SYSTEMS OF LINEAR
EQUATIONS
A system of n linear equations
in n unknowns has (hopefully) one solution.
Example :
5x + 3y - z = 8
x - y + 2z = 5
2x - 3y + 4z = 8
METHODS TO FIND THE
SOLUTION
TRANSFORMS PRESERVING SOLUTION
:
- Exchange two rows
- Multiply a row by a constant
- Add a multiple of one row to
another
Transform rows until each has only 1 variable.
This means "diagonal"
elements 1, all other numbers except right column are 0.
TRIANGULARIZATION :
Transform equations until lower
triangle is all zeroes, then apply "back-substitution".
have,
5x + 3y - z = 8
x - y + 2z = 5
2x - 3y + 4z = 8
This is represented with an AUGMENTED MATRIX:| 5 3 -1 8 | | 1 -1 2 5 | | 2 -3 4 8 |The augmented matrix is the coefficients of the linear equations, including the right hand sides, written as a matrix with n rows and n+1 columns.
R1=R1/5 | 1 0.6 -0.2 1.6 |
| 1 -1 2 5 |
| 2 -3 4 8 |
| 1 0.6 -0.2 1.6 |
R2=R2-R1 | 0 -1.6 2.2 3.4 |
R3=R3-2R1 | 0 -4.2 4.4 4.8 |
| 1 0.6 -0.2 1.6 |
R2=R2/-1.6 | 0 1 -1.375 -2.125 |
| 0 -4.2 4.4 4.8 |
| 1 0.6 -0.2 1.6 |
| 0 1 -1.375 -2.125 |
R3=R3+4.2R2 | 0 0 -1.375 -4.125 |
| 1 0.6 -0.2 1.6 |
| 0 1 -1.375 -2.125 |
R3=R3/-1.375| 0 0 1 3 |
It is obvious from here that z=3.
Substituting this value into R2 gives y=2. And finally, substituting
these values of y and z into R1 gives x=1.
The final solution is x=1, y=2, z=3.
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