Gauss-Jordan Algorithm

Transcription

1 Gauss-Jordan Algorihm The Gauss-Jordan algorihm is a sep by sep procedure for solving a sysem of linear equaions which may conain any number of variables and any number of equaions. The algorihm is carried ou by performing a series of elemenary row operaions on he rows of a marix. Each row operaion preserves he soluion of he original sysem of equaions. The hree elemenary row operaions are: 1) Swap or exchange wo rows. 2) Muliply a row by a consan. 3) Add a muliple of a row o anoher row. (The row muliplied is no changed. The row added o is changed because a muliple of anoher row is added o i.) The general sraegy is o ransform he sysem of equaions marix so ha he marix elemens on he main diagonal (a 1,1, a 2,2, a 3,3...) equal one and he marix elemens above and below he main diagonal equal zero. Saring wih column one and working lef o righ, apply elemenary row operaions ha ransform he column ino a uni column. When every column bu he righ column of he marix is a uni column, any soluion can be deermined by examining he row-reduced echelon marix. See handou on inerpreing row-reduced echelon marices. Example: Solve he sysem of equaions: 2x + y + 2z = 9 x + 2y + 2z = 10 2x + 2y + z = 1 Iniial Marix. Column 1 is now a uni column R 1 R 2-2R 1 + R 2-2R 1 + R 3 Column 2 is now a uni column -1/3R 2-2R 2 + R 1 2R 2 + R 3 Column 3 is now a uni column and he marix is in row-reduced echelon form. -3/5R 3-2/3R 3 + R 1-2/3R 3 + R 2 I is now easy o see ha he soluion o he sysem of equaions is x = -2, y = -1, and z = 7.

2 How o Inerpre Row-Reduced Echelon Marices Afer a marix ha represens a sysem of linear equaions has been ransformed o he row-reduced echelon form, one needs o know how o read and inerpre he marix. A sysem of equaions will have a unique soluion, no soluions, or infiniely many soluions which can be described by a se of parameric equaions. Example 1 : Two variable linear sysem wih exacly one soluion. Original marix Example 2 : Two variable linear sysem wih no soluion. (All main diagonal elemens equal 1.) Unique soluion: x = -7 and y = Commen: You should graph he equaions in he original marix and observe ha heir graphs inersec a (-7, -0.5). The second row of he RREF marix implies 0*x + 0*y = 45 which is clearly impossible. Therefore he sysem of equaions has no soluions. Commen: You should graph he equaions in he original marix and observe ha he graphs are parallel lines. Example 3 : Two variable linear sysem wih infiniely many soluions. Commen: These equaions are called he parameric form of he equaion of he line x = 3y + 5. You should graph hese parameric equaions on your graphing calculaor. Since he original equaions describe he same line, he graph of he parameric equaions equals he graph of he original equaions. The second row of he RREF marix implies 0*x + 0*y = 0 which is rue for all (x,y) pairs. The firs row of he RREF marix ells us ha x 3y = 5 ===> x = 3y + 5. Leing y =, he parameric equaions x = and y = describe all soluions o he sysem of equaions. Any real value of generaes a soluion o he sysem of equaions. See he commen o he lef. Example 4 : Three variable linear sysem wih exacly one soluion. Example 5 : Three variable linear sysem wih no soluion. (All main diagonal elemens equal 1.) Unique soluion: x = 2, y = 6 and z = 9 Commen: You should graph he equaions in he original marix and observe ha heir graphs inersec a (2, 6, 9). You can use EquaionGrapher's 3D poin ploing feaure o plo he poin (2, 6, 9) in 3D space. The hird row of he RREF marix implies 0*x + 0*y + 0*z = -2 which is clearly impossible. Therefore he sysem of equaions has no soluions. Commen: You should graph he equaions in he original marix and observe ha he hree graphs do no have any common inersecion poins. Use he mouse o roae he graphs in 3D space so ha you can view he graphs of he planes from differen view poins..

3 Example 6 : Three variable linear sysem wih infiniely many soluions. The hird row of he RREF marix implies 0*x. + 0*y + 0*z = 0 which is rue of all (x,y,z). The firs row of he RREF marix ells us x + 10/7z = 11/7 ===> x = -10/7z + 11/7. The second row of he RREF marix ells us y + (- 6/7)z = -1/7 ====> y = 6/7z - 1/7. Since boh x and y depend on z, we can form a se of parameric equaions by replacing he variable z wih he parameric variable. The graph is a line in 3D space. Commen: You should graph he equaions of he hree planes in he original marix. Then graph he parameric equaions of he 3D line and observe ha he graph of he parameric equaions equals he inersecion of he hree planes. Also plo poins on he line by picking specific values of and plo he poins corresponding o he values of. Make a simple able for he variables, x, y, and z and use EquaionGrapher's 3D poin ploing feaure o plo he (x,y,z) poins from your able in 3D space. { x = -10/7 + 11/7, y = 6/7 1/7, z = } Any real value of generaes a soluion of he sysem of equaions. The coordinaes of poins on he line have heir own privae formula.. See commen o he lef. Example 7: Four variable linear sysem wih infiniely many soluions. Original marix The firs row of he RREF marix ells us x 1 = -11x 3 / /6. The second row of he RREF marix ells us x 2 = -x 3 / 3 + 2/3. The hird row of he RREF marix ells us x 4 = ½. x 1 = -11/3 + 17/6, x 2 = -/3 + 2/3 x 3 = and x 4 = ½. Any real value of generaes a soluion of he sysem of equaions. Boh x 1 and x 2 depend on x 3. x 4 does no depend on any variable. The parameric equaions ha describe all soluions are formed by replacing x 3 wih he parameric variable. Example 8: Five variable linear sysem wih infiniely many soluions. This is why we need wo parameric variables o describe he soluions. The firs row of he RREF marix implies x 1 = 21 x 3 24x > x 1 depends on x 3 and x 5 The second row of he RREF marix implies x 2 = 2x 3 + 8x > x 2 depends on x 3 and x 5 The hird row of he RREF marix implies x 4 = 3 2x > x 4 depends on x 5 If we le x 3 = he parameric variable s and x 5 = he parameric variable, he general soluion can be described as all ordered 5-uples of he form (21 s 24, 2s + 8-7, s, 3 2, ) where s and can be any real number.

4 y z x x How o Find he Equaion of a Line in 3D Space Given Two Poins A basic posulae of elemenary geomery saes ha wo poins deermine a unique line. In he 2D x-y coordinae plane, lines are described by equaions of he form x = k, y = k, y = mx + b, or Ax + By = k where k, m, b, A, and B are consans. In 3D space, lines are described by a se of parameric equaions of he form { x = a + b, y = c + d, z = e + f } where a, b, c, d, e, and f are real number consans. The x, y, and z coordinaes of a poin on a line depends on a fourh parameric variable. Each coordinae of a poin on a line has is own privae equaion which is similar o he slope-inercep form of he equaion of a line in 2D space. Example 1 : Find he equaion of he line in 3D space ha conains he poins (4, -3, 2) and (-2, 7, 10). By choosing = 0 o correspond o (4, -3, 2) and = 1 o correspond o (-2, 7, 10), i is a simple ask o find a se of parameric equaions of a line ha conains he given poins. x = a + b y = c + d z = e + f a = slope = -6/1 = -6 c = slope = 10/1 = 10 e = slope = 8/1 = 8 b = inercep = 4 d = inercep = -3 f = inercep = 2 The se of parameric equaions ha describes he line is { x = , y = 10 3, z = }. Every real number value of will generae a (x, y, z) poin on he line. The poins corresponding o = 0, 1, 2, 3, 4, and 5 are (4,-3,2), (-2,7,10), (-8,17,18), (-14,27,26), (-20,37,34), and (-26,47,42). Noe: The se of parameric equaions for a line is no unique. Leing values of oher han 0 and 1 correspond o (4, -3, 2) and (-2, 7, 10) would produce a differen se of parameric line equaions, bu boh ses of parameric equaions would describe he same line. For example, if we le = 5 correspond o (4, -3, 2) and = 10 correspond o (-2, 7, 10), he parameric se of equaions for he line would equal { x = -6/5 + 10, y = 2-13, z = 8/5 6 } Example 2 : Find he equaion of he line in 3D space ha conains he poins (0, 6, -2) and (7, -4, 3). By choosing = 0 o correspond o (0, 6, -2) and = 1 o correspond o (7, -4, 3), i is easy o find a se of parameric equaions of a line ha conains he given poins.. x = a + b y = c + d z = e + f +7 y z a = slope = 7/1 = 7 c = slope = -10/1 = -10 e = slope = 5/1 = 5. b = inercep = 0 d = inercep = 6 f = inercep = -2 The se of parameric equaions for he line is { x = 7, y = , z = 5-2 } Every real number value of would generae a (x,y,z) poin on he line

5 How o Find he Equaion of he Line of Inersecion of Two Planes in 3D Space Example 1 : From elemenary geomery we know ha he inersecion of wo planes is a line. Suppose he equaions of wo planes in 3D space are: -2x + 3y z = 12 and x 4y 7z = 14 Original marix In he firs row of he RREF marix we have x + 5z = -18 ===> x = -5z In he second row of he RREF marix we have y + 3z = -8 ===> y = -3z 8. Afer replacing z wih, he equaion of he inersecion of he wo planes is he se of parameric equaions { x = -5-18, y = -3-8, z = }. Any real value of generaes a poin in he inersecion of he wo planes. The poins on he line of inersecion corresponding o = 0, 1, 2, 3, 4, and 5 are (-18, - 8, 0), (-23, -11, 1), (-28, -14, 2), (-33, -17, 3), (-38, -20, 4) and (-43, -23, 5). Example 2 : Find he equaion of he inersecion of wo planes in 3D space. From elemenary geomery we know ha he inersecion of wo planes is a line. Suppose he equaions of wo planes in 3D space are: x + y = 6 and x y + 4z = 18 Original marix From he firs row of he RREF marix we have x + 2z = 12 ===> x = -2z From he second row of he RREF marix we have y - 2z = -6 ===> y = 2z 6. Afer replacing z wih, he equaion of he inersecion of he wo planes is he se of parameric equaions { x = , y = 2-6, z = }. Any real value of generaes a poin in he inersecion of he wo planes. The poins on he line of inersecion corresponding o = 0, 1, 2, 3, 4, and 5 are (12, -6, 0), (10,-4, 1), (8, -2, 2), (6, 0, 3), (4, 2, 4) and (2, 4, 5).