Calculate the algebraic complement of a matrix online. Minors and algebraic complements
determinant by elements of a row or column
Further properties are related to the concepts of minor and algebraic complement
Definition. Minor element is called a determinant made up of elements remaining after crossing outi-th drains andjth column at the intersection of which this element is located. Minor of the element of the determinant n-th order has order ( n- 1). We will denote it by .
Example 1. Let 
, Then 
.
This minor is obtained from A by crossing out the second row and third column.
Definition.
Algebraic complement
element is called the corresponding minor, multiplied by nat.e 
, Wherei–line number andj-columns at the intersection of which this element is located.
VІІІ. (Decomposition of the determinant into elements of a certain string). The determinant is equal to the sum of the products of the elements of a certain row and their corresponding algebraic complements.
.
Example 2. Let it be then
.
Example 3. Let's find the determinant of the matrix by expanding it into the elements of the first row.
Formally, this theorem and other properties of determinants are applicable only for determinants of matrices of no higher than third order, since we have not considered other determinants. The following definition will allow us to extend these properties to determinants of any order.
Definition. Determinant matrices A nth order is a number calculated by sequential application of the expansion theorem and other properties of determinants.
You can check that the result of the calculations does not depend on the order in which the above properties are applied and for which rows and columns. Using this definition, the determinant is uniquely found.
Although this definition does not contain an explicit formula for finding the determinant, it allows one to find it by reducing it to the determinants of matrices of lower order. Such definitions are called recurrent.
Example 4. Calculate the determinant: .
Although the factorization theorem can be applied to any row or column of a given matrix, fewer computations are obtained by factoring along the column that contains as many zeros as possible.
Since the matrix does not have zero elements, we obtain them using property 7). Multiply the first line sequentially by the numbers (–5), (–3) and (–2) and add it to the 2nd, 3rd and 4th lines and get:
Let's expand the resulting determinant along the first column and get:
(we take (–4) from the 1st line, (–2) from the 2nd line, (–1) from the 3rd line according to property 4) 
(since the determinant contains two proportional columns).
§ 1.3. Some types of matrices and their determinants
Definition. Square m an matrix with zero elements below or above the main diagonal(=0 at i j, or =0 at i j) calledtriangular .
Let's continue the conversation about actions with matrices. Namely, during the study of this lecture you will learn how to find the inverse matrix. Learn. Even if math is difficult.
What is an inverse matrix? Here we can draw an analogy with inverse numbers: consider, for example, the optimistic number 5 and its inverse number. The product of these numbers is equal to one: . Everything is similar with matrices! The product of a matrix and its inverse matrix is equal to – identity matrix, which is the matrix analogue of the numerical unit. However, first things first – let’s first solve an important practical issue, namely, learn how to find this very inverse matrix.
What do you need to know and be able to do to find the inverse matrix? You must be able to decide qualifiers. You must understand what it is matrix and be able to perform some actions with them.
There are two main methods for finding the inverse matrix:
by using algebraic additions And using elementary transformations.
Today we will study the first, simpler method.
Let's start with the most terrible and incomprehensible. Let's consider square matrix. The inverse matrix can be found using the following formula:
Where is the determinant of the matrix, is the transposed matrix of algebraic complements of the corresponding elements of the matrix.
The concept of an inverse matrix exists only for square matrices, matrices “two by two”, “three by three”, etc.
Designations: As you may have already noticed, the inverse matrix is denoted by a superscript
Let's start with the simplest case - a two-by-two matrix. Most often, of course, “three by three” is required, but, nevertheless, I strongly recommend studying a simpler task in order to understand the general principle of the solution.
Example:
Find the inverse of a matrix
Let's decide. It is convenient to break down the sequence of actions point by point.
1) First we find the determinant of the matrix.
If your understanding of this action is not good, read the material How to calculate the determinant?
Important! If the determinant of the matrix is equal to ZERO– inverse matrix DOES NOT EXIST.
In the example under consideration, as it turned out, , which means everything is in order.
2) Find the matrix of minors.
To solve our problem, it is not necessary to know what a minor is, however, it is advisable to read the article How to calculate the determinant.
The matrix of minors has the same dimensions as the matrix, that is, in this case.
The only thing left to do is find four numbers and put them in place of the stars.
Let's return to our matrix
Let's look at the top left element first:
How to find it minor?
And this is done like this: MENTALLY cross out the row and column in which this element is located:
The remaining number is minor of this element, which we write in our matrix of minors:
Consider the following matrix element:
Mentally cross out the row and column in which this element appears:
What remains is the minor of this element, which we write in our matrix:
Similarly, we consider the elements of the second row and find their minors:
Ready.
It's simple. In the matrix of minors you need CHANGE SIGNS two numbers:
These are the numbers that I circled!
– matrix of algebraic additions of the corresponding elements of the matrix.
And just...
4) Find the transposed matrix of algebraic additions.
– transposed matrix of algebraic complements of the corresponding elements of the matrix.
5) Answer.
Let's remember our formula
Everything has been found!
So the inverse matrix is: 
It is better to leave the answer as is. NO NEED divide each element of the matrix by 2, since the result is fractional numbers. This nuance is discussed in more detail in the same article. Actions with matrices.
How to check the solution?
You need to perform matrix multiplication or
Examination: 
Received already mentioned identity matrix is a matrix with ones by main diagonal and zeros in other places.
Thus, the inverse matrix is found correctly.
If you carry out the action, the result will also be an identity matrix. This is one of the few cases where matrix multiplication is commutative, more details can be found in the article Properties of operations on matrices. Matrix Expressions. Also note that during the check, the constant (fraction) is brought forward and processed at the very end - after the matrix multiplication. This is a standard technique.
Let's move on to a more common case in practice - the three-by-three matrix:
Example:
Find the inverse of a matrix 
The algorithm is exactly the same as for the “two by two” case.
We find the inverse matrix using the formula: , where is the transposed matrix of algebraic complements of the corresponding elements of the matrix.
1) Find the determinant of the matrix.
Here the determinant is revealed on the first line.
Also, don’t forget that, which means everything is fine - inverse matrix exists.
2) Find the matrix of minors.
The matrix of minors has a dimension of “three by three” 
, and we need to find nine numbers.
I'll look at a couple of minors in detail:
Consider the following matrix element: 
MENTALLY cross out the row and column in which this element is located: 
We write the remaining four numbers in the “two by two” determinant. 
This two-by-two determinant and is the minor of this element. It needs to be calculated: 
That’s it, the minor has been found, we write it in our matrix of minors: 
As you probably guessed, you need to calculate nine two-by-two determinants. The process, of course, is tedious, but the case is not the most severe, it can be worse.
Well, to consolidate – finding another minor in the pictures: 
Try to calculate the remaining minors yourself.
Final result: 
– matrix of minors of the corresponding elements of the matrix.
The fact that all the minors turned out to be negative is purely an accident.
3) Find the matrix of algebraic additions.
In the matrix of minors it is necessary CHANGE SIGNS strictly for the following elements: 
In this case:
We do not consider finding the inverse matrix for a “four by four” matrix, since only a sadistic teacher can give such a task (for the student to calculate one “four by four” determinant and 16 “three by three” determinants). In my practice, there was only one such case, and the customer of the test paid quite dearly for my torment =).
In a number of textbooks and manuals you can find a slightly different approach to finding the inverse matrix, but I recommend using the solution algorithm outlined above. Why? Because the likelihood of getting confused in calculations and signs is much less.
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 mхn. 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:
.
Definition. If in the nth order determinant we choose arbitrarily k rows and k columns, then the elements at the intersection of these rows and columns form a square matrix of order k. The determinant of such a square matrix is called minor of kth order .
Denoted by Mk. If k=1, then the first-order minor is an element of the determinant.
The elements at the intersection of the remaining (n-k) rows and (n-k) columns form a square matrix of order (n-k). The determinant of such a matrix is called a minor, additional to minor M k . Denoted by Mn-k.
Algebraic complement of the minor M k we will call it an additional minor, taken with a “+” or “-” sign, depending on whether the sum of the numbers of all rows and columns in which the minor M k is located is even or odd.
If k=1, then the algebraic complement to the element a ik calculated by the formula
A ik =(-1) i+k M ik, where M ik- minor (n-1) order.
Theorem. The product of a kth order minor and its algebraic complement is equal to the sum of a certain number of terms of the determinant D n.
Proof
1. Let's consider a special case. Let the minor M k occupy the upper left corner of the determinant, that is, located in lines numbered 1, 2, ..., k, then the minor M n-k will occupy lines k+1, k+2, ..., n.
Let us calculate the algebraic complement to the minor M k . A-priory,
A n-k =(-1) s M n-k, where s=(1+2+...+k) +(1+2+...+k)= 2(1+2+...+k), then
(-1)s=1 and A n-k = M n-k. We get
M k A n-k = M k M n-k. (*)
We take an arbitrary term of the minor M k
, (1)
where s is the number of inversions in the substitution
and an arbitrary minor term M n-k
where s * is the number of inversions in the substitution
(4)
Multiplying (1) and (3), we get
The product consists of n elements located in different rows and columns of the determinant D. Consequently, this product is a member of the determinant D. The sign of the product (5) is determined by the sum of inversions in substitutions (2) and (4), and the sign of a similar product in the determinant D is determined number of inversions s k in the substitution
It is obvious that s k =s+s * .
Thus, returning to equality (*), we obtain that the product M k A n-k consists only of the terms of the determinant.
2. Let minor M k located in rows with numbers i 1 , i 2 , ..., i k and in columns with numbers j 1, j 2, ..., j k, and i 1< i 2 < ...< i k And j 1< j 2 < ...< j k .
Using the properties of determinants, using transpositions we will move the minor to the upper left corner. We obtain the determinant D ¢, in which the minor M k occupies the upper left corner, and the additional minor M¢ n-k is the lower right corner, then, according to what was proven in point 1, we obtain that the product M k M¢ n-k is the sum of a certain number of elements of the determinant D ¢, taken with their own sign. But D¢ is obtained from D using ( i 1 -1)+(i 2 -2)+ ...+(i k -k)=(i 1 + i 2 + ...+ i k)-(1+2+...+k) string transpositions and ( j 1 -1)+(j 2 -2)+ ...+(j k -k)=(j 1 + j 2 + ...+ j k)- (1+2+...+k) column transpositions. That is, everything was done
(i 1 + i 2 + ...+ i k)-(1+2+...+k)+ (j 1 + j 2 + ...+ j k)- (1+2+...+k )= (i 1 + i 2 + ...+ i k)+ (j 1 + j 2 + ...+ j k)- 2(1+2+...+k)=s-2(1+2 +...+k). Therefore, the terms of the determinants D and D ¢ differ in sign (-1) s-2(1+2+...+k) =(-1) s, therefore, the product (-1) s M k M¢ n-k will consist of a certain number of terms of the determinant D, taken with the same signs as they have in this determinant.
Laplace's theorem. If in the nth order determinant we choose arbitrarily k rows (or k columns) 1£k£n-1, then the sum of the products of all kth order minors contained in the selected rows and their algebraic complements is equal to the determinant D.
Proof
Let's choose random lines i 1 , i 2 , ..., i k and we will prove that
It was previously proven that all elements on the left side of the equality are contained as terms in the determinant D. Let us show that each term in the determinant D falls into only one of the terms. Indeed, everything ts looks like t s =. if in this product we note the factors whose first indices i 1 , i 2 , ..., i k, and compose their product, then you can notice that the resulting product belongs to the kth order minor. Consequently, the remaining terms, taken from the remaining n-k rows and n-k columns, form an element belonging to the complementary minor, and, taking into account the sign, to the algebraic complement, therefore, any ts falls into only one of the products, which proves the theorem.
Consequence(theorem on the expansion of the determinant in a row) . The sum of the products of the elements of a certain row of the determinant and the corresponding algebraic complements is equal to the determinant.
(Proof as an exercise.)
Theorem. The sum of the products of the elements of the i-th row of the determinant by the corresponding algebraic complements to the elements of the j-th row (i¹j) is equal to 0.
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’s 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. The 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. $$
