02-03-03 - Lecture 3 Andrei Antonenko February 03, 2003 1...

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Lecture 3 Andrei Antonenko February 03, 2003 1 Linear systems and their solutions This lecture we’re going to speak about the most important and boring part of linear algebra — about general linear systems — we will learn how to solve and analyze them. Definition 1.1. Linear system is a bunch of linear equations considered together: a 11 x 1 + a 12 x 2 + ··· + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + ··· + a 2 n x n = b 2 ... a m 1 x 1 + a m 2 x 2 + ··· + a mn x n = b m (1) Here we have again 3 types of letters: x i ’s are variables which have to be determined, letter a ij represents the coefficient in the i -th equation before the variable x j , and letter b i is the constant term in the i -th equation. Both a ij ’s and b i ’s are given numbers. The solution of the system is an n -tuple of numbers such that it is a solution of each of the system’s equation. To solve the system means to find a set of all its solutions, i.e. solution set. Definition 1.2. The systems are called equivalent if they have the same solution sets. In other words, two systems are equivalent if any solution of the first system is a solution for the second system and any solution of the second system is a solution for the first system. 2 Elementary operations: three ways to get equivalent systems The following three operations can be applied to any linear system to get an equivalent system. They are called elementary operations. 1
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1. Interchanging of the equations. If we interchange two equations, i.e. rewrite them in different order we’ll get the equivalent system — it’s obvious. Example 2.1. Let’s consider 2 systems: ( x 1 + 2 x 2 = 4 2 x 1 - x 2 = 3 and ( 2 x 1 - x 2 = 3 x 1 + 2 x 2 = 4 The second system is got from the first by interchanging its equations. It is obvious that they have same solutions — x 1 = 2 and x 2 = 1 . 2. Multiplication. We can multiply any equation by any number not equal to zero. Example 2.2. Let’s consider 2 systems: ( x 1 + 2 x 2 = 4 2 x 1 - x 2 = 3 and ( 2 x 1 + 4 x 2 = 8 2 x 1 - x 2 = 3 The second system is obtained from the first by multiplying the first equation by 2. It is obvious that they have same solutions — x 1 = 2 and x 2 = 1 . 3.
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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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02-03-03 - Lecture 3 Andrei Antonenko February 03, 2003 1...

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