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02-03-03

# 02-03-03 - Lecture 3 Andrei Antonenko 1 Linear systems and...

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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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02-03-03 - Lecture 3 Andrei Antonenko 1 Linear systems and...

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