02-05-03 - Lecture 4 Andrei Antonenko February 05, 2003 1...

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Lecture 4 Andrei Antonenko February 05, 2003 1 Analysis of the system Now, we can formulate the main theoretical result about the system. This theoretical result follows directly from our practical method of solving them. Case 1 If during our procedures we got an equation of the form 0 = b , where b 6 = 0, then we should stop and deduce that the system has no solution. Case 2 If in the row echelon form we have no equation of the form 0 = b , where b 6 = 0, and we have free variables, than the system has infinitely many solutions, since we can assign to them any (real) values. Case 3 If in the row echelon form we have no equation of the form 0 = b , where b 6 = 0, and we don’t have free variables, than the solution can be determined uniquely, so the corresponding system has only 1 solution. Theorem 1.1. The system of linear equations can have 0, 1 or infinitely many solutions. Last lecture we provided an example of a linear system which had free variables in its REF. Now we will provide an example of a system without free variables, so it would have a unique numerical solution. Example 1.2. Let’s reduce the following system to the row echelon form and then solve it. x 1 + x 2 + x 3 = 4 x 1 + 2 x 2 - x 3 = 1 2 x 1 - x 2 + x 3 = 3 (1) First variable with nonzero coefficient is x 1 , and the first equation has a nonzero coefficient before it too. So, we don’t have to interchange the equations. Now, we’ll subtract the first 1
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equation from the second one, then multiply it by 2 and subtract from the third one. We’ll get the following system (we should not omit the first equation!!!): x 1 + x 2 + x 3 = 4 x 2 - 2 x 3 = - 3 - 3 x 2 - x 3 = - 5 (2) Now, we’ll apply the same steps to equations from the 2nd to the 3rd. So, we multiply the second equation by 3 and add it to the third one (again, do not omit any equations!!!): x 1 + x 2 + x 3 = 4 x 2 - 2 x 3 = - 3 - 7 x 3 = - 14 (3) Now we have a system in a row echelon form. Now it possible to solve it using the back substitu- tion process. From the last equation we can determine the value for
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This note was uploaded on 05/28/2011 for the course AMS 2010 taught by Professor Andant during the Spring '03 term at SUNY Stony Brook.

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02-05-03 - Lecture 4 Andrei Antonenko February 05, 2003 1...

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