Let M 0 – set of solutions of a homogeneous system (4) linear equations.
Definition 6.12. Vectors With 1 ,With 2 , …, with p, which are solutions of a homogeneous system of linear equations are called fundamental set of solutions(abbreviated FNR), if
1) vectors With 1 ,With 2 , …, with p linearly independent (i.e., none of them can be expressed in terms of the others);
2) any other solution to a homogeneous system of linear equations can be expressed in terms of solutions With 1 ,With 2 , …, with p.
Note that if With 1 ,With 2 , …, with p– any f.n.r., then the expression k 1× With 1 + k 2× With 2 + … + k p× with p you can describe the whole set M 0 solutions to system (4), so it is called general view of the system solution (4).
Theorem 6.6. Any indeterminate homogeneous system of linear equations has a fundamental set of solutions.
The way to find the fundamental set of solutions is as follows:
Find common decision homogeneous system of linear equations;
Build ( n – r) partial solutions of this system, while the values of the free unknowns must form identity matrix;
Write out general form solutions included in M 0 .
Example 6.5. Find a fundamental set of solutions to the following system:
Solution. Let's find a general solution to this system.
~ 
~ 
~ ~ 
Þ 
Þ Þ 
There are five unknowns in this system ( n= 5), of which there are two main unknowns ( r= 2), there are three free unknowns ( n – r), that is, the fundamental solution set contains three solution vectors. Let's build them. We have x 1 and x 3 – main unknowns, x 2 , x 4 , x 5 – free unknowns
Values of free unknowns x 2 , x 4 , x 5 form the identity matrix E third order. Got that vectors With 1 ,With 2 , With 3 form f.n.r. of this system. Then the set of solutions of this homogeneous system will be M 0 = {k 1× With 1 + k 2× With 2 + k 3× With 3 , k 1 , k 2 , k 3 О R).
Let us now find out the conditions for the existence of nonzero solutions of a homogeneous system of linear equations, in other words, the conditions for the existence of a fundamental set of solutions.
A homogeneous system of linear equations has non-zero solutions, that is, it is uncertain if
1) the rank of the main matrix of the system is less than the number of unknowns;
2) in a homogeneous system of linear equations, the number of equations is less than the number of unknowns;
3) if in a homogeneous system of linear equations the number of equations is equal to the number of unknowns, and the determinant of the main matrix is equal to zero (i.e. | A| = 0).
Example 6.6. At what parameter value a homogeneous system of linear equations 
has non-zero solutions?
Solution. Let's compose the main matrix of this system and find its determinant: = = 1×(–1) 1+1 × = – A– 4. The determinant of this matrix is equal to zero at a = –4.
Answer: –4.
7. Arithmetic n-dimensional vector space
Basic Concepts
In previous sections we have already encountered the concept of a set of real numbers located in in a certain order. This is a row matrix (or column matrix) and a solution to a system of linear equations with n unknown. This information can be summarized.
Definition 7.1. n-dimensional arithmetic vector called an ordered set of n real numbers.
Means A= (a 1 , a 2 , …, a n), where a iО R, i = 1, 2, …, n– general view of the vector. Number n called dimension vectors, and numbers a i are called his coordinates.
For example: A= (1, –8, 7, 4, ) – five-dimensional vector.
All set n-dimensional vectors are usually denoted as Rn.
Definition 7.2. Two vectors A= (a 1 , a 2 , …, a n) And b= (b 1 , b 2 , …, b n) of the same dimension equal if and only if their corresponding coordinates are equal, i.e. a 1 = b 1 , a 2 = b 2 , …, a n= b n.
Definition 7.3.Amount two n-dimensional vectors A= (a 1 , a 2 , …, a n) And b= (b 1 , b 2 , …, b n) is called a vector a + b= (a 1 + b 1, a 2 + b 2, …, a n+ b n).
Definition 7.4. The work real number k to vector A= (a 1 , a 2 , …, a n) is called a vector k× A = (k×a 1, k×a 2 , …, k×a n)
Definition 7.5. Vector O= (0, 0, …, 0) is called zero(or null vector).
It is easy to verify that the actions (operations) of adding vectors and multiplying them by a real number have the following properties: " a, b, c Î Rn, " k, lО R:
1) a + b = b + a;
2) a + (b+ c) = (a + b) + c;
3) a + O = a;
4) a+ (–a) = O;
5) 1× a = a, 1 О R;
6) k×( l× a) = l×( k× a) = (l× k)× a;
7) (k + l)× a = k× a + l× a;
8) k×( a + b) = k× a + k× b.
Definition 7.6. A bunch of Rn with the operations of adding vectors and multiplying them by a real number given on it is called arithmetic n-dimensional vector space.
Given matrices
Find: 1) aA - bB,
Solution: 1) We find it sequentially, using the rules of multiplying a matrix by a number and adding matrices..
2. Find A*B if
Solution: We use the matrix multiplication rule
Answer: 
3. For a given matrix, find the minor M 31 and calculate the determinant.
Solution: Minor M 31 is the determinant of the matrix that is obtained from A
after crossing out line 3 and column 1. We find
1*10*3+4*4*4+1*1*2-2*4*10-1*1*4-1*4*3 = 0.
Let's transform matrix A without changing its determinant (let's make zeros in row 1)
| -3*, -, -4* | |||
| -10 | -15 | ||
| -20 | -25 | ||
| -4 | -5 |
Now we calculate the determinant of matrix A by expansion along row 1
Answer: M 31 = 0, detA = 0
Solve using the Gauss method and Cramer method.
2x 1 + x 2 + x 3 = 2
x 1 + x 2 + 3x 3 = 6
2x 1 + x 2 + 2x 3 = 5
Solution: Let's check
You can use Cramer's method
Solution of the system: x 1 = D 1 / D = 2, x 2 = D 2 / D = -5, x 3 = D 3 / D = 3
Let's apply the Gaussian method.
Let us reduce the extended matrix of the system to triangular form.
For ease of calculation, let's swap the lines:
Multiply the 2nd line by (k = -1 / 2 = -1 / 2 ) and add to the 3rd:
| 1 / 2 | 7 / 2 |
Multiply the 1st line by (k = -2 / 2 = -1 ) and add to the 2nd:
Now the original system can be written as:
x 1 = 1 - (1/2 x 2 + 1/2 x 3)
x 2 = 13 - (6x 3)
From the 2nd line we express
From the 1st line we express
The solution is the same.
Answer: (2; -5; 3)
Find the general solution of the system and the FSR
13x 1 – 4x 2 – x 3 - 4x 4 - 6x 5 = 0
11x 1 – 2x 2 + x 3 - 2x 4 - 3x 5 = 0
5x 1 + 4x 2 + 7x 3 + 4x 4 + 6x 5 = 0
7x 1 + 2x 2 + 5x 3 + 2x 4 + 3x 5 = 0
Solution: Let's apply the Gaussian method. Let us reduce the extended matrix of the system to triangular form.
| -4 | -1 | -4 | -6 | |
| -2 | -2 | -3 | ||
| x 1 | x 2 | x 3 | x 4 | x 5 |
Multiply the 1st line by (-11). Multiply the 2nd line by (13). Let's add the 2nd line to the 1st:
| -2 | -2 | -3 | ||
Multiply the 2nd line by (-5). Let's multiply the 3rd line by (11). Let's add the 3rd line to the 2nd:
Multiply the 3rd line by (-7). Let's multiply the 4th line by (5). Let's add the 4th line to the 3rd:
The second equation is a linear combination of the others
Let's find the rank of the matrix.
| -18 | -24 | -18 | -27 | |
| x 1 | x 2 | x 3 | x 4 | x 5 |
The selected minor has the highest order (of possible minors) and is non-zero (it is equal to the product of the elements on the reverse diagonal), therefore rang(A) = 2.
This minor is basic. It includes coefficients for the unknowns x 1 , x 2 , which means that the unknowns x 1 , x 2 are dependent (basic), and x 3 , x 4 , x 5 are free.
The system with the coefficients of this matrix is equivalent to the original system and has the form:
18x 2 = 24x 3 + 18x 4 + 27x 5
7x 1 + 2x 2 = - 5x 3 - 2x 4 - 3x 5
Using the method of eliminating unknowns, we find common decision:
x 2 = - 4 / 3 x 3 - x 4 - 3 / 2 x 5
x 1 = - 1 / 3 x 3
We find a fundamental system of solutions (FSD), which consists of (n-r) solutions. In our case, n=5, r=2, therefore, the fundamental system of solutions consists of 3 solutions, and these solutions must be linearly independent.
For the rows to be linearly independent, it is necessary and sufficient that the rank of the matrix composed of row elements be equal to the number of rows, that is, 3.
It is enough to give the free unknowns x 3 , x 4 , x 5 values from the lines of the 3rd order determinant, non-zero, and calculate x 1 , x 2 .
The simplest non-zero determinant is the identity matrix.
But it’s more convenient to take here
We find using the general solution:
a) x 3 = 6, x 4 = 0, x 5 = 0 Þ x 1 = - 1 / 3 x 3 = -2, x 2 = - 4 / 3 x 3 - x 4 - 3 / 2 x 5 = - 4 Þ
I decision of the FSR: (-2; -4; 6; 0;0)
b) x 3 = 0, x 4 = 6, x 5 = 0 Þ x 1 = - 1 / 3 x 3 = 0, x 2 = - 4 / 3 x 3 - x 4 - 3 / 2 x 5 = - 6 Þ
II FSR solution: (0; -6; 0; 6;0)
c) x 3 = 0, x 4 = 0, x 5 = 6 Þ x 1 = - 1/3 x 3 = 0, x 2 = - 4/3 x 3 - x 4 - 3/2 x 5 = -9 Þ
III decision of the FSR: (0; - 9; 0; 0;6)
Þ FSR: (-2; -4; 6; 0;0), (0; -6; 0; 6;0), (0; - 9; 0; 0;6)
6. Given: z 1 = -4 + 5i, z 2 = 2 – 4i. Find: a) z 1 – 2z 2 b) z 1 z 2 c) z 1 /z 2
Solution: a) z 1 – 2z 2 = -4+5i+2(2-4i) = -4+5i+4-8i = -3i
b) z 1 z 2 = (-4+5i)(2-4i) = -8+10i+16i-20i 2 = (i 2 = -1) = 12 + 26i
Answer: a) -3i b) 12+26i c) -1.4 – 0.3i
We will continue to polish our technology elementary transformations on homogeneous system of linear equations.
Based on the first paragraphs, the material may seem boring and mediocre, but this impression is deceptive. In addition to further development of techniques, there will be a lot of new information, so please try not to neglect the examples in this article.
What is a homogeneous system of linear equations?
The answer suggests itself. A system of linear equations is homogeneous if the free term everyone equation of the system is zero. For example:
It is absolutely clear that a homogeneous system is always consistent, that is, it always has a solution. And, first of all, what catches your eye is the so-called trivial solution 
. Trivial, for those who do not understand the meaning of the adjective at all, means without a show-off. Not academically, of course, but intelligibly =) ...Why beat around the bush, let's find out if this system has any other solutions:
Example 1
Solution: to solve a homogeneous system it is necessary to write system matrix and with the help of elementary transformations bring it to a stepwise form. Please note that here there is no need to write down the vertical bar and the zero column of free terms - after all, no matter what you do with zeros, they will remain zeros: 
(1) The first line was added to the second line, multiplied by –2. The first line was added to the third line, multiplied by –3.
(2) The second line was added to the third line, multiplied by –1.
Dividing the third line by 3 doesn't make much sense.
As a result of elementary transformations, an equivalent homogeneous system is obtained 
, and, using the inverse of the Gaussian method, it is easy to verify that the solution is unique.
Answer: 
Let us formulate an obvious criterion: a homogeneous system of linear equations has just a trivial solution, If system matrix rank(V in this case 3) equal to the number of variables (in this case – 3 pieces).
Let's warm up and tune our radio to the wave of elementary transformations:
Example 2
Solve a homogeneous system of linear equations 
To finally consolidate the algorithm, let’s analyze the final task:
Example 7
Solve a homogeneous system, write the answer in vector form. 
Solution: let’s write down the matrix of the system and, using elementary transformations, bring it to a stepwise form: 
(1) The sign of the first line has been changed. Once again I draw attention to a technique that has been encountered many times, which allows you to significantly simplify the next action.
(1) The first line was added to the 2nd and 3rd lines. The first line, multiplied by 2, was added to the 4th line.
(3) The last three lines are proportional, two of them have been removed.
As a result, a standard step matrix is obtained, and the solution continues along the knurled track:
– basic variables;
– free variables.
Let us express the basic variables in terms of free variables. From the 2nd equation:
– substitute into the 1st equation:
So the general solution is:
Since in the example under consideration there are three free variables, the fundamental system contains three vectors.
Let's substitute a triple of values 
into the general solution and obtain a vector whose coordinates satisfy each equation of the homogeneous system. And again, I repeat that it is highly advisable to check each received vector - it will not take much time, but it will completely protect you from errors.
For a triple of values 
find the vector
And finally for the three 
we get the third vector:
Answer: , Where
Those wishing to avoid fractional values may consider triplets 
and get an answer in equivalent form:
Speaking of fractions. Let's look at the matrix obtained in the problem 
and let us ask ourselves: is it possible to simplify the further solution? After all, here we first expressed the basic variable through fractions, then through fractions the basic variable, and, I must say, this process was not the simplest and not the most pleasant.
Second solution:
The idea is to try choose other basis variables. Let's look at the matrix and notice two ones in the third column. So why not have a zero at the top? Let's carry out one more elementary transformation:
Homogeneous system of linear equations over a field
DEFINITION. The fundamental system of solutions to the system of equations (1) is called a non-empty linear independent system its solutions, the linear span of which coincides with the set of all solutions of system (1).
Note that a homogeneous system of linear equations that has only a zero solution does not have a fundamental system of solutions.
PROPOSAL 3.11. Any two fundamental systems of solutions to a homogeneous system of linear equations consist of the same number decisions.
Proof. In fact, any two fundamental systems of solutions to the homogeneous system of equations (1) are equivalent and linearly independent. Therefore, by Proposition 1.12, their ranks are equal. Consequently, the number of solutions included in one fundamental system is equal to the number of solutions included in any other fundamental system of solutions.
If the main matrix A of the homogeneous system of equations (1) is zero, then any vector from is a solution to system (1); in this case, any set of linearly independent vectors from is a fundamental system of solutions. If the column rank of matrix A is equal to , then system (1) has only one solution - zero; therefore, in this case, the system of equations (1) does not have a fundamental system of solutions.
THEOREM 3.12. If the rank of the main matrix of a homogeneous system of linear equations (1) is less than the number of variables , then system (1) has a fundamental solution system consisting of solutions.
Proof. If the rank of the main matrix A of the homogeneous system (1) is equal to zero or , then it was shown above that the theorem is true. Therefore, below it is assumed that Assuming , we will assume that the first columns of matrix A are linearly independent. In this case, matrix A is rowwise equivalent to a reduced stepwise matrix, and system (1) is equivalent to the following reduced stepwise system of equations:
It is easy to check that any system of values of free variables of system (2) corresponds to one and only one solution to system (2) and, therefore, to system (1). In particular, only the zero solution of system (2) and system (1) corresponds to a system of zero values.
In system (2) we will assign one of the free variables a value equal to 1, and the remaining variables - zero values. As a result, we obtain solutions to the system of equations (2), which we write in the form of rows of the following matrix C:
The row system of this matrix is linearly independent. Indeed, for any scalars from the equality
equality follows
and, therefore, equality
Let us prove that the linear span of the system of rows of the matrix C coincides with the set of all solutions to system (1).
Arbitrary solution of system (1). Then the vector
is also a solution to system (1), and
You can order detailed solution your task!!!To understand what it is fundamental decision system you can watch a video tutorial for the same example by clicking. Now let's move on to the description of the whole necessary work. This will help you understand the essence of this issue in more detail.
How to find the fundamental system of solutions to a linear equation?
Let's take for example the following system of linear equations:
Let's find the solution to this linear system of equations. To begin with, we you need to write out the coefficient matrix of the system.
Let's transform this matrix to a triangular one. We rewrite the first line without changes. And all the elements that are under $a_(11)$ must be made zeros. To make a zero in place of the element $a_(21)$, you need to subtract the first from the second line, and write the difference in the second line. To make a zero in place of the element $a_(31)$, you need to subtract the first from the third line and write the difference in the third line. To make a zero in place of the element $a_(41)$, you need to subtract the first multiplied by 2 from the fourth line and write the difference in the fourth line. To make a zero in place of the element $a_(31)$, you need to subtract the first multiplied by 2 from the fifth line and write the difference in the fifth line. 
We rewrite the first and second lines without changes. And all the elements that are under $a_(22)$ must be made zeros. To make a zero in place of the element $a_(32)$, you need to subtract the second one multiplied by 2 from the third line and write the difference in the third line. To make a zero in place of the element $a_(42)$, you need to subtract the second multiplied by 2 from the fourth line and write the difference in the fourth line. To make a zero in place of the element $a_(52)$, you need to subtract the second multiplied by 3 from the fifth line and write the difference in the fifth line. 
We see that the last three lines are the same, so if you subtract the third from the fourth and fifth, they will become zero. 
According to this matrix write down new system equations.
We see that we have only three linearly independent equations, and five unknowns, so the fundamental system of solutions will consist of two vectors. So we we need to move the last two unknowns to the right.
Now, we begin to express those unknowns that are on the left side through those that are on the right side. We start with the last equation, first we express $x_3$, then we substitute the resulting result into the second equation and express $x_2$, and then into the first equation and here we express $x_1$. Thus, we expressed all the unknowns that are on the left side through the unknowns that are on the right side. 
Then, instead of $x_4$ and $x_5$, we can substitute any numbers and find $x_1$, $x_2$ and $x_3$. Each five of these numbers will be the roots of our original system of equations. To find the vectors that are included in FSR we need to substitute 1 instead of $x_4$, and substitute 0 instead of $x_5$, find $x_1$, $x_2$ and $x_3$, and then vice versa $x_4=0$ and $x_5=1$.