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Tutorial Math 204 Tutorial 1 System of linear equations Gauss / Gauss - Jordan reduction by
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# Tutorial Math 204 Tutorial 1 System of linear equations Gauss / Gauss - Jordan reduction

Ph. D Candiate: Jesús Rogelio Pérez Buendía.   If you want to send me an email please use Moodle email service. You can Find me on The Math Help Centre LB 912. You will hae acces to this matherial on Moodle Meta Site. You will have a 15 min Quiz based on one (similar) exercise on this asignment  and/or Tutorial problems for next week.

## Problem 1

### Gauss/Gauss Jordan Reduction

There are three types of elementary row operations which may be performed on the rows of a matrix: Type 1: Swap the positions of two rows. Type 2: Multiply a row by a nonzero scalar. Type 3: Add to one row a scalar multiple of another.

### The system with zeros

The Augmented Matrix is:, Now we use the algorithm, Second operation, 3rd operation, 4th operation, 5th operation, 6th operation, 7th operation, 8th operation, 9th operation, 10th operation

### Reduce echalon form

Specifically, a matrix is in row echelon form if All nonzero rows (rows with at least one nonzero element) are above any rows of all zeroes (all zero rows, if any, belong at the bottom of the matrix). The leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it. All entries in a column below a leading entry are zeroes (implied by the first two criteria).[1] Some texts add the condition that the leading coefficient of any nonzero row must be 1.[2]

What is it?

So…? No equation of this system has a form zero = nonzero; Therefore, the system is consistent. The leading entries in the matrix have been highlighted in yellow. A leading entry on the (i,j) position indicates that the j-th unknown will be determined using the i-th equation.

Then… Those columns in the coefficient part of the matrix that do not contain leading entries, correspond to unknowns that will be arbitrary. The system has infinitely many solutions:

### Is this the best way to solve it?

For a computer YES (almost)

For Humans NO… Let's try to solve it easier., Go to this site: