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General Research Into the System of Linear Algebraic Equations

3.3. Linear homogeneous systems

Linear homogeneous equations are the equations, free terms of which are equal to zero:

 a 11 x 1 + a 12 x 2 + ... + a 1n x n = 0,


a 21 x 1 + a 22 x 2 + ... + a 2n x n = 0,
(3.4)
 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .,


a m1 x 1 + a m2 x 2 + ... + a mn x n = 0.
The system (3.4) is always compatible, because it has a zero solution: x 1 = 0; x 2 = 0; . . . ;
x n = 0. We are going to find out whether this solution is the only one or system (3.4) have other
(non-zero) solutions.

Theorem 3.2.
In order for system (3.4) to have non-zero solutions, it is necessary and sufficient that a rank of
its matrix is less than n (a number of unknowns). ⋆

Indeed, supposing r = n, from Kronecker-Capelli’s theorem we can make a conclusion that
system (3.4) has the only solution — a zero one.
Supposing r < n, system (3.4) is undetermined (because it can’t be noncompatible), and so,
posseses many solutions.

Theorem 3.3.
In order for a homogeneous system with n linear equations and n unknowns to have non-zero

solutions, it is necessary and sufficient that its determinant equals to zero. ⋆

PROOF. An equality ∆ = 0 is necessary, because, if ∆ ̸= 0, then the given system has the
only non-zero solution. This condition is sufficient either, because, if ∆ = 0, then a rank of a matrix
r < n, and so, a system have many (non-zero) solutions. 2

3.4. Gauss’ method of successive exception of unknowns

In practice, most often Gauss’ method is used to solve the system of linear algebraic equations,
which consists of successive exception of unknowns. To solve the system of linear equations

 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 ,
(3.5)
 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .,

a m1 x 1 + a m2 x 2 + ... + a mn x n = b m ,

the main and extended matrix of this system are written down in one matrix with the help of a
vertical line, which separates the column of free members:
 
a 11 a 12 . . . a 1n b 1
 a 21 a 22 . . . a 2n b 2 
A/A =   .
 . . . . . . . . . . . . . . . 
a m1 a m2 . . . a mn b m
Then basic transformations with rows of this matrix are made in order to get a matrix of
triangle or trapezium shape. If the matrix obtained, has a triangle shape, then the given system

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