Let's consider system of linear algebraic equations(SLAU) relatively n unknown x 1 , x 2 , ..., x n :
This system in a “collapsed” form can be written as follows:
S n i=1 a ij x j = b i , i=1,2, ..., n.
In accordance with the rule of matrix multiplication, the considered system linear equations can be written in matrix form Ax=b, Where
,
,.
Matrix A, the columns of which are the coefficients for the corresponding unknowns, and the rows are the coefficients for the unknowns in the corresponding equation is called matrix of the system. Column matrix b, the elements of which are the right-hand sides of the equations of the system, is called the right-hand side matrix or simply right side of the system. Column matrix x , whose elements are the unknown unknowns, is called system solution.
A system of linear algebraic equations written in the form Ax=b, is matrix equation.
If the system matrix non-degenerate, then she has inverse matrix and then the solution to the system Ax=b is given by the formula:
x=A -1 b.
Example Solve the system 
matrix method.
Solution let's find the inverse matrix for the coefficient matrix of the system
Let's calculate the determinant by expanding along the first line:
Because the Δ ≠ 0 , That A -1 exists.
The inverse matrix was found correctly.
Let's find a solution to the system 
Hence, x 1 = 1, x 2 = 2, x 3 = 3 .
Examination: 
7. The Kronecker-Capelli theorem on the compatibility of a system of linear algebraic equations.
System of linear equations has the form:
a 21 x 1 + a 22 x 2 +... + a 2n x n = b 2, (5.1)
a m1 x 1 + a m1 x 2 +... + a mn x n = b m.
Here a i j and b i (i = ; j = ) are given, and x j are unknown real numbers. Using the concept of product of matrices, we can rewrite system (5.1) in the form:
where A = (a i j) is a matrix consisting of coefficients for the unknowns of system (5.1), which is called matrix of the system, X = (x 1 , x 2 ,..., x n) T , B = (b 1 , b 2 ,..., b m) T are column vectors composed respectively of unknowns x j and free terms b i .
Ordered collection n real numbers (c 1 , c 2 ,..., c n) are called system solution(5.1), if as a result of substituting these numbers instead of the corresponding variables x 1, x 2,..., x n, each equation of the system turns into an arithmetic identity; in other words, if there is a vector C= (c 1 , c 2 ,..., c n) T such that AC B.
System (5.1) is called joint, or solvable, if it has at least one solution. The system is called incompatible, or unsolvable, if it has no solutions.
,
formed by assigning a column of free terms to the matrix A on the right is called extended matrix of the system.
The question of compatibility of system (5.1) is solved by the following theorem.
Kronecker-Capelli theorem . A system of linear equations is consistent if and only if the ranks of matrices A andA coincide, i.e. r(A) = r(A) = r.
For the set M of solutions of system (5.1) there are three possibilities:
1) M = (in this case the system is inconsistent);
2) M consists of one element, i.e. the system has a unique solution (in this case the system is called certain);
3) M consists of more than one element (then the system is called uncertain). In the third case, system (5.1) has an infinite number of solutions.
The system has a unique solution only if r(A) = n. In this case, the number of equations is not less than the number of unknowns (mn); if m>n, then m-n equations are consequences of the others. If 0 To solve an arbitrary system of linear equations, you need to be able to solve systems in which the number of equations is equal to the number of unknowns - the so-called Cramer type systems: a 11 x 1 + a 12 x 2 +... + a 1n x n = b 1, a 21 x 1 + a 22 x 2 +... + a 2n x n = b 2, (5.3) ...
... ... ...
... ... a n1 x 1 + a n1 x 2 +... + a nn x n = b n . Systems (5.3) are solved in one of the following ways: 1) the Gauss method, or the method of eliminating unknowns; 2) according to Cramer's formulas; 3) matrix method. Example 2.12. Explore the system of equations and solve it if it is consistent: 5x 1 - x 2 + 2x 3 + x 4 = 7, 2x 1 + x 2 + 4x 3 - 2x 4 = 1, x 1 - 3x 2 - 6x 3 + 5x 4 = 0. Solution. We write out the extended matrix of the system:
Let's calculate the rank of the main matrix of the system. It is obvious that, for example, the second-order minor in the upper left corner = 7 0; the third-order minors containing it are equal to zero: Consequently, the rank of the main matrix of the system is 2, i.e. r(A) = 2. To calculate the rank of the extended matrix A, consider the bordering minor this means that the rank of the extended matrix r(A) = 3. Since r(A) r(A), the system is inconsistent. In the first part, we looked at some theoretical material, the substitution method, as well as the method of term-by-term addition of system equations. I recommend everyone who accessed the site through this page to read the first part. Perhaps some visitors will find the material too simple, but in the process of solving systems of linear equations, I made a number of very important comments and conclusions regarding the solution of mathematical problems in general. Now we will analyze Cramer’s rule, as well as solving a system of linear equations using an inverse matrix (matrix method). All materials are presented simply, in detail and clearly; almost all readers will be able to learn how to solve systems using the above methods. First, we will take a closer look at Cramer's rule for a system of two linear equations in two unknowns. For what? – After all, the simplest system can be solved using the school method, the method of term-by-term addition! The fact is that, albeit sometimes, such a task occurs - to solve a system of two linear equations with two unknowns using Cramer's formulas. Secondly, a simpler example will help you understand how to use Cramer's rule for a more complex case - a system of three equations with three unknowns. In addition, there are systems of linear equations with two variables, which are advisable to solve using Cramer’s rule! Consider the system of equations At the first step, we calculate the determinant, it is called main determinant of the system. Gauss method. If , then the system has a unique solution, and to find the roots we must calculate two more determinants: In practice, the above qualifiers can also be denoted by a Latin letter. We find the roots of the equation using the formulas: Example 7 Solve a system of linear equations Solution: We see that the coefficients of the equation are quite large; on the right side there are decimal fractions with a comma. The comma is a rather rare guest in practical tasks in mathematics; I took this system from an econometric problem. How to solve such a system? You can try to express one variable in terms of another, but in this case you will probably end up with terrible fancy fractions that are extremely inconvenient to work with, and the design of the solution will look simply terrible. You can multiply the second equation by 6 and subtract term by term, but the same fractions will arise here too. What to do? In such cases, Cramer's formulas come to the rescue. ; ; Answer: , Both roots have infinite tails and are found approximately, which is quite acceptable (and even commonplace) for econometrics problems. Comments are not needed here, since the task is solved using ready-made formulas, however, there is one caveat. When using this method, compulsory A fragment of the task design is the following fragment: “This means that the system has a unique solution”. Otherwise, the reviewer may punish you for disrespect for Cramer's theorem. It would not be superfluous to check, which can be conveniently carried out on a calculator: we substitute approximate values into the left side of each equation of the system. As a result, with a small error, you should get numbers that are on the right sides. Example 8 Present the answer in ordinary improper fractions. Do a check. This is an example for you to solve on your own (an example of the final design and the answer at the end of the lesson). Let's move on to consider Cramer's rule for a system of three equations with three unknowns: We find the main determinant of the system: If , then the system has infinitely many solutions or is inconsistent (has no solutions). In this case, Cramer's rule will not help; you need to use the Gauss method. If , then the system has a unique solution and to find the roots we must calculate three more determinants: And finally, the answer is calculated using the formulas: As you can see, the “three by three” case is fundamentally no different from the “two by two” case; the column of free terms sequentially “walks” from left to right along the columns of the main determinant. Example 9 Solve the system using Cramer's formulas. Solution: Let's solve the system using Cramer's formulas. Answer: Actually, here again there is nothing special to comment on, due to the fact that the solution follows ready-made formulas. But there are a couple of comments. It happens that as a result of calculations, “bad” irreducible fractions are obtained, for example: . 1) There may be an error in the calculations. As soon as you encounter a “bad” fraction, you immediately need to check Is the condition rewritten correctly?. If the condition is rewritten without errors, then you need to recalculate the determinants using expansion in another row (column). 2) If no errors are identified as a result of checking, then most likely there was a typo in the task conditions. In this case, calmly and CAREFULLY work through the task to the end, and then be sure to check and we draw it up on a clean sheet after the decision. Of course, checking a fractional answer is an unpleasant task, but it will be a disarming argument for the teacher, who really likes to give a minus for any bullshit like . How to handle fractions is described in detail in the answer to Example 8. If you have a computer at hand, then use an automated program to check, which can be downloaded for free at the very beginning of the lesson. By the way, it is most profitable to use the program right away (even before starting the solution); you will immediately see the intermediate step where you made a mistake! The same calculator automatically calculates the solution of the system using the matrix method. Second remark. From time to time there are systems in the equations of which some variables are missing, for example: Example 10 Solve the system using Cramer's formulas. This is an example for an independent solution (a sample of the final design and the answer at the end of the lesson). For the case of a system of 4 equations with 4 unknowns, Cramer’s formulas are written according to similar principles. You can see a live example in the lesson Properties of Determinants. Reducing the order of the determinant - five 4th order determinants are quite solvable. Although the task is already very reminiscent of a professor’s shoe on the chest of a lucky student. The inverse matrix method is essentially a special case matrix equation(See Example No. 3 of the specified lesson). To study this section, you must be able to expand determinants, find the inverse of a matrix, and perform matrix multiplication. Relevant links will be provided as the explanations progress. Example 11 Solve the system using the matrix method Solution: Let's write the system in matrix form: Please look at the system of equations and matrices. I think everyone understands the principle by which we write elements into matrices. The only comment: if some variables were missing from the equations, then zeros would have to be placed in the corresponding places in the matrix. We find the inverse matrix using the formula: First, let's look at the determinant: Here the determinant is expanded on the first line. Attention! If , then the inverse matrix does not exist, and it is impossible to solve the system using the matrix method. In this case, the system is solved by the method of eliminating unknowns (Gauss method). Now we need to calculate 9 minors and write them into the minors matrix Reference: It is useful to know the meaning of double subscripts in linear algebra. The first digit is the number of the line in which the element is located. The second digit is the number of the column in which the element is located: Let there be a square matrix of nth order Matrix A -1 is called inverse matrix in relation to matrix A, if A*A -1 = E, where E is the identity matrix of the nth order. Identity matrix- such a square matrix in which all the elements along the main diagonal, passing from the upper left corner to the lower right corner, are ones, and the rest are zeros, for example: inverse matrix may exist only for square matrices those. for those matrices in which the number of rows and columns coincide. In order for a matrix to have an inverse matrix, it is necessary and sufficient that it be non-singular. The matrix A = (A1, A2,...A n) is called non-degenerate, if the column vectors are linearly independent. The number of linearly independent column vectors of a matrix is called the rank of the matrix. Therefore, we can say that in order for an inverse matrix to exist, it is necessary and sufficient that the rank of the matrix is equal to its dimension, i.e. r = n. For matrix A, find the inverse matrix A -1 Solution: We write matrix A and assign the identity matrix E to the right. Using Jordan transformations, we reduce matrix A to the identity matrix E. The calculations are given in Table 31.1. Let's check the correctness of the calculations by multiplying the original matrix A and the inverse matrix A -1. As a result of matrix multiplication, the identity matrix was obtained. Therefore, the calculations were made correctly. Answer: Matrix equations can look like: AX = B, HA = B, AXB = C, where A, B, C are the specified matrices, X is the desired matrix. Matrix equations are solved by multiplying the equation by inverse matrices.
For example, to find the matrix from the equation, you need to multiply this equation by on the left. Therefore, to find a solution to the equation, you need to find the inverse matrix and multiply it by the matrix on the right side of the equation. Other equations are solved similarly. Solve the equation AX = B if Solution: Since the inverse matrix is equal to (see example 1) Along with others, they are also used matrix methods. These methods are based on linear and vector-matrix algebra. Such methods are used for the purposes of analyzing complex and multidimensional economic phenomena. Most often, these methods are used when it is necessary to make a comparative assessment of the functioning of organizations and their structural divisions. In the process of applying matrix analysis methods, several stages can be distinguished. At the first stage a system of economic indicators is being formed and on its basis a matrix of initial data is compiled, which is a table in which system numbers are shown in its individual rows (i = 1,2,....,n), and in vertical columns - numbers of indicators (j = 1,2,....,m). At the second stage For each vertical column, the largest of the available indicator values is identified, which is taken as one. After this, all amounts reflected in this column are divided by the largest value and a matrix of standardized coefficients is formed. At the third stage all components of the matrix are squared. If they have different significance, then each matrix indicator is assigned a certain weight coefficient k. The value of the latter is determined by expert opinion. On the last one, fourth stage found rating values Rj are grouped in order of their increase or decrease. The matrix methods outlined should be used, for example, in a comparative analysis of various investment projects, as well as in assessing other economic indicators of the activities of organizations. (sometimes this method is also called the matrix method or the inverse matrix method) requires preliminary familiarization with such a concept as the matrix form of notation of SLAE. The inverse matrix method is intended for solving those systems of linear algebraic equations in which the determinant of the system matrix is different from zero. Naturally, this assumes that the matrix of the system is square (the concept of a determinant exists only for square matrices). The essence of the inverse matrix method can be expressed in three points: Any SLAE can be written in matrix form as $A\cdot X=B$, where $A$ is the matrix of the system, $B$ is the matrix of free terms, $X$ is the matrix of unknowns. Let the matrix $A^(-1)$ exist. Let's multiply both sides of the equality $A\cdot X=B$ by the matrix $A^(-1)$ on the left: $$A^(-1)\cdot A\cdot X=A^(-1)\cdot B.$$ Since $A^(-1)\cdot A=E$ ($E$ is the identity matrix), the equality written above becomes: $$E\cdot X=A^(-1)\cdot B.$$ Since $E\cdot X=X$, then: $$X=A^(-1)\cdot B.$$ Example No. 1 Solve the SLAE $ \left \( \begin(aligned) & -5x_1+7x_2=29;\\ & 9x_1+8x_2=-11. \end(aligned) \right.$ using the inverse matrix. $$ A=\left(\begin(array) (cc) -5 & 7\\ 9 & 8 \end(array)\right);\; B=\left(\begin(array) (c) 29\\ -11 \end(array)\right);\; X=\left(\begin(array) (c) x_1\\ x_2 \end(array)\right). $$ Let's find the inverse matrix to the system matrix, i.e. Let's calculate $A^(-1)$. In example No. 2 $$ A^(-1)=-\frac(1)(103)\cdot\left(\begin(array)(cc) 8 & -7\\ -9 & -5\end(array)\right) . $$ Now let's substitute all three matrices ($X$, $A^(-1)$, $B$) into the equality $X=A^(-1)\cdot B$. Then we perform matrix multiplication $$ \left(\begin(array) (c) x_1\\ x_2 \end(array)\right)= -\frac(1)(103)\cdot\left(\begin(array)(cc) 8 & -7\\ -9 & -5\end(array)\right)\cdot \left(\begin(array) (c) 29\\ -11 \end(array)\right)=\\ =-\frac (1)(103)\cdot \left(\begin(array) (c) 8\cdot 29+(-7)\cdot (-11)\\ -9\cdot 29+(-5)\cdot (- 11) \end(array)\right)= -\frac(1)(103)\cdot \left(\begin(array) (c) 309\\ -206 \end(array)\right)=\left( \begin(array) (c) -3\\ 2\end(array)\right). $$ So, we got the equality $\left(\begin(array) (c) x_1\\ x_2 \end(array)\right)=\left(\begin(array) (c) -3\\ 2\end(array )\right)$. From this equality we have: $x_1=-3$, $x_2=2$. Answer: $x_1=-3$, $x_2=2$. Example No. 2 Solve SLAE $ \left\(\begin(aligned) & x_1+7x_2+3x_3=-1;\\ & -4x_1+9x_2+4x_3=0;\\ & 3x_2+2x_3=6. \end(aligned)\right .$ using the inverse matrix method. Let us write down the matrix of the system $A$, the matrix of free terms $B$ and the matrix of unknowns $X$. $$ A=\left(\begin(array) (ccc) 1 & 7 & 3\\ -4 & 9 & 4 \\0 & 3 & 2\end(array)\right);\; B=\left(\begin(array) (c) -1\\0\\6\end(array)\right);\; X=\left(\begin(array) (c) x_1\\ x_2 \\ x_3 \end(array)\right). $$ Now it’s the turn to find the inverse matrix to the system matrix, i.e. find $A^(-1)$. In example No. 3 on the page dedicated to finding inverse matrices, the inverse matrix has already been found. Let's use the finished result and write $A^(-1)$: $$ A^(-1)=\frac(1)(26)\cdot \left(\begin(array) (ccc) 6 & -5 & 1 \\ 8 & 2 & -16 \\ -12 & - 3 & 37\end(array)\right). $$ Now let's substitute all three matrices ($X$, $A^(-1)$, $B$) into the equality $X=A^(-1)\cdot B$, and then perform matrix multiplication on the right side of this equality. $$ \left(\begin(array) (c) x_1\\ x_2 \\ x_3 \end(array)\right)= \frac(1)(26)\cdot \left(\begin(array) (ccc) 6 & -5 & 1 \\ 8 & 2 & -16 \\ -12 & -3 & 37\end(array) \right)\cdot \left(\begin(array) (c) -1\\0\ \6\end(array)\right)=\\ =\frac(1)(26)\cdot \left(\begin(array) (c) 6\cdot(-1)+(-5)\cdot 0 +1\cdot 6 \\ 8\cdot (-1)+2\cdot 0+(-16)\cdot 6 \\ -12\cdot (-1)+(-3)\cdot 0+37\cdot 6 \end(array)\right)=\frac(1)(26)\cdot \left(\begin(array) (c) 0\\-104\\234\end(array)\right)=\left( \begin(array) (c) 0\\-4\\9\end(array)\right) $$ So, we got the equality $\left(\begin(array) (c) x_1\\ x_2 \\ x_3 \end(array)\right)=\left(\begin(array) (c) 0\\-4\ \9\end(array)\right)$. From this equality we have: $x_1=0$, $x_2=-4$, $x_3=9$. This is a concept that generalizes all possible operations performed with matrices. Mathematical matrix - table of elements. About a table where m lines and n columns, this matrix is said to have the dimension m on n. General view of the matrix: For matrix solutions it is necessary to understand what a matrix is and know its main parameters. Main elements of the matrix: Main types of matrices: The matrix can be symmetrical with respect to the main and secondary diagonals. That is, if a 12 = a 21, a 13 =a 31,….a 23 =a 32…. a m-1n =a mn-1, then the matrix is symmetrical about the main diagonal. Only square matrices can be symmetric. Almost all matrix solving methods consist in finding its determinant n-th order and most of them are quite cumbersome. To find the determinant of the 2nd and 3rd order there are other, more rational methods. To calculate the determinant of a matrix A 2nd order, it is necessary to subtract the product of the elements of the secondary diagonal from the product of the elements of the main diagonal: Below are the rules for finding the 3rd order determinant. Simplified rule of triangle as one of matrix solving methods, can be depicted this way: In other words, the product of elements in the first determinant that are connected by straight lines is taken with a “+” sign; Also, for the 2nd determinant, the corresponding products are taken with the “-” sign, that is, according to the following scheme: At solving matrices using Sarrus' rule, to the right of the determinant, add the first 2 columns and the products of the corresponding elements on the main diagonal and on the diagonals that are parallel to it are taken with a “+” sign; and the products of the corresponding elements of the secondary diagonal and the diagonals that are parallel to it, with the sign “-”: Decomposing the determinant in a row or column when solving matrices. The determinant is equal to the sum of the products of the elements of the row of the determinant and their algebraic complements. Typically, the row/column that contains zeros is selected. The row or column along which the decomposition is carried out will be indicated by an arrow. Reducing the determinant to triangular form when solving matrices. At solving matrices method of reducing the determinant to a triangular form, they work like this: using the simplest transformations on rows or columns, the determinant becomes triangular in form and then its value, in accordance with the properties of the determinant, will be equal to the product of the elements that are on the main diagonal. Laplace's theorem for solving matrices. When solving matrices using Laplace's theorem, you need to know the theorem itself. Laplace's theorem: Let Δ
- this is a determinant n-th order. We select any k rows (or columns), provided k≤
n - 1. In this case, the sum of the products of all minors k-th order contained in the selected k rows (columns), by their algebraic complements will be equal to the determinant. Sequence of actions for inverse matrix solutions: For solutions of matrix systems The Gaussian method is most often used. The Gauss method is a standard method for solving systems of linear algebraic equations (SLAEs) and it consists in the fact that variables are sequentially eliminated, i.e., with the help of elementary changes, the system of equations is brought to an equivalent triangular system and from it, sequentially, starting from the latter (by number), find each element of the system. Gauss method is the most versatile and best tool for finding matrix solutions. If a system has an infinite number of solutions or the system is incompatible, then it cannot be solved using Cramer’s rule and the matrix method. The Gauss method also implies direct (reducing the extended matrix to a stepwise form, i.e., obtaining zeros under the main diagonal) and reverse (obtaining zeros above the main diagonal of the extended matrix) moves. The forward move is the Gauss method, the reverse move is the Gauss-Jordan method. The Gauss-Jordan method differs from the Gauss method only in the sequence of eliminating variables.
.
And
, 
, 
, 
, which means the system has a unique solution.
.
I recommend the following “treatment” algorithm. If you don’t have a computer at hand, do this:
Here in the first equation there is no variable , in the second there is no variable . In such cases, it is very important to correctly and CAREFULLY write down the main determinant: 
– zeros are placed in place of missing variables.
By the way, it is rational to open determinants with zeros according to the row (column) in which the zero is located, since there are noticeably fewer calculations.
Solving a system using an inverse matrix
, Where 
, where is the transposed matrix of algebraic complements of the corresponding elements of the matrix.
That is, a double subscript indicates that the element is in the first row, third column, and, for example, the element is in the 3rd row, 2nd column
Theorem for the existence condition of an inverse matrix
Algorithm for finding the inverse matrix
Example 1 
Solving matrix equations
Matrix method in economic analysis
Methods for solving matrices.
Finding 2nd order determinants.
Methods for finding 3rd order determinants.
Solving the inverse matrix.
Solving matrix systems.
