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# cha1 practice - Chapter 1 LINEAR EQUATIONS 1.1 Introduction...

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Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1 , x 2 , ··· , x n is an equation of the form a 1 x 1 + a 2 x 2 + + a n x n = b, where a 1 , a 2 ,...,a n , b are given real numbers. For example, with x and y instead of x 1 and x 2 , the linear equation 2 x + 3 y = 6 describes the line passing through the points (3 , 0) and (0 , 2). Similarly, with x, y and z instead of x 1 , x 2 and x 3 , the linear equa- tion 2 x + 3 y + 4 z = 12 describes the plane passing through the points (6 , 0 , 0) , (0 , 4 , 0) , (0 , 0 , 3). A system of m linear equations in n unknowns x 1 , x 2 , , x n is a family of linear equations 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 . We wish to determine if such a system has a solution, that is to ﬁnd out if there exist numbers x 1 , x 2 , , x n which satisfy each of the equations simultaneously. We say that the system is consistent if it has a solution. Otherwise the system is called inconsistent . 1

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2 CHAPTER 1. LINEAR EQUATIONS Note that the above system can be written concisely as n X j =1 a ij x j = b i , i = 1 , 2 , ··· ,m. The matrix a 11 a 12 a 1 n a 21 a 22 a 2 n . . . . . . a m 1 a m 2 a mn is called the coefcient matrix of the system, while the matrix a 11 a 12 a 1 n b 1 a 21 a 22 a 2 n b 2 . . . . . . . . . a m 1 a m 2 a mn b m is called the augmented matrix of the system. Geometrically, solving a system of linear equations in two (or three) unknowns is equivalent to determining whether or not a family of lines (or planes) has a common point of intersection. EXAMPLE 1.1.1 Solve the equation 2 x + 3 y = 6 . Solution . The equation 2 x + 3 y = 6 is equivalent to 2 x = 6 - 3 y or x = 3 - 3 2 y , where y is arbitrary. So there are inﬁnitely many solutions. EXAMPLE 1.1.2 Solve the system x + y + z = 1 x - y + z = 0 . Solution . We subtract the second equation from the ﬁrst, to get 2 y = 1 and y = 1 2 . Then x = y - z = 1 2 - z , where z is arbitrary. Again there are inﬁnitely many solutions. EXAMPLE 1.1.3 Find a polynomial of the form y = a 0 + a 1 x + a 2 x 2 + a 3 x 3 which passes through the points ( - 3 , - 2) , ( - 1 , 2) , (1 , 5) , (2 , 1).
1.1. INTRODUCTION TO LINEAR EQUATIONS 3 Solution . When x has the values - 3 , - 1 , 1 , 2, then y takes corresponding values - 2 , 2 , 5 , 1 and we get four equations in the unknowns a 0 , a 1 , a 2 , a 3 : a 0 - 3 a 1 + 9 a 2 - 27 a 3 = - 2 a 0 - a 1 + a 2 - a 3 = 2 a 0 + a 1 + a 2 + a 3 = 5 a 0 + 2 a 1 + 4 a 2 + 8 a 3 = 1 . This system has the unique solution a 0 = 93 / 20 , a 1 = 221 / 120 , a 2 = - 23 / 20 , a 3 = - 41 / 120. So the required polynomial is y = 93 20 + 221 120 x - 23 20 x 2 - 41 120 x 3 . In [26, pages 33–35] there are examples of systems of linear equations which arise from simple electrical networks using KirchhoF’s laws for elec- trical circuits. Solving a system consisting of a single linear equation is easy. However if we are dealing with two or more equations, it is desirable to have a systematic method of determining if the system is consistent and to ﬁnd all solutions.

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## This test prep was uploaded on 01/29/2008 for the course MATH 3355 taught by Professor Britt during the Spring '08 term at LSU.

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cha1 practice - Chapter 1 LINEAR EQUATIONS 1.1 Introduction...

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