred (-1) & \ boldred (2) & \ boldred (0) & \ boldred (-2) & \ boldred (-14) \\ 3 & \ boldred (-17) & -3 & 19 & \ boldred (29) \\ \ boldred (5) & \ boldred (-6) & \ boldred (8) & \ boldred (-9) & \ boldred (41) \ \ -5 & \ boldred (11) & 16 & -20 & \ boldred (-98) \\ -7 & \ boldred (10) & 14 & -36 & \ boldred (79) \ end (array) \ right) ; \; M "= \ left | \ begin (array) (ccc) 3 & -3 & 19 \\ -5 & 16 & -20 \\ -7 & 14 & -36 \ end (array) \ right |. $$

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

Algebraic complement to a minor$ M $ of a square matrix $ A_ (n \ times n) $ is the expression $ (- 1) ^ (\ alpha) \ cdot M "$, where $ \ alpha $ is the sum of the numbers of rows and columns 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 an additional minor of the third order: $ M "= \ left | \ begin (array) (ccc) 3 & -3 & 19 \\ -5 & 16 & -20 \\ -7 & 14 & -36 \ end (array) \ right | $. Denote the algebraic complement of the minor $ M $ as $ M ^ * $. Then, by definition:

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

The $ \ alpha $ parameter is the sum of the row and column numbers on which the $ M $ minor is located. This minor is located at the intersection of rows # 1, # 3 and columns # 2, # 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 # 2 from the topic of calculating determinants of the second and third orders, you can bring the calculations to the end, getting the value $ M ^ * $:

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