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Discrete Mathematics Demystified
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Figure 71 Typical (empty) intersection of three lines in the plane
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Matters are more complicated for systems of three equations in three unknowns, or more generally k equations in k unknowns It becomes dif cult to keep track of the information, and the attendant calculations can be daunting In this section we introduce the method of Gaussian elimination, which gives a straightforward and virtually mechanical method for solving systems of linear equations In fact it is straightforward to implement the method of Gaussian elimination on a computer; for a system of k equations in k unknowns it takes about k 3 calculations to nd the solution So this is a robust and ef cient technique We concentrate our efforts on systems of k equations in k unknowns because such a system will generically have a unique solution (that is, a single value for each of the variables that solves the system) When there are fewer unknowns than equations then the system will typically have no solutions1 When there are more unknowns than equations then the system will typically have an entire space of solutions For example, the solution of the system x+z =4 z=3
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1 Think of intersecting lines in the plane Two lines in the plane will usually have a point of intersection as long as they are not parallel Three lines in the plane will usually not have a mutual point of intersection see Fig 71
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are all points of the form (1, y, 3) In other words, the set of solutions forms an entire line It requires ideas from linear algebra to handle these matters properly, and we cannot treat them here So we will focus our attention on k equations in k unknowns In the method of Gaussian elimination we will typically have a system of equations
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1 1 1 a1 x 1 + a2 x 2 + a3 x 3 + + ak 1 x k = 2 2 2 2 a1 x 1 + a2 x 2 + a3 x 3 + + ak 2 x k = 2
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k k k a1 x 1 + a2 x 2 + a3 x 3 + + ak k x k = k
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We will study this system by creating the associated augmented matrix 1 a1 a 2 1 k a1
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1 a2 2 a2 k a2 1 a3 2 a3 k a3
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1 ak 1 2 ak 2 k k ak
Now there are certain allowable operations on the augmented matrix that we use to reduce it to a normalized form The main diagonal of this matrix is the line of k 1 2 terms given by a1 , a2 , , ak We want all the terms below this main diagonal to be equal to 0 And we want the terms on the main diagonal to all be 1s The three allowable operations are: 1 Switch the position of two rows 2 Multiply any row by a nonzero constant 3 Add a multiple of one row to another It turns out that these three simple moves are always adequate to do the job The ideas are best illustrated with some examples EXAMPLE 75 Let us solve the system x 2y + 3z = 4 x + y + z = 2 x + 2y + z = 1
Discrete Mathematics Demystified
using Gaussian elimination We begin by writing the augmented matrix: 1 1 1 2 3 1 1 2 1 4 2 1
Keeping in mind that our aim is to produce all zeros below the main diagonal, we subtract the rst row from the second This yields 1 2 0 3 1 2 3 2 1 4 6 1
This produces a zero below the main diagonal Next we add the rst row to the last The result is 1 0 0 2 3 3 2 0 4 4 6 5
Finally let us multiply the second row by 1/3 and the third row by 1/4 The end result is 1 0 0 2 3 1 2/3 0 1 4 2 5/4
This is the normalized form that we seek Writing our information again as a linear system, we have x 2y + 3z = 4 0x + 1y 2/3z = 2 0x + 0y + z = 5/4 We may immediately read off from the last equation that z = 5/4 Substituting this information into the second equation gives y = 7/6 Lastly, putting both these values into the rst equation yields x = 25/12 Thus the solution of our system is x = 25/12, y = 7/6, z = 5/4 or ( 25/12, 7/6, 5/4)