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Theorem on the existence of a system of linear equations solution

(3.1):

 a 11 x + a 12 x + ... + a 1n x = b 1 ,
′
′
′
 1 2 n
a 21 x + a 22 x + ... + a 2n x = b 2 ,

′
′
′
1 2 n (3.2)
 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .,

 ′ ′ ′
a m1 x + a m2 x + ... + a mn x = b m
1 2 n
and perform under an extended matrix A the following elementary transformations: the last column
of this matrix to be added to the first one multiplied into −x ; then the second one, multiplied into
′
1
− →
′
′
−x etc; the column n, multiplied into −x . Then, using (3.2) we will get an equivalent matrix A =
′
n
2
 
0
a 11 a 12 . . . a 1n
 0 
a 21 a 22 . . . a 2n
  .
. . . . . . . . . . . . . . .
 
0
a m1 a m2 . . . a mn
It is known that elementary transformations don’t change a rank of a matrix. Therefore r(A) =
′ ′
r(A ). Because of the last column consisting with zero-members: r(A ) = r(A). So, r(A) = r(A).
Sufficiency. Supposing that a non-zero n-order determinant is situated in the left upper corner:

a 11 a 12 . . . a 1n

a 21 a 22 . . . a 2n
̸= 0
. . . . . . . . . . . .

a n1 a n2 . . . a nn
and r(A) = r(A) = r. Then the first r rows of a matrix A are linear independent. Moreover,
because of a rank of this matrix equals to zero, the rest rows of matrix A can be written with the help
of the first ones. It means that the first r equations of system (3.1) are linear independent, and the
rest of (m − r) equations are their linear combinations, thus are their consequences. So, a given
system, indeed, contains only r linear independent equations. That’s why it is sufficient to solve these
equations. The solution found is automatically a solution for the rest of (m − r) equations.
Let’s analyse two following cases:
1. r = n. Then the system

 a 11 x 1 + a 12 x 2 + ... + a 1n x r = b 1 ,


a 21 x 1 + a 22 x 2 + ... + a 2n x r = b 2 ,
 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .,


a r1 x 1 + a r2 x 2 + ... + a rr x r = b r
canbesolvedbyCramer’sformulas( forexample). Inthiscasethesystemhastheonlysolution.
The system is compatible and determined.
2. r < n. Let’s consider the first r equations of a given system and, having left in their left sides
the first r unknowns, the rest ones are to be taken to the right sides:

 a 11 x 1 + a 12 x 2 + ... + a 1n x r = b 1 − a 1r+1 x r+1 − ... − a 1n x n ,


a 21 x 1 + a 22 x 2 + ... + a 2n x r = b 2 − a 2r+1 x r+1 − ... − a 2n x n ,
(3.3)
 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .,


a m1 x 1 + a m2 x 2 + ... + a mn x r = b m − a mr+1 x r+1 − ... − a mn x n .
Unknowns x r+1 , x r+2 , . . . , x n that are called “free unknowns”, can take any real values. At the
same time, unknowns x 1 , x 2 , . . . , x r will take some values, dependent of free unknowns. In this case
the given system is compatible, but nondetermined and so, posseses many solutions.
Note, if r(A) ̸= r(A), then the system is noncompatible. 2

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