# The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard.

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The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard A. Medeiros next Lesson 21 Part 2

Higher Dimensional Linear Systems of Equations Higher Dimensional Linear Systems of Equations © 2007 Herbert I. Gross next

© 2007 Herbert I. Gross The technique for solving the 2-dimensional system of linear equations generalizes very nicely to the case of a linear system of equations with any number of unknowns. next More specifically: if we have n linear equations in n unknowns, we may use this method to eliminate the first variable, in every equation after the first; the second variable, in every equation after the second; the third variable, in every equation after the third, etc., until the last equation has only one variable.

© 2007 Herbert I. Gross As an illustration of our add the opposite method, lets show how to find the values of x, y, and z for which… next 2x + 3y + 4z = 8 3x + 4y + 8z = 9 4x + 2y + 9z = 7

next © 2007 Herbert I. Gross What we will do is develop a systematic way to replace… …by an equivalent but an easier-to-solve system. The technique that we will use requires nothing much more than what weve already done. 2x + 3y + 4z = 8 3x + 4y + 8z = 9 4x + 2y + 9z = 7

next © 2007 Herbert I. Gross More specifically, we shall first replace next That is, in the equivalent system the last two equations will involve only y and z. In this way, we will have reduced a system of three equations in three unknowns to an equivalent system with two of the equations having only two unknowns. 2x + 3y + 4z = 8 3x + 4y + 8z = 9 4x + 2y + 9z = 7 by an equivalent system in which the variable, x, appears only in the first equation.

next © 2007 Herbert I. Gross It is a common practice in mathematics to replace (or at least try to replace) an unsolved problem by a sequence of equivalent but previously solved problems. next To accomplish this, notice that the coefficients of x in are 2, 3, and 4; and the least common multiple of these numbers is 12. 2x + 3y + 4z = 8 3x + 4y + 8z = 9 4x + 2y + 9z = 7

next © 2007 Herbert I. Gross Hence, our first step will be to rewrite our system in an equivalent form in which the coefficient of x in the first equation will be 12; but in the second and third equations, it will be - 12. next To this end, we first multiply both sides of the top equation by 6 to obtain the equivalent equation… 2x + 3y + 4z = 8 3x + 4y + 8z = 9 4x + 2y + 9z = 7 6(2x + 3y+ 4z) = 6(8)12x + 18y + 24z = 48 or next 12x + 18y + 24z = 48 next

© 2007 Herbert I. Gross We may then multiply both sides of the middle equation by - 4 to obtain the equivalent equation… next Finally, we may then multiply both sides of the bottom equation by - 3 to obtain the equivalent equation… - 3(4x + 2y+ 9z) = - 3(7) or - 4(3x + 4y+ 8z) = - 4(9) or 12x + 18y + 24z = 48 3x + 4y + 8z = 9 4x + 2y + 9z = 7 next - 12x + - 16y + - 32z = - 36 - 12x + - 6y + - 27z = - 21

next © 2007 Herbert I. Gross Since multiplying both sides of an equation by the same (non-zero) number doesnt change the solution set of the system, what we have shown so far is that the above system is equivalent to our original system… next 2x + 3y + 4z = 8 3x + 4y + 8z = 9 4x + 2y + 9z = 7 - 12x + - 6y + - 27z = - 21 - 12x + - 16y + - 32z = - 36 12x + 18y + 24z = 48

next © 2007 Herbert I. Gross Next: in the above system, using the equals added to equals principle, we replace the second equation by the second plus the first (which gives us a new second equation which has no x term). next - 12x + - 6y + - 27z = - 21 - 12x + - 16y + - 32z = - 36 12x + 18y + 24z = 48 - 12x + - 16y + - 32z = - 36 12x + 18y + 24z = 48 2y + - 8z = 12

next © 2007 Herbert I. Gross Then, we replace the third equation by the third plus the first (for the same reason). next - 12x + - 6y + - 27z = - 21 12x + 18y + 24z = 48 2y + - 8z = 12 - 12x + - 6y + - 27z = - 21 12x + 18y+ 24z = 48 12y + - 3z = 27

next © 2007 Herbert I. Gross 12x + 18y + 24z = 48 2y + - 8z = 12 12y + - 3z = 27 2x + 3y + 4z = 8 next In order to work with smaller numbers in our new system… 2x + 3y + 4z = 8 we may divide both sides of 12x + 18y + 24z = 48 by 6 and replace it by the simpler equation…

next © 2007 Herbert I. Gross next 2x + 3y + 4z = 8 2y + - 8z = 12 12y + - 3z = 27 y + - 4z = 6 4y + - z = 9 next We then divide both sides of 2y + - 8z = 12 by 2 and replace it by the simpler equation… Finally, we divide both sides of 12y + - 3z = 27 by 3 and replace it by the simpler equation…

© 2007 Herbert I. Gross In summary… next is equivalent to… - 12x + - 6y + - 27z = - 21 - 12x + - 16y + - 32z = - 36 12x + 18y+ 24z = 48 2x + 3y + 4z = 8 y + - 4z = 6 4y + - z = 9

next © 2007 Herbert I. Gross The beauty of this equivalent system, is that the last two equations have been reduced to a system of two equations in two unknowns; a system that we already know how to solve. 2x + 3y + 4z = 8 y + - 4z = 6 4y + - z = 9 However, the main purpose of this presentation is to reinforce a more general approach. next

© 2007 Herbert I. Gross Namely, to proceed further, we can rewrite the system above so that y does not appear below the middle equation. In that case, the bottom equation will contain z as its only variable. 2x + 3y + 4z = 8 y + - 4z = 6 4y + - z = 9 We can then solve the bottom equation for z; then replace z by that value in the middle equation to solve for y; and once we know the values for z and y, we can return to the top equation and solve for x. next

© 2007 Herbert I. Gross So lets next multiply the middle equation of our system (y + - 4z = 6) by - 4 to obtain… 2x + 3y + 4z = 8 y + - 4z = 6 4y + - z = 9 and then, replace the bottom equation of our system by the sum of the bottom equation and the middle equation (this eliminates y from the bottom equation). That is… 2x + 3y + 4z = 8 - 4y + 16z = - 24 15z = - 15 next -4-4 + 16z - 24

© 2007 Herbert I. Gross We can then simplify the above system of equations by dividing each term in the middle equation by - 4, and each term in the bottom equation by 15 to obtain… next 2x + 3y + 4z = 8 y + - 4z = 6 z = - 1 2x + 3y + 4z = 8 - 4y + 16z = - 24 15z = - 15

© 2007 Herbert I. Gross The beauty of this system is that we can solve it virtually by inspection. next 2x + 3y + 4z = 8 y + - 4z = 6 z = - 1 Namely, the bottom equation tells us that… We can then replace z by - 1 in the middle equation to see that y + - 4( - 1) = 6 or y + 4 = 6; or y = 2 next

© 2007 Herbert I. Gross Finally, we may replace y by 2 and z by - 1 in the top equation of our system to see that… 2x + 3(2) + 4( - 1) = 8 or next 2x + 3y + 4z = 8 y = 2 z = - 1 x = 3 2x + 6 – 4 = 8 or 2x + 2 = 8 or 2x = 6; hence…

© 2007 Herbert I. Gross Since the systems above are equivalent, it means that they have the same solution set; namely, { (x,y,z) : x = 3, y = 2, z = - 1} = {(3, 2, - 1)} next x = 3 y = 2 z = - 1 As a check notice that… next 2x + 3y + 4z = 83x + 4y + 8z = 9 4x + 2y + 9z = 7 2(3) + 3(2) + 4( - 1) = 8 3(3) + 4(2) + 8( - 1) = 9 4(3) + 2(2) + 9( - 1) = 7

© 2007 Herbert I. Gross There is a tendency to think that x = 3, y = 2, and z = - 1 is nothing more than the answer to a system of linear equations. However, it is, in fact, itself a system of linear equations. next More formally, it is an abbreviation for… x + 0y + 0z = 3 0x + y + 0z = 2 0x + 0y + z = - 1 And for this reason we often refer to it as the trivial form. next Thus, in summary, what we did here was to replace a given system of equations by its equivalent trivial form. Aside

© 2007 Herbert I. Gross As the number of variables increases it becomes more advantageous for us to develop some shorthand notation. So lets begin by describing the new notation. next Some New Notation

© 2007 Herbert I. Gross Rather than use a different letter to denote each variable, we use the same letter but with different subscripts. next For example, instead of using x, y and z we might use x 1, x 2, and x 3. In a similar way if there were four variables, we would denote them by x 1, x 2, x 3 and x 4 ; and if there were n variables, we would denote them by x 1, x 2, x 3,..., and x n

© 2007 Herbert I. Gross Then, in a manner similar to how we use 234 as an abbreviation for 2(100) + 3(10) + 4(1) when we are dealing with place value; we could use 2 3 4 as an abbreviation for 2x 1 + 3x 2 + 4x 3 when we are dealing with linear expressions. next In this way we do not have to keep rewriting the variables in each equation. For example, instead of writing 2x + 3y + 4z = 8, we write 2x 1 + 3x 2 + 4x 3 = 8.

© 2007 Herbert I. Gross next In the more advanced course, Linear Algebra, we introduce the notion of a matrix to extend our shorthand further, and we rewrite… 2x 1 + 3x 2 + 4x 3 = 8 3x 1 + 4x 2 + 8x 3 = 9 4x 1 + 2x 2 + 9x 3 = 7 in the form… 2 3 4 8 3 4 8 9 4 2 9 7

© 2007 Herbert I. Gross next 2 3 4 8 3 4 8 9 4 2 9 7 In this notation, the first column holds the place of x 1 ; the second column holds the place of x 2, and the third column holds the place of x 3. The vertical line holds the place of the equal sign. next Actually, the matrix appears without such a vertical line. However, we prefer using it in order to separate the algebraic expression from the number it equals.

© 2007 Herbert I. Gross This configuration is referred to as a 3 by 4 matrix to indicate that there are 3 rows and 4 columns. More generally, an m by n matrix would be a rectangular array consisting of m rows and n columns. next 2 3 4 8 3 4 8 9 4 2 9 7 The steps that took us from… next 2x + 3y + 4z = 8 y – 4z = 6 z = - 1 2x + 3y + 4z = 8 3x + 4y + 8z = 9 4x + 2y + 9z = 7 to… in matrix notation, would appear as…

© 2007 Herbert I. Gross next 2 3 4 8 3 4 8 9 4 2 9 7 Multiply the top row by 6; next 12 18 24 48 - 12 - 16 - 32 - 36 - 12 - 6 - 27 - 21 2 3 4 8 3 4 8 9 4 2 9 7 the middle row by - 4; and the bottom row by - 3. next 2x + 3y + 4z = 8 3x + 4y + 8z = 9 4x + 2y + 9z = 7

© 2007 Herbert I. Gross Replace the middle row by the middle row + top row next 12 18 24 48 - 12 - 16 - 32 36 - 12 - 6 - 27 - 21 12 18 24 48 - 12 - 16 - 32 36 - 12 - 6 - 27 - 21 12 - 3 27 2 - 8 12 12 18 24 48 12 - 3 27 - 12 - 6 - 27 - 21 12 18 24 48 - 12 - 16 - 32 36 2 - 8 12 next and the bottom row by the bottom row + top row.

© 2007 Herbert I. Gross Divide the top row by 6, next 12 18 24 48 0 2 - 8 12 0 12 - 3 27 2 3 4 8 0 1 - 4 6 0 4 - 1 9 Multiply the middle row by - 4. 12 18 24 48 0 2 - 8 12 0 12 - 3 27 the middle row by 2, and the bottom row by 3. 2 3 4 8 0 4 - 1 9 0 1 - 4 6 0 - 4 - 16 - 24 next

© 2007 Herbert I. Gross Replace the bottom row by the bottom row + middle row. next 2 3 4 8 0 - 4 16 - 24 0 4 - 1 9 0 0 - 15 - 15 0 0 15 - 15 0 - 4 16 - 24 0 4 - 1 9

© 2007 Herbert I. Gross Divide the middle row by - 4 next 2 3 4 8 0 - 4 16 - 24 0 0 15 - 15 and the bottom row by 15. 2 3 4 8 0 - 4 16 - 24 0 0 15 - 15 0 1 - 4 6 0 0 1 - 1 The above matrix translates into the system… 2x + 3y + 4z = 8 y – 4z = 6 z = - 1 next

© 2007 Herbert I. Gross This method becomes even more convenient as the number of equations in the system increases. For example, suppose we were asked to solve the following system of 4 linear equations in 4 unknowns… next x 1 + x 2 + x 3 + x 4 = 4 2x 1 + 3x 2 + 4x 3 + - x 4 = - 8 3x 1 + 2x 2 + 5x 3 + 2x 4 = 0 4x 1 + 5x 2 + 6x 3 + 3x 4 = 8

© 2007 Herbert I. Gross next x 1 + x 2 + x 3 + x 4 = 4 2x 1 + 3x 2 + 4x 3 + - x 4 = - 8 3x 1 + 2x 2 + 5x 3 + 2x 4 = 0 4x 1 + 5x 2 + 6x 3 + 3x 4 = 8 next In matrix notation we start with the matrix… 2x 1 + 3x 2 + 4x 3 + - x 4 = - 8 3x 1 + 2x 2 + 5x 3 + 2x 4 = 0 4x 1 + 5x 2 + 6x 3 + 3x 4 = 8 1 1 1 1 4 2 3 4 - 1 - 8 3 2 5 2 0 4 5 6 3 8

© 2007 Herbert I. Gross Our goal is to create all 0s below the first entry in the first column. To do this we want to make sure that the coefficients of x 1 in all rows below the first are the opposite of the coefficient of x 1 in the first row. next 1 1 1 1 4 2 3 4 - 1 - 8 3 2 5 2 0 4 5 6 3 8 The least common multiple of 1, 2, 3, and 4 Is 12. next

© 2007 Herbert I. Gross next Thus, the sequence of steps might be… 1 1 1 1 4 2 3 4 - 1 - 8 3 2 5 2 0 4 5 6 3 8 Multiply the 1st row by 12,the 2nd by - 6, the 3rd by - 4,and the 4 th by - 3. 12 12 12 12 48 - 12 - 18 - 24 6 48 - 12 - 8 - 20 - 8 0 - 12 - 15 - 18 - 9 - 24 1 1 1 1 4 2 3 4 - 1 - 8 3 2 5 2 0 4 5 6 3 8

© 2007 Herbert I. Gross Replace row 2 by row 2 + row 1… next 12 12 12 12 48 - 12 - 18 - 24 6 48 - 12 - 8 - 20 - 8 0 - 12 - 15 - 18 - 9 - 24 12 12 12 12 48 - 12 - 18 - 24 6 48 0 - 6 - 12 18 96

© 2007 Herbert I. Gross Replace row 3 by row 3 + row 1… next 12 12 12 12 48 - 12 - 8 - 20 - 8 0 12 12 12 12 48 0 - 6 - 12 18 96 - 12 - 8 - 20 - 8 0 - 12 - 15 - 18 - 9 - 24 0 4 - 8 4 48

© 2007 Herbert I. Gross Replace row 4 by row 4 + row 1… next 12 12 12 12 48 - 12 - 15 - 18 - 9 - 24 12 12 12 12 48 0 - 6 - 12 18 96 0 4 - 8 4 48 - 12 - 15 - 18 - 9 - 24 0 - 3 - 6 3 24

© 2007 Herbert I. Gross next In order to work with smaller numbers, divide the 1st row by 12, the 2nd row by - 6, the 3rd row by 4, and the 4th row by - 3. 12 12 12 12 48 0 - 6 - 12 18 96 0 4 - 8 4 48 0 - 3 - 6 3 24 1 1 1 1 4 12 12 12 12 48 0 - 6 - 12 18 96 0 4 - 8 4 48 0 - 3 - 6 3 24 0 1 2 - 3 - 16 0 1 - 2 1 12 0 1 2 - 1 - 8

© 2007 Herbert I. Gross At this point notice that… next 1 1 1 1 4 0 1 2 - 3 - 16 0 1 - 2 1 12 0 1 2 - 1 - 8 essentially allows us to solve the system of 4 linear equations by first solving a system of 3 linear equations. More specifically, the last 3 rows represent the system of equations… x 2 + 2x 3 + - 3x 4 = - 16 x 2 + - 2x 3 + x 4 = 12 x 2 + 2x 3 + - x 4 = - 8 next

© 2007 Herbert I. Gross Knowing how to solve a system of 3 linear equations, we may now solve the system below for the values of x 2, x 3 and x 4, and we could then substitute these values into any one of the original equations to solve for x 1. next x 2 + 2x 3 + - 3x 4 = - 16 x 2 + - 2x 3 + x 4 = 12 x 2 + 2x 3 + - x 4 = - 8

© 2007 Herbert I. Gross To accomplish this, we want the coefficients of x 2 in the rows below the second row to be the opposite of the coefficient of x 2 in the second row. So we multiply the 3rd and 4th rows by - 1. next 1 1 1 1 4 0 1 2 - 3 - 16 0 1 - 2 1 12 0 1 2 - 1 - 8 0 - 1 2 - 1 - 12 0 - 1 - 2 1 8 next 1 1 1 1 4 0 1 2 - 3 - 16 0 1 - 2 1 12 0 1 2 - 1 - 8

© 2007 Herbert I. Gross Replace row 3 by row 3 + row 2… next 1 1 1 1 4 0 1 2 - 3 - 16 0 - 1 2 - 1 - 12 0 - 1 - 2 1 8 0 1 2 - 3 - 16 0 - 1 2 - 1 - 12 0 0 4 - 4 - 28

© 2007 Herbert I. Gross Replace row 4 by row 4 + row 2… next 1 1 1 1 4 0 1 2 - 3 - 16 0 0 4 - 4 - 28 0 - 1 - 2 1 - 8 0 1 2 - 3 - 16 0 - 1 - 2 1 8 0 0 0 - 2 - 8

© 2007 Herbert I. Gross next In order to work with smaller numbers, divide the 3rd row by 4, and the 4th row by - 2. 1 1 1 1 4 0 1 2 - 3 - 16 0 0 4 - 4 - 28 0 0 0 - 2 - 8 1 1 1 1 4 0 1 2 - 3 - 16 0 0 4 - 4 - 28 0 0 0 - 2 - 8 0 0 0 1 4 0 0 1 - 1 - 7 next

© 2007 Herbert I. Gross By coincidence, the entry in the 4th row 3rd column is already 0. If it hadnt been 0 we would have had to work with the last two rows and convert the matrix to that form. next 1 1 1 1 4 0 1 2 - 3 - 16 0 0 1 - 1 - 7 0 0 0 1 4

© 2007 Herbert I. Gross next The matrix above codes the equivalent system of equations… 1 1 1 1 4 0 1 2 - 3 - 16 0 0 1 - 1 - 7 0 0 0 1 4 next x 1 + x 2 + x 3 + x 4 = 4 x 2 + 2x 3 – 3x 4 = - 16 x 3 – x 4 = - 7 x 4 = 4

© 2007 Herbert I. Gross Starting with the fourth (bottom) equation we see at once that… next x 1 + x 2 + x 3 + x 4 = 4 x 2 + 2x 3 + - 3x 4 = - 16 x 3 + - x 4 = - 7 x 4 = 4 We next replace x 4 by 4 in the third equation to get x 3 + - 4 = - 7, and we see that… x 3 = - 3 next

© 2007 Herbert I. Gross We then replace x 3 by - 3 and x 4 by 4 in the second equation to get… x 2 + 2( - 3) + - 3(4) = - 16; or next x 1 + x 2 + x 3 + x 4 = 4 x 2 + 2x 3 + - 3x 4 = - 16 x 3 + - x 4 = - 7 x 4 = 4 x 2 = 2 next x 2 + - 6 + - 12 = - 16; or x 2 + - 18 = - 16; hence…

© 2007 Herbert I. Gross Finally, in the first equation we replace x 2 by 2, x 3 by - 3 and x 4 by 4 to obtain… next x 1 = 1 next x 1 + 2 + - 3 + 4 = 4; or x 1 + 3 = 4; hence… x 1 + x 2 + x 3 + x 4 = 4 x 2 + 2x 3 + - 3x 4 = - 16 x 3 + - x 4 = - 7 x 4 = 4

© 2007 Herbert I. Gross Thus, the solution of the equivalent system… next x 1 + x 2 + x 3 + x 4 = 4 2x 1 + 3x 2 + 4x 3 + - x 4 = - 8 3x 1 + 2x 2 + 5x 3 + 2x 4 = 0 4x 1 + 5x 2 + 6x 3 + 3x 4 = 8 is given by… x 1 = 1, x 2 = 2, x 3 = - 3, and x 4 = 4

© 2007 Herbert I. Gross As a check, we see that each of the equations below is a true statement… next 1 + 2 + - 3 + 4 = 4 2(1) + 3(2) + 4( - 3) – 4 = - 8 3(1) + 2(2) + 5( - 3) + 2(4) = 0 4(1) + 5(2) + 6( - 3) + 3(4) = 8 x 1 + x 2 + x 3 + x 4 = 4 2x 1 + 3x 2 + 4x 3 – x 4 = - 8 3x 1 + 2x 2 + 5x 3 + 2x 4 = 0 4x 1 + 5x 2 + 6x 3 + 3x 4 = 8 next

© 2007 Herbert I. Gross While this procedure becomes more tedious as the number of variables increases, it is very systematic and lends itself nicely to computer techniques. next Final Note

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