A system of n* linear* equations
in n unknowns has (hopefully) one solution.

**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".

**EXAMPLE**

have,

5x + 3y - z = 8 x - y + 2z = 5 2x - 3y + 4z = 8This is represented with an

| 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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