Calculate the inverse of the matrix. Algorithm for calculating the inverse matrix using algebraic complements: the adjoint (adjoint) matrix method. Matrix method in economic analysis
inverse matrix
Inverse matrix is called a matrix that, when multiplied both on the right and on the left by this matrix, gives the identity matrix.
Let us denote the matrix inverse to the matrix AND through, then according to the definition we get:
where E Is the identity matrix.
Square matrix called non-special (non-degenerate) if its determinant is not zero. Otherwise, it is called special (degenerate) or singular.
The following theorem holds: every nonsingular matrix has an inverse.
The operation of finding the inverse matrix is \u200b\u200bcalled appeal matrices. Consider the matrix inversion algorithm. Let there be given a nonsingular matrix n-th order:
where Δ \u003d det A ≠ 0.
Algebraic Complement of an Elementmatrices n -th order AND the determinant of the matrix ( n –1) th order obtained by deleting i-th line and jth column of the matrix AND:
Let's compose the so-called attached matrix:
where are the algebraic complements of the corresponding elements of the matrix AND.
Note that the algebraic complements of the elements of the rows of the matrix AND are placed in the corresponding columns of the matrix Ã
, that is, the matrix is \u200b\u200btransposed at the same time.
Dividing all the elements of the matrix Ã
by Δ - the value of the determinant of the matrix AND, we get the inverse matrix as a result:
We note a number of special properties of the inverse matrix:
1) for a given matrix AND its inverse matrix
is the only one;
2) if there is an inverse matrix, then right reverse and left reverse matrices coincide with it;
3) a special (degenerate) square matrix has no inverse matrix.
Main properties of the inverse matrix:
1) the determinant of the inverse matrix and the determinant of the original matrix are reciprocal values;
2) the inverse matrix of the product of square matrices is equal to the product of inverse matrices of factors, taken in reverse order:
3) the transposed inverse of the matrix is \u200b\u200bequal to the inverse of the given transposed matrix:
PRI me r. Calculate the inverse of the given matrix.
For any nondegenerate matrix A, there exists and, moreover, a unique matrix A -1 such that
A * A -1 \u003d A -1 * A \u003d E,
where E is the identity matrix of the same orders as A. The matrix A -1 is called the inverse of the matrix A.
In case someone forgot, in the identity matrix, except for the diagonal filled with ones, all other positions are filled with zeros, an example of the identity matrix:
Finding the inverse matrix by the adjoint matrix method
The inverse matrix is \u200b\u200bdefined by the formula:
where A ij are elements a ij.
Those. to calculate the inverse matrix, you need to calculate the determinant of this matrix. Then find the algebraic complements for all its elements and compose a new matrix from them. Next, you need to transport this matrix. And divide each element of the new matrix by the determinant of the original matrix.
Let's look at some examples.
Find A -1 for Matrix
Solution. Let us find A -1 by the adjoint matrix method. We have det A \u003d 2. Let us find the algebraic complements of the elements of the matrix A. In this case, the algebraic complements of the matrix elements are the corresponding elements of the matrix itself, taken with a sign in accordance with the formula
We have A 11 \u003d 3, A 12 \u003d -4, A 21 \u003d -1, A 22 \u003d 2. We form the adjoint matrix
We transport the matrix A *:
We find the inverse matrix by the formula:
We get:
Find A -1 using the adjoint matrix method if
Solution. First of all, we calculate the definition of the given matrix to make sure that the inverse matrix exists. We have
Here we have added to the elements of the second row the elements of the third row, multiplied previously by (-1), and then expanded the determinant on the second row. Since the given matrix is \u200b\u200bdetermined to be nonzero, the inverse matrix exists. To construct the adjoint matrix, we find the algebraic complements of the elements of this matrix. We have
According to the formula
transport the matrix A *:
Then by the formula
Finding the inverse matrix by the method of elementary transformations
In addition to the method for finding the inverse matrix that follows from the formula (the method of the adjoint matrix), there is a method for finding the inverse matrix, called the method of elementary transformations.
Elementary matrix transformations
The following transformations are called elementary matrix transformations:
1) permutation of rows (columns);
2) multiplying a row (column) by a nonzero number;
3) adding to the elements of a row (column) the corresponding elements of another row (column), previously multiplied by a certain number.
To find the matrix A -1, we construct a rectangular matrix B \u003d (A | E) of orders (n; 2n), assigning to the matrix A on the right the identity matrix E through the dividing line:
Let's look at an example.
Using the method of elementary transformations, find A -1 if
Solution. Let us form the matrix B:
Let us denote the rows of matrix B by α 1, α 2, α 3. Let us perform the following transformations on the rows of the matrix B.
Definition 1: a matrix is \u200b\u200bcalled degenerate if its determinant is zero.
Definition 2: a matrix is \u200b\u200bcalled nondegenerate if its determinant is not zero.
The matrix "A" is called inverse matrixif the condition A * A-1 \u003d A-1 * A \u003d E (identity matrix) is satisfied.
A square matrix is \u200b\u200binvertible only if it is non-degenerate.
Inverse matrix calculation scheme:
1) Calculate the determinant of the matrix "A" if ∆ A \u003d 0, then the inverse matrix does not exist.
2) Find all the algebraic complements of the matrix "A".
3) Create a matrix of algebraic complements (Aij)
4) Transpose the matrix of algebraic complements (Aij) T
5) Multiply the transposed matrix by the inverse of the determinant of this matrix.
6) Check:
At first glance, it may seem that it is difficult, but in fact, everything is very simple. All solutions are based on simple arithmetic operations, the main thing when deciding is not to get confused with the signs "-" and "+", and not to lose them.
Now let's solve a practical task together with you by calculating the inverse matrix.
Task: find the inverse matrix "A" shown in the picture below:
1. The first thing to do is find the determinant of the matrix "A":
Explanation:
We have simplified our qualifier by taking advantage of its core functionality. First, we added to rows 2 and 3 the elements of the first row, multiplied by one number.
Secondly, we changed columns 2 and 3 of the determinant, and according to its properties, we changed the sign in front of it.
Thirdly, we moved out the common factor (-1) of the second line, thereby changing the sign again and it became positive. We've also simplified line 3 just like at the very beginning of the example.
We have obtained a triangular determinant, in which the elements below the diagonal are equal to zero, and by property 7, it is equal to the product of the elements of the diagonal. As a result, we got ∆ A \u003d 26, hence the inverse exists.
A11 \u003d 1 * (3 + 1) \u003d 4
A12 \u003d -1 * (9 + 2) \u003d -11
A13 \u003d 1 * 1 \u003d 1
A21 \u003d -1 * (- 6) \u003d 6
A22 \u003d 1 * (3-0) \u003d 3
A23 \u003d -1 * (1 + 4) \u003d -5
A31 \u003d 1 * 2 \u003d 2
A32 \u003d -1 * (- 1) \u003d -1
A33 \u003d 1+ (1 + 6) \u003d 7
3. The next step is to compile a matrix from the resulting additions:
5. Multiply this matrix by the inverse of the determinant, that is, by 1/26:
6. Well, now we just need to check:
During the check, we received the identity matrix, therefore, the solution was carried out absolutely correctly.
2 way to calculate the inverse matrix.
1. Elementary matrix transformation
2. Inverse matrix through an elementary transformer.
Elementary matrix transformation includes:
1. Multiplication of a string by a nonzero number.
2. Adding to any line another string multiplied by a number.
3. Swapping the rows of the matrix.
4. Applying a chain of elementary transformations, we obtain another matrix.
AND -1 = ?
1. (A | E) ~ (E | A -1 )
2.A -1 * A \u003d E
Let's look at a practical example with real numbers.
The task: Find the inverse of the matrix.
Decision:
Let's check:
A little clarification on the solution:
First, we rearranged rows 1 and 2 of the matrix, then multiplied the first row by (-1).
After that, the first row was multiplied by (-2) and added to the second row of the matrix. Then we multiplied the 2nd row by 1/4.
The final stage of the transformation was multiplying the second line by 2 and adding from the first. As a result, on the left, we got the identity matrix, therefore, the inverse is the matrix on the right.
After checking, we made sure that the solution was correct.
As you can see, calculating the inverse matrix is \u200b\u200bvery easy.
In conclusion of this lecture, I would also like to devote a little time to the properties of such a matrix.
ALGEBRAIC SUPPLEMENTS AND MINORS
Let us have a determinant of the third order: 
.
Minorcorresponding to this element a ij determinant of the third order, is called the determinant of the second order, obtained from the given by deleting the row and column at the intersection of which the given element stands, i.e. i-th line and jth column. Minors corresponding to a given element a ij will denote M ij.
for example, minor M 12corresponding to element a 12, there will be a determinant 
, which is obtained by deleting the 1st row and 2nd column from the given determinant.
Thus, the formula defining the determinant of the third order shows that this determinant is equal to the sum of the products of the elements of the 1st row by the corresponding minors; the minor corresponding to the element a 12, taken with a “-” sign, i.e. we can write that
| . | (1) |
Similarly, we can introduce definitions of minors for determinants of the second order and higher orders.
Let's introduce one more concept.
Algebraic complementelement a ij determinant is called its minor M ijmultiplied by (–1) i + j.
Algebraic Complement of an Element a ij denoted A ij.
From the definition, we find that the connection between the algebraic complement of an element and its minor is expressed by the equality A ij \u003d (–1) i + j M ij.
For example,
Example. A determinant is given. To find A 13, A 21, A 32.
It is easy to see that using the algebraic complements of elements, formula (1) can be written in the form:
Similarly to this formula, you can get the decomposition of the determinant into the elements of any row or column.
For example, the factorization of the determinant by the elements of the 2nd line can be obtained as follows. According to property 2 of the determinant, we have:
Let us expand the resulting determinant by the elements of the 1st row.
 .
| (2) |
From here 
since the determinants of the second order in formula (2) are the minors of the elements a 21, a 22, a 23... Thus, i.e. we got the decomposition of the determinant in terms of the elements of the 2nd row.
Similarly, you can get the factorization of the determinant by the elements of the third row. Using property 1 of determinants (about transposition), one can show that similar expansions are also valid for the expansion in terms of column elements.
Thus, the following theorem is true.
Theorem (on the expansion of a determinant in a given row or column). The determinant is equal to the sum of the products of elements of any of its rows (or columns) by their algebraic complements.
All of the above is also true for determinants of any higher order.
Examples.
INVERSE MATRIX
The concept of an inverse matrix is \u200b\u200bintroduced only for square matrices.
If a A Is a square matrix, then reverse for it, a matrix is \u200b\u200ba matrix denoted A -1 and satisfying the condition. (This definition is introduced by analogy with multiplication of numbers)
In this article, we will talk about the matrix method for solving a system of linear algebraic equations, find its definition and give examples of the solution.
Definition 1
Inverse matrix method is a method used to solve SLAEs in the event that the number of unknowns is equal to the number of equations.
Example 1
Find a solution to a system of n linear equations with n unknowns:
a 11 x 1 + a 12 x 2 +. ... ... + a 1 n x n \u003d b 1 a n 1 x 1 + a n 2 x 2 +. ... ... + a n n x n \u003d b n
Matrix recording : A × X \u003d B
where А \u003d а 11 а 12 ⋯ а 1 n а 21 а 22 ⋯ а 2 n ⋯ ⋯ ⋯ ⋯ а n 1 а n 2 ⋯ а n n - matrix of the system.
X \u003d x 1 x 2 ⋮ x n - column of unknowns,
B \u003d b 1 b 2 ⋮ b n - column of free coefficients.
From the equation we got, you need to express X. To do this, you need to multiply both sides of the matrix equation on the left by A - 1:
A - 1 × A × X \u003d A - 1 × B.
Since A - 1 × A \u003d E, then E × X \u003d A - 1 × B or X \u003d A - 1 × B.
Comment
The inverse matrix to the matrix A has the right to exist only if the condition d e t A is not equal to zero. Therefore, when solving SLAE by the inverse matrix method, first of all, d e t A.
In the event that d e t A is not equal to zero, the system has only one solution: using the inverse matrix method. If d e t А \u003d 0, then the system cannot be solved by this method.
An example of solving a system of linear equations using the inverse matrix method
Example 2We solve the SLAE by the inverse matrix method:
2 x 1 - 4 x 2 + 3 x 3 \u003d 1 x 1 - 2 x 2 + 4 x 3 \u003d 3 3 x 1 - x 2 + 5 x 3 \u003d 2
How to solve?
- We write the system in the form of a matrix equation A X \u003d B, where
A \u003d 2 - 4 3 1 - 2 4 3 - 1 5, X \u003d x 1 x 2 x 3, B \u003d 1 3 2.
- We express from this equation X:
- Find the determinant of the matrix A:
det A \u003d 2 - 4 3 1 - 2 4 3 - 1 5 \u003d 2 × (- 2) × 5 + 3 × (- 4) × 4 + 3 × (- 1) × 1 - 3 × (- 2) × 3 - - 1 × (- 4) × 5 - 2 × 4 - (- 1) \u003d - 20 - 48 - 3 + 18 + 20 + 8 \u003d - 25
d e t А is not equal to 0, therefore the inverse matrix solution method is suitable for this system.
- Find the inverse matrix A - 1 using the union matrix. We calculate the algebraic complements A i j to the corresponding elements of the matrix A:
A 11 \u003d (- 1) (1 + 1) - 2 4 - 1 5 \u003d - 10 + 4 \u003d - 6,
A 12 \u003d (- 1) 1 + 2 1 4 3 5 \u003d - (5 - 12) \u003d 7,
A 13 \u003d (- 1) 1 + 3 1 - 2 3 - 1 \u003d - 1 + 6 \u003d 5,
A 21 \u003d (- 1) 2 + 1 - 4 3 - 1 5 \u003d - (- 20 + 3) \u003d 17,
A 22 \u003d (- 1) 2 + 2 2 3 3 5 - 10 - 9 \u003d 1,
A 23 \u003d (- 1) 2 + 3 2 - 4 3 - 1 \u003d - (- 2 + 12) \u003d - 10,
A 31 \u003d (- 1) 3 + 1 - 4 3 - 2 4 \u003d - 16 + 6 \u003d - 10,
A 32 \u003d (- 1) 3 + 2 2 3 1 4 \u003d - (8 - 3) \u003d - 5,
A 33 \u003d (- 1) 3 + 3 2 - 4 1 - 2 \u003d - 4 + 4 \u003d 0.
- We write down the union matrix A *, which is composed of the algebraic additions of the matrix A:
A * \u003d - 6 7 5 17 1 - 10 - 10 - 5 0
- We write the inverse matrix according to the formula:
A - 1 \u003d 1 d e t A (A *) T: A - 1 \u003d - 1 25 - 6 17 - 10 7 1 - 5 5 - 10 0,
- We multiply the inverse matrix A - 1 by the column of free terms B and get the solution to the system:
X \u003d A - 1 × B \u003d - 1 25 - 6 17 - 10 7 1 - 5 5 - 10 0 1 3 2 \u003d - 1 25 - 6 + 51 - 20 7 + 3 - 10 5 - 30 + 0 \u003d - 1 0 1
Answer : x 1 \u003d - 1; x 2 \u003d 0; x 3 \u003d 1
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