Matrix method for solving systems of linear equations

Consider a system of linear equations of the following form:

$\left\(\begin(array)(c) (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) ) \\ (...) \\ (a_ (n1) x_(1) +a_(n2) x_(2) +...+a_(nn) x_(n) =b_(n) ) \end(array)\right .$.

The numbers $a_(ij) (i=1..n,j=1..n)$ are the coefficients of the system, the numbers $b_(i) (i=1..n)$ are the free terms.

Definition 1

In the case when all free terms are equal to zero, the system is called homogeneous, otherwise it is called inhomogeneous.

Each SLAE can be associated with several matrices and the system can be written in the so-called matrix form.

Definition 2

The matrix of system coefficients is called the system matrix and is usually denoted by the letter $A$.

The column of free terms forms a column vector, which is usually denoted by the letter $B$ and is called the matrix of free terms.

The unknown variables form a column vector, which is usually denoted by the letter $X$ and is called the matrix of unknowns.

The matrices described above have the form:

$A=\left(\begin(array)(cccc) (a_(11) ) & (a_(12) ) & (...) & (a_(1n) ) \\ (a_(21) ) & ( a_(22) ) & (...) & (a_(2n) ) \\ (...) & (...) & (...) & (...) \\ (a_(n1) ) & (a_(n2) ) & (...) & (a_(nn) ) \end(array)\right),B=\left(\begin(array)(c) (b_(1) ) \ \ (b_(2) ) \\ (...) \\ (b_(n) ) \end(array)\right),X=\left(\begin(array)(c) (x_(1) ) \\ (x_(2) ) \\ (...) \\ (x_(n) ) \end(array)\right).$

Using matrices, the SLAE can be rewritten as $A\cdot X=B$. This notation is often called a matrix equation.

Generally speaking, any SLAE can be written in matrix form.

Examples of solving a system using an inverse matrix

Example 1

Given SLAE: $\left\(\begin(array)(c) (3x_(1) -2x_(2) +x_(3) -x_(4) =3) \\ (x_(1) -12x_(2 ) -x_(3) -x_(4) =7) \\ (2x_(1) -3x_(2) +x_(3) -3x_(4) =5) \end(array)\right $. system in matrix form.

Solution:

$A=\left(\begin(array)(cccc) (3) & (-2) & (1) & (-1) \\ (1) & (-12) & (-1) & (-1 ) \\ (2) & (-3) & (1) & (-3) \end(array)\right),B=\left(\begin(array)(c) (3) \\ (7) \\ (5) \end(array)\right),X=\left(\begin(array)(c) (x_(1) ) \\ (x_(2) ) \\ (x_(3) ) \ end(array)\right).$

$\left(\begin(array)(cccc) (3) & (-2) & (1) & (-1) \\ (1) & (-12) & (-1) & (-1) \ \ (2) & (-3) & (1) & (-3) \end(array)\right)\cdot \left(\begin(array)(c) (x_(1) ) \\ (x_( 2) ) \\ (x_(3) ) \end(array)\right)=\left(\begin(array)(c) (3) \\ (7) \\ (5) \end(array)\ right)$

In the case when the matrix of the system is square, the SLAE can be solved using the matrix method.

Having a matrix equation $A\cdot X=B$, we can express $X$ from it in the following way:

$A^(-1) \cdot A\cdot X=A^(-1) \cdot B$

$A^(-1) \cdot A=E$ (matrix product property)

$E\cdot X=A^(-1) \cdot B$

$E\cdot X=X$ (matrix product property)

$X=A^(-1) \cdot B$

Algorithm for solving a system of algebraic equations using an inverse matrix:

• write the system in matrix form;
• calculate the determinant of the system matrix;
• if the determinant of the system matrix is ​​different from zero, then we find the inverse matrix;
• We calculate the solution of the system using the formula $X=A^(-1) \cdot B$.

If the matrix of a system has a determinant that is not equal to zero, then this system has a unique solution that can be found using the matrix method.

If the matrix of a system has a determinant equal to zero, then this system cannot be solved using the matrix method.

Example 2

Given SLAE: $\left\(\begin(array)(c) (x_(1) +3x_(3) =26) \\ (-x_(1) +2x_(2) +x_(3) =52) \\ (3x_(1) +2x_(2) =52) \end(array)\right $. Solve the SLAE using the inverse matrix method, if possible.

Solution:

$A=\left(\begin(array)(ccc) (1) & (0) & (3) \\ (-1) & (2) & (1) \\ (3) & (2) & ( 0) \end(array)\right),B=\left(\begin(array)(c) (26) \\ (52) \\ (52) \end(array)\right),X=\left (\begin(array)(c) (x_(1) ) \\ (x_(2) ) \\ (x_(3) ) \end(array)\right). $

Finding the determinant of the system matrix:

$\begin(array)(l) (\det A=\left|\begin(array)(ccc) (1) & (0) & (3) \\ (-1) & (2) & (1) \\ (3) & (2) & (0) \end(array)\right|=1\cdot 2\cdot 0+0\cdot 1\cdot 3+2\cdot (-1)\cdot 3-3 \cdot 2\cdot 3-2\cdot 1\cdot 1-0\cdot (-1)\cdot 0=0+0-6-18-2-0=-26\ne 0) \end(array)$ Since the determinant is not equal to zero, the matrix of the system has an inverse matrix and, therefore, the system of equations can be solved by the inverse matrix method. The resulting solution will be unique.

Let's solve the system of equations using the inverse matrix:

$A_(11) =(-1)^(1+1) \cdot \left|\begin(array)(cc) (2) & (1) \\ (2) & (0) \end(array) \right|=0-2=-2; A_(12) =(-1)^(1+2) \cdot \left|\begin(array)(cc) (-1) & (1) \\ (3) & (0) \end(array) \right|=-(0-3)=3;$

$A_(13) =(-1)^(1+3) \cdot \left|\begin(array)(cc) (-1) & (2) \\ (3) & (2) \end(array )\right|=-2-6=-8; A_(21) =(-1)^(2+1) \cdot \left|\begin(array)(cc) (0) & (3) \\ (2) & (0) \end(array)\ right|=-(0-6)=6; $

$A_(22) =(-1)^(2+2) \cdot \left|\begin(array)(cc) (1) & (3) \\ (3) & (0) \end(array) \right|=0-9=-9; A_(23) =(-1)^(2+3) \cdot \left|\begin(array)(cc) (1) & (0) \\ (3) & (2) \end(array)\ right|=-(2-0)=-2;$

$A_(31) =(-1)^(3+1) \cdot \left|\begin(array)(cc) (0) & (3) \\ (2) & (1) \end(array) \right|=0-6=-6; A_(32) =(-1)^(3+2) \cdot \left|\begin(array)(cc) (1) & (3) \\ (-1) & (1) \end(array) \right|=-(1+3)=-4;$

$A_(33) =(-1)^(3+3) \cdot \left|\begin(array)(cc) (1) & (0) \\ (-1) & (2) \end(array )\right|=2-0=2$

The required inverse matrix:

$A^(-1) =\frac(1)(-26) \cdot \left(\begin(array)(ccc) (-2) & (6) & (-6) \\ (3) & ( -9) & (-4) \\ (-8) & (-2) & (2) \end(array)\right)=\frac(1)(26) \cdot \left(\begin(array) (ccc) (2) & (-6) & (6) \\ (-3) & (9) & (4) \\ (8) & (2) & (-2) \end(array)\right )=\left(\begin(array)(ccc) (\frac(2)(26) ) & (\frac(-6)(26) ) & (\frac(6)(26) ) \\ (\ frac(-3)(26) ) & (\frac(9)(26) ) & (\frac(4)(26) ) \\ (\frac(8)(26) ) & (\frac(2) (26) ) & (\frac(-2)(26) ) \end(array)\right)=\left(\begin(array)(ccc) (\frac(1)(13) ) & (-\ frac(3)(13) ) & (\frac(3)(13) ) \\ (-\frac(3)(26) ) & (\frac(9)(26) ) & (\frac(2) (13) ) \\ (\frac(4)(13) ) & (\frac(1)(13) ) & (-\frac(1)(13) ) \end(array)\right).$

Let's find a solution to the system:

$X=\left(\begin(array)(ccc) (\frac(1)(13) ) & (-\frac(3)(13) ) & (\frac(3)(13) ) \\ ( -\frac(3)(26) ) & (\frac(9)(26) ) & (\frac(2)(13) ) \\ (\frac(4)(13) ) & (\frac(1 )(13) ) & (-\frac(1)(13) ) \end(array)\right)\cdot \left(\begin(array)(c) (26) \\ (52) \\ (52 ) \end(array)\right)=\left(\begin(array)(c) (\frac(1)(13) \cdot 26-\frac(3)(13) \cdot 52+\frac(3 )(13) \cdot 52) ​​\\ (-\frac(3)(26) \cdot 26+\frac(9)(26) \cdot 52+\frac(2)(13) \cdot 52) ​​\\ (\frac(4)(13) \cdot 26+\frac(1)(13) \cdot 52-\frac(1)(13) \cdot 52) ​​\end(array)\right)=\left(\ begin(array)(c) (2-12+12) \\ (-3+18+8) \\ (8+4-4) \end(array)\right)=\left(\begin(array) (c) (2) \\ (23) \\ (8) \end(array)\right)$

$X=\left(\begin(array)(c) (2) \\ (23) \\ (8) \end(array)\right)$ is the desired solution to the system of equations.

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 matrix multiplication rule, the considered system of 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 , the elements of which 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 it has an inverse matrix and then the solution to the system is 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 right side of the matrix A 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 andA 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 (mn); 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:

Therefore, 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.

This online calculator solves a system of linear equations using the matrix method. A very detailed solution is given. To solve a system of linear equations, select the number of variables. Choose a method for calculating the inverse matrix. Then enter the data in the cells and click on the "Calculate" button.

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Data entry instructions. Numbers are entered as integers (examples: 487, 5, -7623, etc.), decimals (ex. 67., 102.54, etc.) or fractions. The fraction must be entered in the form a/b, where a and b are integers or decimals. Examples 45/5, 6.6/76.4, -7/6.7, etc.

Matrix method for solving systems of linear equations

Consider the following system of linear equations:

Given the definition of an inverse matrix, we have A −1 A=E, Where E- identity matrix. Therefore (4) can be written as follows:

Thus, to solve the system of linear equations (1) (or (2)), it is enough to multiply the inverse of A matrix per constraint vector b.

Examples of solving a system of linear equations using the matrix method

Example 1. Solve the following system of linear equations using the matrix method:

Let's find the inverse of matrix A using the Jordan-Gauss method. On the right side of the matrix A Let's write the identity matrix:

Let's exclude the elements of the 1st column of the matrix below the main diagonal. To do this, add lines 2,3 with line 1, multiplied by -1/3, -1/3, respectively:

Let's exclude the elements of the 2nd column of the matrix below the main diagonal. To do this, add line 3 with line 2 multiplied by -24/51:

Let's exclude the elements of the 2nd column of the matrix above the main diagonal. To do this, add line 1 with line 2 multiplied by -3/17:

Separate the right side of the matrix. The resulting matrix is ​​the inverse matrix of A :

Matrix form of writing a system of linear equations: Ax=b, Where

Let's calculate all algebraic complements of the matrix A:

,

,

,

,

,

Where A ij − algebraic complement of a matrix element A, located at the intersection i-th line and j-th column, and Δ is the determinant of the matrix A.

Using the inverse matrix formula, we get:

Purpose of the service. Using this online calculator, unknowns (x 1, x 2, ..., x n) are calculated in a system of equations. The decision is carried out inverse matrix method. Wherein:

• the determinant of the matrix A is calculated;
• through algebraic additions the inverse matrix A -1 is found;
• a solution template is created in Excel;

The decision is carried out directly on the website (online) and is free. The calculation results are presented in a report in Word format.

Instructions. To obtain a solution using the inverse matrix method, you need to specify the dimension of the matrix. Next, in a new dialog box, fill in the matrix A and the vector of results B.

Recall that a solution to a system of linear equations is any set of numbers (x 1, x 2, ..., x n), the substitution of which into this system instead of the corresponding unknowns turns each equation of the system into an identity.
A system of linear algebraic equations is usually written as (for 3 variables): See also Solving matrix equations.

Solution algorithm

1. The determinant of the matrix A is calculated. If the determinant is zero, then the solution is over. The system has an infinite number of solutions.
2. When the determinant is different from zero, the inverse matrix A -1 is found through algebraic additions.
3. The solution vector X =(x 1, x 2, ..., x n) is obtained by multiplying the inverse matrix by the result vector B.

Example No. 1. Find a solution to the system using the matrix method. Let's write the matrix in the form:

Algebraic additions.

A 1,1 = (-1) 1+1

1 2 0 -2

∆ 1,1 = (1 (-2)-0 2) = -2

A 1,2 = (-1) 1+2

3 2 1 -2

∆ 1,2 = -(3 (-2)-1 2) = 8

A 1.3 = (-1) 1+3

3 1 1 0

∆ 1,3 = (3 0-1 1) = -1

A 2,1 = (-1) 2+1

-2 1 0 -2

∆ 2,1 = -(-2 (-2)-0 1) = -4

A 2,2 = (-1) 2+2

2 1 1 -2

∆ 2,2 = (2 (-2)-1 1) = -5

A 2,3 = (-1) 2+3

2 -2 1 0

∆ 2,3 = -(2 0-1 (-2)) = -2

A 3.1 = (-1) 3+1

-2 1 1 2

∆ 3,1 = (-2 2-1 1) = -5

A 3.2 = (-1) 3+2

2 1 3 2

∆ 3,2 = -(2 2-3 1) = -1

·

3

-2

-1

X T = (1,0,1)
x 1 = -21 / -21 = 1
x 2 = 0 / -21 = 0
x 3 = -21 / -21 = 1
Examination:
2 1+3 0+1 1 = 3
-2 1+1 0+0 1 = -2
1 1+2 0+-2 1 = -1

Example No. 2. Solve SLAE using the inverse matrix method.
2 x 1 + 3x 2 + 3x 3 + x 4 = 1
3 x 1 + 5x 2 + 3x 3 + 2x 4 = 2
5 x 1 + 7x 2 + 6x 3 + 2x 4 = 3
4 x 1 + 4x 2 + 3x 3 + x 4 = 4

Let's write the matrix in the form:

Vector B:
B T = (1,2,3,4)
Main determinant
Minor for (1,1):

= 5 (6 1-3 2)-7 (3 1-3 2)+4 (3 2-6 2) = -3
Minor for (2,1):

= 3 (6 1-3 2)-7 (3 1-3 1)+4 (3 2-6 1) = 0
Minor for (3,1):

= 3 (3 1-3 2)-5 (3 1-3 1)+4 (3 2-3 1) = 3
Minor for (4,1):

= 3 (3 2-6 2)-5 (3 2-6 1)+7 (3 2-3 1) = 3
Determinant of minor
∆ = 2 (-3)-3 0+5 3-4 3 = -3

Example No. 4. Write the system of equations in matrix form and solve using the inverse matrix.
Solution:xls

Example No. 5. A system of three linear equations with three unknowns is given. Required: 1) find its solution using Cramer's formulas; 2) write the system in matrix form and solve it using matrix calculus.
Guidelines. After solving by Cramer's method, find the "Solving by inverse matrix method for source data" button. You will receive the appropriate solution. Thus, you will not have to fill in the data again.
Solution. Let us denote by A the matrix of coefficients for unknowns; X - matrix-column of unknowns; B - matrix-column of free members:

-1 3 0 3 -2 1 2 1 -1

Vector B:
B T =(4,-3,-3)
Taking into account these notations, this system of equations takes the following matrix form: A*X = B.
If the matrix A is non-degenerate (its determinant is different from zero, then it has an inverse matrix A -1. Multiplying both sides of the equation by A -1, we get: A -1 *A*X = A -1 *B, A -1 * A=E.
This equality is called matrix notation of the solution to a system of linear equations. To find a solution to the system of equations, it is necessary to calculate the inverse matrix A -1.
The system will have a solution if the determinant of the matrix A is nonzero.
Let's find the main determinant.
∆=-1 (-2 (-1)-1 1)-3 (3 (-1)-1 0)+2 (3 1-(-2 0))=14
So, the determinant is 14 ≠ 0, so we continue the solution. To do this, we find the inverse matrix through algebraic additions.
Let us have a non-singular matrix A:
We calculate algebraic complements.

A 1,1 =(-1) 1+1

-2 1 1 -1

∆ 1,1 =(-2 (-1)-1 1)=1

A 1,2 =(-1) 1+2

3 1 0 -1

∆ 1,2 =-(3 (-1)-0 1)=3

A 1.3 =(-1) 1+3

3 -2 0 1

∆ 1,3 =(3 1-0 (-2))=3

A 2,1 =(-1) 2+1

3 2 1 -1

∆ 2,1 =-(3 (-1)-1 2)=5

A 2,2 =(-1) 2+2

-1 2 0 -1

∆ 2,2 =(-1 (-1)-0 2)=1

A 2,3 =(-1) 2+3

-1 3 0 1

∆ 2,3 =-(-1 1-0 3)=1

A 3.1 =(-1) 3+1

3 2 -2 1

∆ 3,1 =(3 1-(-2 2))=7
·

4

-3

-3

X=1/14

-3))

Main determinant
∆=4 (0 1-3 (-2))-2 (1 1-3 (-1))+0 (1 (-2)-0 (-1))=16
Transposed matrix
∆ 1,1 =(0 1-(-2 3))=6

A 1,2 =(-1) 1+2

1 3 -1 1

∆ 1,2 =-(1 1-(-1 3))=-4

A 1.3 =(-1) 1+3

1 0 -1 -2

∆ 1,3 =(1 (-2)-(-1 0))=-2

A 2,1 =(-1) 2+1

2 0 -2 1

∆ 2,1 =-(2 1-(-2 0))=-2

A 2,2 =(-1) 2+2

4 0 -1 1

∆ 2,2 =(4 1-(-1 0))=4

A 2,3 =(-1) 2+3

4 2 -1 -2

∆ 2,3 =-(4 (-2)-(-1 2))=6

A 3.1 =(-1) 3+1

2 0 0 3

∆ 3,1 =(2 3-0 0)=6

A 3.2 =(-1) 3+2

4 0 1 3

∆ 3,2 =-(4 3-1 0)=-12

A 3.3 =(-1) 3+3

1/16

6 -4 -2 -2 4 6 6 -12 -2

E=A*A -1 =

(4 6)+(1 (-2))+(-1 6)	(4 (-4))+(1 4)+(-1 (-12))	(4 (-2))+(1 6)+(-1 (-2))
(2 6)+(0 (-2))+(-2 6)	(2 (-4))+(0 4)+(-2 (-12))	(2 (-2))+(0 6)+(-2 (-2))
(0 6)+(3 (-2))+(1 6)	(0 (-4))+(3 4)+(1 (-12))	(0 (-2))+(3 6)+(1 (-2))

=1/16

16 0 0 0 16 0 0 0 16

A*A -1 =

1 0 0 0 1 0 0 0 1

Example No. 7. Solving matrix equations.
Let's denote:

A=

3 0 5 2 1 4 -1 3 0

Algebraic additions

A 1,1 = (-1) 1+1

1 3 4 0

∆ 1,1 = (1*0 - 4*3) = -12

A 1,2 = (-1) 1+2

0 3 5 0

∆ 1,2 = -(0*0 - 5*3) = 15

A 1.3 = (-1) 1+3

0 1 5 4

∆ 1,3 = (0*4 - 5*1) = -5

A 2,1 = (-1) 2+1

2 -1 4 0

∆ 2,1 = -(2*0 - 4*(-1)) = -4

A 2,2 = (-1) 2+2

3 -1 5 0

∆ 2,2 = (3*0 - 5*(-1)) = 5

A 2,3 = (-1) 2+3

3 2 5 4

∆ 2,3 = -(3*4 - 5*2) = -2

A 3.1 = (-1) 3+1

2 -1 1 3

∆ 3,1 = (2*3 - 1*(-1)) = 7
· 1/-1

-12	15	-5
-4	5	-2
7	-9	3

= Vector B:
B T =(31,13,10)

X T =(4.05,6.13,7.54)
x 1 = 158 / 39 =4.05
x 2 = 239 / 39 =6.13
x 3 = 294 / 39 =7.54
Examination.
-2 4.05+-1 6.13+6 7.54=31
1 4.05+-1 6.13+2 7.54=13
2 4.05+4 6.13+-3 7.54=10

Example No. 9. Let us denote by A the matrix of coefficients for unknowns; X - matrix-column of unknowns; B - matrix-column of free members:

-2 1 6 1 -1 2 2 4 -3

Vector B:
B T =(31,13,10)

X T =(5.21,4.51,6.15)
x 1 = 276 / 53 =5.21
x 2 = 239 / 53 =4.51
x 3 = 326 / 53 =6.15
Examination.
-2 5.21+1 4.51+6 6.15=31
1 5.21+-1 4.51+2 6.15=13
2 5.21+4 4.51+-3 6.15=10

Example No. 10. Solving matrix equations.
Let's denote:

Algebraic additions
A 11 = (-1) 1+1 ·-3 = -3; A 12 = (-1) 1+2 ·3 = -3; A 21 = (-1) 2+1 ·1 = -1; A 22 = (-1) 2+2 ·2 = 2;
Inverse matrix A -1 .
· 1/-9

-3	-3
-1	2

=

1	-2
1	1

Answer:

X =

1 -2 1 1

6.4. Some applications of the dot product

11. Expression of the scalar product of a vector through the coordinates of the factors. Theorem.

12. Length of a vector, length of a segment, angle between vectors, condition of perpendicularity of vectors.

13. Vector product of vectors, its properties. Area of ​​a parallelogram.

14. Mixed product of vectors, its properties. Condition for vector coplanarity. Volume of a parallelepiped. Volume of the pyramid.

15. Methods for defining a straight line on a plane.

16. Normal equation of a line on a plane (derivation). Geometric meaning of coefficients.

17. Equation of a straight line on a plane in segments (derivation).

Reducing the general equation of the plane to the equation of the plane in segments.

18. Equation of a straight line on a plane with an angular coefficient (derivation).

19. Equation of a straight line on a plane passing through two points (derivation).

20. Angle between straight lines on a plane (output).

21. Distance from a point to a straight line on a plane (output).

22. Conditions for parallelism and perpendicularity of lines on a plane (derivation).

23. Equation of a plane. Normal plane equation (derivation). Geometric meaning of coefficients.

24. Equation of a plane in segments (derivation).

25. Equation of a plane passing through three points (derivation).

26. Angle between planes (output).

27. Distance from a point to a plane (output).

28. Conditions for parallelism and perpendicularity of planes (conclusion).

29. Equations of a line in r3. Equations of a line passing through two fixed points (derivation).

30. Canonical equations of a straight line in space (derivation).

Drawing up canonical equations of a straight line in space.

Special cases of canonical equations of a straight line in space.

Canonical equations of a line passing through two given points in space.

Transition from the canonical equations of a line in space to other types of equations of a line.

31. Angle between straight lines (output).

32. Distance from a point to a straight line on a plane (output).

Distance from a point to a straight line on a plane - theory, examples, solutions.

The first way to find the distance from a given point to a given straight line on a plane.

The second method allows you to find the distance from a given point to a given straight line on a plane.

Solving problems of finding the distance from a given point to a given straight line on a plane.

Distance from a point to a line in space - theory, examples, solutions.

The first way to find the distance from a point to a line in space.

The second method allows you to find the distance from a point to a line in space.

33. Conditions for parallelism and perpendicularity of lines in space.

34. The relative position of lines in space and a line with a plane.

35. Classical ellipse equation (derivation) and its construction. The canonical equation of an ellipse has the form where are positive real numbers, and. How to construct an ellipse?

36. Classic hyperbola equation (derivation) and its construction. Asymptotes.

37. Canonical parabola equation (derivation) and construction.

38. Function. Basic definitions. Graphs of basic elementary functions.

39. Number sequences. Limit of number sequence.

40. Infinitely small and infinitely large quantities. Theorem about the connection between them, properties.

41. Theorems on actions on variables having finite limits.

42. Number e.

Content

Determination methods

Properties

Story

Approximations

43. Determination of the limit of a function. Uncovering uncertainties.

44. Remarkable limits, their conclusion. Equivalent infinitesimal quantities.

Content

The first wonderful limit

Second wonderful limit

45. One-sided limits. Continuity and discontinuities of function. One-sided limits

Left and right limits of a function

Discontinuity point of the first kind

Discontinuity point of the second kind

Removable break point

46. ​​Definition of derivative. Geometrical meaning, mechanical meaning of derivative. Tangent and normal equations for a curve and a point.

47. Theorems on the derivative of inverse, complex functions.

48. Derivatives of the simplest elementary functions.

49. Differentiation of parametric, implicit and power-exponential functions.

21. Differentiation of implicit and parametrically defined functions

21.1. Implicit function

21.2. Parametrically defined function

50. Higher order derivatives. Taylor's formula.

51. Differential. Application of differential to approximate calculations.

52. Theorems of Rolle, Lagrange, Cauchy. L'Hopital's rule.

53. Theorem on necessary and sufficient conditions for the monotonicity of a function.

54. Determination of the maximum and minimum of a function. Theorems on necessary and sufficient conditions for the existence of an extremum of a function.

Theorem (necessary condition for extremum)

55. Convexity and concavity of curves. Inflection points. Theorems on necessary and sufficient conditions for the existence of inflection points.

Proof

57. Determinants of the nth order, their properties.

58. Matrices and actions on them. Matrix rank.

Definition

Related definitions

Properties

Linear transformation and matrix rank

59. Inverse matrix. Theorem on the existence of an inverse matrix.

60. Systems of linear equations. Matrix solution of systems of linear equations. Cramer's rule. Gauss method. Kronecker-Capelli theorem.

Solving systems of linear algebraic equations, solution methods, examples.

Definitions, concepts, designations.

Solving elementary systems of linear algebraic equations.

Solving systems of linear equations using Cramer's method.

Solving systems of linear algebraic equations using the matrix method (using an inverse matrix).

Solving systems of linear equations using the Gauss method.

Solving systems of linear algebraic equations of general form.

Kronecker–Capelli theorem.

Gauss method for solving systems of linear algebraic equations of general form.

Writing a general solution to homogeneous and inhomogeneous linear algebraic systems using vectors of the fundamental system of solutions.

Solving systems of equations that reduce to slough.

Examples of problems that reduce to solving systems of linear algebraic equations.

Solving systems of linear algebraic equations using the matrix method (using an inverse matrix).

Let the system of linear algebraic equations be given in matrix form , where the matrix A has dimension n on n and its determinant is nonzero.

Since , then the matrix A– is invertible, that is, there is an inverse matrix. If we multiply both sides of the equality to the left, we obtain a formula for finding a matrix-column of unknown variables. This is how we obtained a solution to a system of linear algebraic equations using the matrix method.

matrix method.

Let's rewrite the system of equations in matrix form:

Because then the SLAE can be solved using the matrix method. Using the inverse matrix, the solution to this system can be found as .

Let's construct an inverse matrix using a matrix from algebraic complements of matrix elements A(if necessary, see the article methods for finding the inverse matrix):

It remains to calculate the matrix of unknown variables by multiplying the inverse matrix to a matrix-column of free members (if necessary, see the article operations on matrices):

or in another post x 1 = 4, x 2 = 0, x 3 = -1 .

The main problem when finding solutions to systems of linear algebraic equations using the matrix method is the complexity of finding the inverse matrix, especially for square matrices of order higher than third.

For a more detailed description of the theory and additional examples, see the article matrix method for solving systems of linear equations.

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Solving systems of linear equations using the Gauss method.

Suppose we need to find a solution to the system from n linear equations with n unknown variables the determinant of the main matrix of which is different from zero.

The essence of the Gauss method consists of sequentially eliminating unknown variables: first eliminating x 1 from all equations of the system, starting from the second, is further excluded x 2 from all equations, starting with the third, and so on, until only the unknown variable remains in the last equation x n. This process of transforming system equations to sequentially eliminate unknown variables is called direct Gaussian method. After completing the forward progression of the Gaussian method, from the last equation we find x n, using this value from the penultimate equation we calculate x n-1, and so on, from the first equation we find x 1 . The process of calculating unknown variables when moving from the last equation of the system to the first is called inverse of the Gaussian method.

Let us briefly describe the algorithm for eliminating unknown variables.

We will assume that , since we can always achieve this by rearranging the equations of the system. Eliminate the unknown variable x 1 from all equations of the system, starting from the second. To do this, to the second equation of the system we add the first, multiplied by, to the third equation we add the first, multiplied by, and so on, to nth to the equation we add the first one multiplied by. The system of equations after such transformations will take the form where and .

We would arrive at the same result if we expressed x 1 through other unknown variables in the first equation of the system and the resulting expression was substituted into all other equations. So the variable x 1 excluded from all equations, starting from the second.

Next, we proceed in a similar way, but only with part of the resulting system, which is marked in the figure

To do this, to the third equation of the system we add the second, multiplied by, to the fourth equation we add the second, multiplied by, and so on, to nth to the equation we add the second, multiplied by. The system of equations after such transformations will take the form where and . So the variable x 2 excluded from all equations starting from the third.

Next we proceed to eliminating the unknown x 3 , in this case we act similarly with the part of the system marked in the figure

So we continue the direct progression of the Gaussian method until the system takes the form

From this moment we begin the reverse of the Gaussian method: we calculate x n from the last equation as, using the obtained value x n we find x n-1 from the penultimate equation, and so on, we find x 1 from the first equation.

Solve system of linear equations Gauss method.

Eliminate the unknown variable x 1 from the second and third equations of the system. To do this, to both sides of the second and third equations we add the corresponding parts of the first equation, multiplied by and respectively:

Now let's exclude from the third equation x 2 , adding to its left and right sides the left and right sides of the second equation, multiplied by:

This completes the forward stroke of the Gauss method; we begin the reverse stroke.

From the last equation of the resulting system of equations we find x 3 :

From the second equation we get .

From the first equation we find the remaining unknown variable and thereby complete the reverse of the Gauss method.

x 1 = 4, x 2 = 0, x 3 = -1 .

For more detailed information and additional examples, see the section on solving elementary systems of linear algebraic equations using the Gauss method.

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