Chapter 1.2

1.2 Notes (Class 3 & Textbook)


Gaussian elimination & Gauss–Jordan elimination example:

Reduced echelon form: leading variable need to be one

Class 4

Example 1: Gauss-elimination:

$ \begin{cases} 3x_1-3x_2+9x_3=24\\
2x_1-2x_2+7x_3=17\\
-x_1+2x_2-4x_3=-11 \end{cases} $

Check: plug $x_1=3, x_2=-2, x_3=1$ into left sides of orginal equations.

Example 2

$\begin{cases} x_1+x_2=2\\
2x_1+3x_2=3\\
x_1+3x_2=0\\
3x_1+6x_2=4 \end{cases} $

No solution: inconsistent.

Example 3: Hamogeneous

Hamogeneous: always consistent, all x = 0 is a solution.

$x_1$ and $x_2$ are local variables.

$x_3$ and $x_4$ are free variables.

Let $x_3 = S_1 $ and $x_4 = S_2$ as free parameters.

Solution:


Theorem: A system of equations either has no solutions, exactly one solution, or infinitely many solutions.

Given any system of equations, form the augmented matrix and perform Gauss-Jordan elimination: getting unique equivalent matrix in reduced echelon form.

Case:

1. Row with zeros ending with constant c != 0 --- No solution (See Textbook Notes above)
	0 0 ... 0 | c 	
2. No such rows
	a) There is no free variable, then there is a unique solution.
	b) There is one or more free variables, then there are solution sets (infinitly many solutions).