/ x+2y+3z=5 | x-y+6z=1 | 3x-2y=4 \ y+4z=8 / x+y+3z-4t=12 \ 3x+y-2z-t=0 / 2x+3y+4z=5 | x-y+2z=6 \ 3x-5y-z=0Each of these systems can be expressed by the help of matrices.
[ 1 2 3] [x] [5]
[ 1 -1 6] . [y] = [1]
[ 3 -2 0] [z] [4]
[ 0 1 4] [8]
[x]
[ 1 1 3 -4] [y] [12]
[ 3 1 -2 -1] . [z] = [ 0]
[t]
[ 2 3 4] [x] [5]
[ 1 -1 2] . [y] = [6]
[ 3 -5 -1] [z] [0]
The first matrix of each representation of a linear system contains the coefficients appearing in the system. This matrix is called the coefficient matrix.[ 1 2 3 5] [ 1 -1 6 1] [ 3 -2 0 4] [ 0 1 4 8]This matrix is called the enlarged matrix of the system.
Action 1 :
Multiply one of the equations with a real number different from 0.
This is equivalent with :
Multiply one of the rows, of the enlarged matrix of the system, with a real number different from 0.
Action 2 :
Multiply one of the equations with a real number, and add the result to another equation, leaving the original equation unchanged.
This is equivalent with :
Multiply one of the rows, of the enlarged matrix of the system, with a real number, and add the result to another row, leaving the original row unchanged.
Action 3 :
Interchange two equations .
This is equivalent with :
Interchange two rows of the enlarged matrix of the system.
Action 4 :
Delete an equation that is equivalent with 0 = 0
This is equivalent with :
Delete a row, of the enlarged matrix of the system, that contains only zero's.
Action 5 :
If an equation of a system is equivalent with 0=k, with k not zero, then the system has no solutions.
This is equivalent with :
If a row, of the enlarged matrix of the system, is composed of zero's, except for the last number, then the system has no solutions.
Each of the actions transforming a system to an equivalent system, is equivalent to a transformation of the rows of the enlarged matrix of the system. These are the so called 'row transformations'.
Now, it can be shown that, by a appropriate consequence of only these 5 actions, we can solve any system of linear equations. The method, to do this in a efficient way, is called the method of Gauss.
/ 2x-3y+2z=21 | x+4y-z=1 \ -x+2y+z=17or equivalently
[2 -3 2 21] [1 4 -1 1] [-1 2 1 17]First I'll show the row transformations, then I'll state the method in words.
R1 <-> R2
[1 4 -1 1]
[2 -3 2 21]
[-1 2 1 17]
R2 + 2.R3
[1 4 -1 1]
[0 1 4 55]
[-1 2 1 17]
R3 + R1
[1 4 -1 1]
[0 1 4 55]
[0 6 0 18]
(1/6).R3
[1 4 -1 1]
[0 1 4 55]
[0 1 0 3]
R3-R2
[1 4 -1 1]
[0 1 4 55]
[0 0 -4 -52]
(-1/4).R3
[1 4 -1 1]
[0 1 4 55]
[0 0 1 13]
Now we shall return to the system.
/ x+4y-z=1 | y+4z=55 \ z=13Working from z to x, we find easily z=13 ; y=3 ; x=2.
Now we state the method.
Only use the row transformations described above and work from top-row to bottom-row.
1. Make in each row, the first non-zero element (called the main element) equal to 1
2. Transform, beneath this main element, all the elements to 0
3. If there comes a row of all zero's,delete this row
4. If there comes a row composed of zero's, except for the last number, then the system has no solutions.
Remark:
All these matrices are called row-equivalent.
The resulting matrix is called an 'echelon' matrix.
It is also possible to transform, the matrix such that each main element is the only non zero element on its column.
In that case the resulting matrix is called row-canonical.
/ 3x-4y+5z-4t=12 | x-y +z -2t = 0 | 2x+y+2z+3t=52 | 2x-2y+2z-4t=0 \ 2x-3y+2z-t=4or equivalently
[3 -4 5 -4 12]
[1 -1 1 -2 0]
[2 1 2 3 52]
[2 -2 2 -4 0]
[2 -3 2 -1 4]
R1-R4
[1 -2 3 0 12]
[1 -1 1 -2 0]
[2 1 2 3 52]
[2 -2 2 -4 0]
[2 -3 2 -1 4]
R2-R1;R3-2R1;R4-2R1;R5-2R1
[1 -2 3 0 12]
[0 1 -2 -2 -12]
[0 5 -4 3 28]
[0 2 -4 -4 -24]
[0 1 -4 -1 -20]
R3-5R2;R4-2R2;R5-R2;
[1 -2 3 0 12]
[0 1 -2 -2 -12]
[0 0 6 13 88]
[0 0 0 0 0]
[0 0 -2 1 -8]
delete R4;R3<->R4;
[1 -2 3 0 12]
[0 1 -2 -2 -12]
[0 0 -2 1 -8]
[0 0 6 13 88]
R4+3R3;(1/16)R4;R3/(-2);
[1 -2 3 0 12]
[0 1 -2 -2 -12]
[0 0 1 -1/2 4]
[0 0 0 1 4]
This is again an echelon matrix. Thus return to the system and calculate
the unknowns from bottom to top.
/ x-2y+3z =12 | y -2z -2t =-12 | z-0.5t=4 \ t=4 x=10 ; y=8 ; z=6 ; t=4
/ x+y-4z+10t=24
\ 3x-2y-2z+6t=15
[1 1 -4 10 24]
[3 -2 -2 6 15]
R2-3R1
[1 1 -4 10 24]
[0 -5 10 -24 -57]
r2/(-5)
[1 1 -4 10 24 ]
[0 1 -2 4.8 11.4]
Back to the system
/ x + y -4z +10t =24
\ y -2z +4.8t=11.4
The fundamental difference between proceeding examples is that, in this case, it is impossible to calculate the unknowns.
Therefore, we write the system as follows.
/ x + y = 4z -10t + 24 \ y = 2z -4.8t+ 11.4 <=> / x = 2z - 5.2t + 12.6 \ y = 2z -4.8t+ 11.4For each chosen value of z and of t, we can calculate one solution of the system.
S = {2z - 5.2t + 12.6 ,2z -4.8t + 11.4 , z, t} with z, t in R
/ x-3y=21
| 4x+2y=14
\ 3x+3y=7
[1 -3 21]
[4 2 14]
[3 3 7]
(1/2).R2 ; R3-3.R1
[1 -3 21]
[2 1 7]
[0 12 -56]
R2-2.R1
[1 -3 21]
[0 7 -35]
[0 12 -56]
(1/7)R2 ; (1/4)R3
[1 -3 21]
[0 1 -5]
[0 3 -14]
R3 -3.R2
[1 -3 21]
[0 1 -5]
[0 0 1]
Because of the last row, the system has no solutions.