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Math 21 notes1

# Math 21 notes1 - Chapter 1 Linear Equations 1.1...

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Chapter 1 Linear Equations 1.1 Introduction A linear equation (over the real numbers R ) in the variables x 1 , x 2 , · · · , x n is a mathematical equation of the form (1.1) a 1 x 1 + a 2 x 2 + · · · + a n x n = c, with c and the coefficients a 1 , · · · , a n some arbitrary real numbers. If c = 0 then the equation is homogeneous , otherwise it is inhomogeneous . A system of linear equations is a set consisting of a finite number of linear equations, and a solution to such a system is any set of specific values for the x i which make each equation a true statement. When we have found all possible solutions we say, appropriately enough, that the system has been solved . You may recall that either substitution or elimination can be used to solve systems of linear equations: Example 1 Solve the following system of equations: 2 x + 3 y = - 6 5 x + 2 y = 7 . We use elimination, say by multiplying the first equation by 5, the second by - 2, 5(2 x + 3 y = - 6) 10 x + 15 y = - 30 - 2(5 x + 2 y = 7) → - 10 x - 4 y = - 14 , and then adding the transformed equations. This yields the single equation 11 y = - 44, so y = - 44 11 = - 4, and substituting this value for y back into either one of the original equations, e.g. 2 x + 3 y = - 6, gives us an equation in x , e.g. 2 x + 3( - 4) = - 6. Thus x = - 6+12 2 = 3, and we have found a unique solution ( x, y ) = (3 , - 4) to the system of equations. Some systems have more than one solution: 1

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Example 2 Find all solutions to the system 2 x - y = 5 8 x - 4 y = 20 . If we multiply the equation 2 x - y = 5 by 4 and subtract the second equation, we obtain the equation 0 = 0, in other words each equation has the same set of solutions, so our elimination trick (which yields the intersection of the two solution sets) gives us no new information. Using, e.g, the first equation, we may define the solution set as { ( x, 2 x - 5) | x R } . And finally, the intersection of the solution sets may well be the empty set : Example 3 Solve the system 2 x - 3 y = 1 4 x - 6 y = 6 . If we multiply the first equation by - 2 and add the resulting equations, we obtain the ridiculous statement 0 = 4, i.e. assuming there is a pair ( x, y ) common to each solution set leads to a logical contradiction. Thus the system has no solution. Classically, some of the earliest motivation for developing the theory of abstract linear algebra came from calculus, specifically the study of differential equations. Example 4 Find all real polynomials f ( t ) of degree less than 3 such that f (1) = 1 , f (3) = 3 , and f (2) = 3 . (This is 1.1, #34 in your text.) The generic polynomial of degree 2 has the form f ( t ) = a 2 t 2 + a 1 t + a 0 , for arbitrary real numbers a 0 , a 1 , a 2 . Thinking of these as variables, the given information yields three linear equations: f (1) = a 2 + a 1 + a 0 = 1 f (3) = 9 a 2 + 3 a 1 + a 0 = 3 f (2) = 2 a 2 · 2 + a 1 = 4 a 2 + a 1 = 3 .
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