The Bisection method or BOLZANO s method or Interval halving method: Find the positive root of x 3 x = 1 correct to four decimal places by bisection method Let f x = x 3 x 1 Here f 0 = 1 = ve, f 1 = ve, f 2 = 5 = +ve Hence a root lies between 1 and 2 Here a = 1 and b = 2 a + b Here x 0 = = 1 + 2 = 15, f x 2 2 0 = f(15) = 0875 = +ve Since f a is negative and f(x 0 ) is positive The root lies between a and x 0 ie 1 and 15, put b = x 0 x 1 = 1 + 15 = 125, f 125 = 02969 = ve 2 The root lies between x 1 and b ie 125 and 15, put a = x 1 125 + 15 x 2 = = 1375, f 1375 = 02246 = +ve 2 Therefore the root lies between a and x 2 ie 125 and 1375 Proceeding like this and these values are put in a tabular form as follows i a b x n f(x n ) 0 1 2 15 0875 1 1 15 125-02969 2 125 15 1375 02246 3 125 1375 13125-00515 4 13125 1375 13438 00828 5 13125 13438 13282 00149 6 13125 13282 13204-00183 7 13204 13282 13243-00018 8 13243 13282 13263 00068 9 13243 13263 13253 00025 10 13243 13253 13248 00003 11 13243 13248 13246-00005 12 13246 13248 13247-00001 13 13247 13248 13248 00003 14 13247 13248 13248 00003 Since the values in 13 th and 14 th iteration are same, the root is 13248
Find the positive root of x cos x = 0 by bisection method Let f x = x cos x Here f 0 = 1 = ve, f 1 = 1 cos 1 = 04597 = +ve Hence a root lies between 0 and 1 Here a = 0 and b = 1 a + b Here x 0 = = 0 + 1 = 05, f x 2 2 0 = f 05 = 03776 = ve Since f x 0 is negative and f(b) is positive The root lies between x 0 and b ie 05 and 1, put a = x 0 x 1 = 05 + 1 = 075, f 075 = 00183 = +ve 2 The root lies between a and x 1 ie 05 and 075, put b = x 1 05 + 075 x 2 = = 0625, f 0625 = 0186 = ve 2 Therefore the root lies between x 2 and b ie 0625 and 075 Proceeding like this and these values are put in a tabular form as follows i a b x n f(x n ) 0 0 1 05-03776 1 05 1 075 00183 2 05 075 0625-0186 3 0625 075 06875-00853 4 06875 075 07188-00338 5 07188 075 07344-00078 6 07344 075 07422 00052 7 07344 07422 07383-00013 8 07383 07422 07403 0002 9 07383 07403 07393 00004 10 07383 07393 07388-00005 11 07388 07393 07391 0 Since the value of the last column is zero in last column of 11 th iteration, the root is 07391 Regula Falsi method or False position method: Solve for a positive root of x 3 4x + 1 = 0 by Regula Falsi method Let f x = x 3 4x + 1 Here f 0 = 1 = +ve, f 1 = 2 = ve A root lies between 0 and 1 Here a = 0 and b = 1
af b bf a x 1 = f b f a = 0 f 1 1 f 0 f 1 f 0 = 1 2 1 = 03333 f x 1 = f 03333 = 02963 = ve Hence the root lies between 0 and 03333 ie a = 0 and b = 03333 (Since f 0 = +ve and f 03333 = ve) x 2 = af b bf a f b f a = 0 f 03333 1 f 0 f 03333 f 0 = 02571 f 02571 = 00114 = ve Hence the root lies between 0 and 03333 ie a = 0 and b = 02571 (Since f 0 = +ve and f 02571 = ve) Continuing in the same way and the values are put in a table below i a b f(a) f(b) x n f(x n ) 1 0 1 1-2 03333-02962 2 0 03333 1-02962 02571-00114 3 0 02571 1-00114 02542-00004 4 0 02542 1-00004 02541 0 Since the value of the last column of 4 th iteration is zero, the root is 02541 Find a positive root of xe x = 2 by the method of false position Let f x = xe x 2 Here f 0 = 2 = ve, f 1 = 07183 = +ve A root lies between 0 and 1 Here a = 0 and b = 1 x 1 = af b bf a f b f a = 0 f 1 1 f 0 f 1 f 0 = 2 07183 + 2 = 03333 f x 1 = f 03333 = 02963 = ve Hence the root lies between 0 and 03333 ie a = 0 and b = 03333 (Since f 0 = +ve and f 03333 = ve) x 2 = af b bf a f b f a = 0 f 03333 1 f 0 f 03333 f 0 = 02571
f 02571 = 00114 = ve Hence the root lies between 0 and 03333 ie a = 0 and b = 02571 (Since f 0 = +ve and f 02571 = ve) Continuing in the same way and the values are put in a table below i a b f(a) f(b) x n f(x n ) 1 0 1-2 07183 07358-04643 2 07358 1-04643 07183 08395-17664 3 08395 1-00564 07183 08512-17881 4 08512 1-00061 07183 08525-17904 5 08525 1-00005 07183 08526-17906 6 08526 1 0 07183 08526-17906 Since the value of the x n column of 5 th and 6 th iteration are same, the root is 08526 up to four decimal places Gauss Jacobi method: Solve by Gauss Jacobi method, the system 12x + y + z = 31,2 x + 8y z = 24, 3x + 4y + 10z = 58 1 + 1 < 12, 2 + 1 < 8, 3 + 4 < 10 Therefore the given equations are diagonally dominant, so we can use Gauss Jacobi method Proceeding in the same manner and writing the solutions in the tabular form we get i x y z 1 25833 3 58 2 185 30792 3825 3 2008 30156 40133 4 19976 29997 39914
5 20007 29995 40008 6 2 29999 4 7 2 3 4 8 2 3 4 Since the values in 7 th and 8 th iteration are same, the solution is x = 2, y = 3, z = 4 up to four decimal places Solve by Gauss Jacobi method, the system 9x y + 2z = 9, x + 10y 2z = 15, 2x 2y 13z = 17 1 + 2 = 3 < 9, 1 + 2 < 10, 2 + 2 < 13 Therefore the given equations are diagonally dominant, so we can use Gauss Jacobi method 1 st iteration x = 1 (9 + y 2z) 9 y = 1 (15 x + 2z) 10 z = 1 (17 + 2x 2y) 13 x (1) = 1 9 9 + 0 2 0 = 1 y (1) = 1 10 15 0 + 2 0 = 15 z (1) = 1 13 17 + 2(0) 2(0) = 13077 2 nd iteration x (2) = 1 9 9 + 15 2 13077 = 08761
y (2) = 1 10 15 15 + 2 13077 = 16615 z (2) = 1 13 17 + 2(1) 2(15) = 12308 Proceeding in the same manner and writing the solutions in the tabular form we get i x y z 1 1 15 13077 2 08761 16615 12308 3 09111 16586 11869 4 09205 16463 11927 5 09179 16465 1196 6 09172 16474 11956 7 09174 16474 11954 8 09174 16473 11954 9 09174 16473 11954 Since the values in 8 th and 9 th iteration are same, the solution is x = 09174, y = 16473, z = 11954 up to four decimal places Gauss Seidel method Solve by Gauss Seidel method, the system 12x + y + z = 31,2 x + 8y z = 24, 3x + 4y + 10z = 58 1 + 1 < 12, 2 + 1 < 8, 3 + 4 < 10 Therefore the given equations are diagonally dominant, so we can use Gauss Seidel method x = 1 12 31 y z y = 1 8 24 2x + z
z = 1 10 58 3x 4y 1 st iteration x = 1 12 31 0 0 = 25833 y = 1 8 24 2 25833 + (0) = 23542 z = 1 10 58 3(25833) 4(23542) = 40833 2 nd iteration x = 1 12 31 23542 40833 = 20469 y = 1 8 z = 1 10 24 2(20469) + 40833 = 29987 58 3(20469) 4(29987) = 39865 Proceeding in the same manner and writing the solutions in the tabular form we get i x y z 1 25833 23542 40833 2 20469 29987 39865 3 20012 2998 40004 4 20001 3 4 5 2 3 4 6 2 3 4 Since the values in 5 th and 6 th iteration are same, the solution is x = 2, y = 3, z = 4 up to four decimal places Solve by Gauss Seidel method, the system 9x y + 2z = 9, x + 10y 2z = 15, 2x 2y 13z = 17
1 + 2 = 3 < 9, 1 + 2 < 10, 2 + 2 < 13 Therefore the given equations are diagonally dominant, so we can use Gauss Seidel method 1 st iteration x = 1 (9 + y 2z) 9 y = 1 (15 x + 2z) 10 z = 1 (17 + 2x 2y) 13 x (1) = 1 9 9 + 0 2 0 = 1 y (1) = 1 10 15 1 + 2 0 = 14 z (1) = 1 13 17 + 2(1) 2(14) = 12462 2 nd iteration x (2) = 1 9 9 + 14 2 12462 = 08786 y (2) = 1 10 15 08786 + 2 12462 = 16614 z (2) = 1 13 17 + 2(08786) 2(16614) = 11873 Proceeding in the same manner and writing the solutions in the tabular form we get i x y z 1 1 14 12462 2 08786 16614 11873 3 09208 16454 11962 4 0917 16475 11953 5 09174 16473 11954 6 09174 16473 11954
Since the values in 5 th and 6 th iteration are same, the solution is x = 09174, y = 16473, z = 11954 up to four decimal places Gauss Elimination method Solve by Gauss Elimination method, the system 12x + y + z = 31,2 x + 8y z = 24, 3x + 4y + 10z = 58 ~ ~ A: B = 12 1 1 0 783 117 0 375 975 12 1 1 0 783 117 0 0 1031 12 1 1 2 8 1 3 4 10 31 1883 5025 31 1883 4124 31 24 58 R 2 R 2 2 12 R 1 R 3 R 3 3 12 R 1 R 3 R 3 375 783 R 2 12x + y + z = 31 783y 117z = 1883 1031z = 4124 z = 4124 1031 = 4 783y 117 4 = 1883 y = 1883 + 117 4 783 = 3 Therefore the solution is x = 2, y = 3, z = 4 Solve by Gauss Elimination method, the system 12x + 3 + 4 = 31 x = 31 7 12 = 2 9x y + 2z = 9, x + 10y 2z = 15, 2x 2y 13z = 17 A: B = 9 1 2 1 10 2 2 2 13 9 15 17
~ 9 1 2 0 101111 22222 0 17778 134444 9 14 19 R 2 R 2 1 9 R 1 R 3 R 3 2 9 R 1 ~ 9 1 2 0 101111 22222 0 0 138352 9 14 165385 R 3 R 3 17778 101111 R 2 9x y + 2z = 9 101111y 22222z = 14 138352z = 4124 z = 165385 138352 = 11954 101111y 22222 11954 = 14 y = 9x 16473 + 2 11954 = 9 x = 14 + 22222 11954 101111 9 + 16473 2 11954 9 = 16473 = 09174 Therefore the solution is x = 09174, y = 16473, z = 11954 Gauss Jordon method Solve by Gauss Jordon method, the system 12x + y + z = 31,2 x + 8y z = 24, 3x + 4y + 10z = 58 ~ ~ ~ A: B = 12 1 1 0 783 117 0 375 975 12 0 115 0 783 117 0 0 1031 12 0 0 0 783 0 0 0 1031 12 1 1 2 8 1 3 4 10 31 1883 5025 286 1883 4124 24 2351 4124 31 24 58 R 2 R 2 2 12 R 1 R 3 R 3 3 12 R 1 R 1 R 1 1 783 R 2 R 3 R 3 375 783 R 2 R 1 R 1 115 1031 R 3 R 2 R 2 117 1031 R 3
Therefore the solution is x = 2, y = 3, z = 4 Solve by Gauss Jordon method, the system 12x = 24 x = 24 12 = 2 783y = 2351 y = 2351 783 = 3 1031z = 4124 z = 4124 1031 = 4 9x y + 2z = 9, x + 10y 2z = 15, 2x 2y 13z = 17 A: B = 9 1 2 1 10 2 2 2 13 9 15 17 ~ 9 1 2 0 101111 22222 0 17778 134444 9 14 19 R 2 R 2 1 9 R 1 R 3 R 3 2 9 R 1 ~ ~ 9 0 17802 0 101111 22222 0 0 138352 9 0 0 0 101111 0 0 0 138352 103846 14 165385 82566 166564 165385 R 1 R 1 1 101111 R 2 R 3 R 3 17778 101111 R 2 R 1 R 1 17802 138352 R 3 R 2 R 2 22222 138352 R 3 9x = 82566 x = 82566 9 = 09174 101111y = 166564 y = 166564 101111 = 16473 138352z = 165385 z = 165385 138352 = 11954 Therefore the solution is x = 09174, y = 16473, z = 11954
NEWTON S METHOD OR NEWTON-RAPHSON METHOD Find the positive root of x 4 x = 10 correct to three decimal places using Newton- Raphson method Solution : Let f x = x 4 x 10 = 0 Now, f 0 = 0 4 0 10 = 10 f 1 = 1 4 1 10 = 10 f 2 = 2 4 2 10 = +4 Therefore the root lies between 1 & 2 Let us take x 0 = 2 {Near to zero} The Newton- Raphson formula is ve ve +ve x n+1 = x n f x n f x n 1 Let f x = x 4 x 10 and f 1 x = 4 x 3 1 x 1 = x 0 f x 0 f x 0 x 1 = 2 f 2 f 2 x 1 = 2 2 4 2 10 4 2 3 1 x 1 = 2 4 31 x 1 = 18709 x 2 = x 1 f x 1 f x 1 x 2 = 2 f 18709 f 18709 x 2 = 18709 18709 4 18709 10 4 18709 3 1 x 2 = 18709 03835 25199 x 2 = 1856 x 3 = x 2 f x 2 f x 2
x 3 = 1856 f 1856 f 1856 x 3 = 1856 1856 4 1856 10 4 1856 3 1 x 3 = 1856 0010 24574 x 3 = 1856 The root of the equation x 4 x = 10 is 1856 Using Newton s iterative method to find the root between 0 and 1 of x 3 = 6x 4 correct to three decimal places Solution : Let f x = x 3 6x + 4 = 0 Now, f 0 = 0 3 6 0 + 4 = +4 f 1 = 1 3 6 1 + 4 = 1 Therefore the root lies between 0 & 1 Let us take x 0 = 1 {Near to zero} The Newton- Raphson formula is +ve ve x n+1 = x n f x n f x n 1 Let f x = x 3 6x + 4 and f 1 x = 3 x 2 6 x 1 = x 0 f x 0 f x 0 x 1 = 1 f 1 f 1 x 1 = 1 1 3 6 1 + 4 3 1 2 6 x 1 = 1 1 3 x 1 = 0666 x 2 = x 1 f x 1 f x 1 x 1 = 0666 f 0666 f 0666
x 2 = 0666 0666 3 6 0666 + 4 3 0666 2 6 x 2 = 0666 028 465 x 1 = 073 x 3 = x 2 f x 2 f x 2 x 3 = 073 f 073 f 073 x 3 = 073 073 3 6 073 + 4 3 073 2 6 x 3 = 073 0009 44013 x 3 = 07320 The root of the equation x 3 6x + 4 = 0 is 0732 Find the real positive root of 3x cos x 1 = 0 correct to six decimal places using Newton method Solution : Let f x = 3x cos x 1 = 0 Now, f 0 = 3 0 cos 0 1 = 2 f 1 = 3 1 cos 1 1 = 1459698 Therefore the root lies between 0 & 1 Let us take x 0 = 1 {Near to zero} The Newton- Raphson formula is ve ve Let x n+1 = x n f x n f x n 1 f x = 3x cos x 1 and f 1 x = 3 + sin x x 1 = x 0 f x 0 f x 0 x 1 = 1 f 1 f 1 x 1 = 1 3 1 cos 1 1 3 + sin 1
x 1 = 062002 x 2 = 062002 x 2 = x 1 f x 1 f x 1 x 2 = 062002 f 062002 f 062002 3 062002 cos 062002 1 3 + sin 062002 x 2 = 060712 x 3 = 060712 x 3 = x 2 f x 2 f x 2 x 3 = 060712 f 060712 f 060712 3 060712 cos 060712 1 3 + sin 060712 x 3 = 06071 The root of the equation x 4 x = 10 is 060712 FIXED POINT ITERATION OR ITERATION METHOD The condition for convergence of a method Let f x = 0 be the given equation whose actual root is r The equation f x = 0 be written as x = g x Let I be the interval containing the root x = r If g x < 1 for all x in I, then the sequence of approximations x 0, x 1, x 2, x n will converge to r, if the initial starting value x 0 is chosen in I Note 1 Since x n r K x n 1 r where K is a constant the convergence is linear and the convergence is of order 1 Note 2 The sufficient condition for the convergence is g x < 1 for all x in I Find the positive root of x 2 2x 3 = 0 by Iteration method Solution : Let f x = x 2 2x 3 = 0 Now, f 0 = 0 2 2 0 3 = 10 ve
f 1 = 0 2 2 0 3 = 10 ve f 2 = 0 2 2 0 3 = +4 +ve Therefore the root lies between 1 & 2 Let us take x 0 = 2 {Near to zero} x 2 2x 3 = 0 x 2 = 2x + 3 x = 2x + 3 Let x 0 = 2 x = g(x) = 2x + 3 x 1 = g x 0 = 2x 0 + 3 = 2 2 + 3 = 26457 x 2 = g x 1 = 2x 1 + 3 = 2 26457 + 3 = 28795 x 3 = g x 2 = 2x 2 + 3 = 2 28795 + 3 = 29595 x 4 = g x 3 = 2x 3 + 3 = 2 29595 + 3 = 29864 x 5 = g x 4 = 2x 4 + 3 = 2 29864 + 3 = 299549 x 6 = g x 5 = 2x 5 + 3 = 2 299549 + 3 = 29985 x 7 = g x 6 = 2x 6 + 3 = 2 29985 + 3 = 29995 x 8 = g x 7 = 2x 7 + 3 = 2 29995 + 3 = 29998 x 9 = g x 8 = 2x 8 + 3 = 2 29998 + 3 = 29999 x 10 = g x 9 = 2x 9 + 3 = 2 29999 + 3 = 29999 Hence the root of the equation is x 2 2x 3 = 0 is 29999 Find the Real root of the equation x 3 + x 2 100 by Fixed point iteration method Let f x = x 3 + x 2 100 = 0 f 0 = 0 3 + 0 2 100 = 100 ve f 1 = 1 3 + 1 2 100 = 98 ve f 2 = 2 3 + 2 2 100 = 88 ve f 3 = 3 3 + 3 2 100 = 64 ve f 4 = 4 3 + 4 2 100 = 20 ve f 5 = 5 3 + 5 2 100 = +50 +ve The root lies between 4 & 5
Now, g x = 10 1 2 Since x 3 + x 2 100 = 0 x 2 x + 1 = 100 x 2 = 100 x + 1 x = g x = 10 x + 1 = 10 x + 1 1 2 x + 1 3 2 = 5 x + 1 ( 3 2 ) g 4 = 5 4 + 1 ( 3 2 ) = < 1 g 5 = 5 5 + 1 ( 3 2 ) = < 1 So that we can use the iteration method Let x 0 = 4 x 1 = g x 0 = 10 x 0 + 1 = 10 4 + 1 = 10 2236 = 44721 x 2 = g x 1 = 10 x 1 + 1 = 10 44721 + 1 = 10 21147 = 42748 x 3 = g x 2 = 10 x 2 + 1 = 10 42748 + 1 = 43541 x 4 = g x 3 = 10 x 3 + 1 = 10 43541 + 1 = 43217 x 5 = g x 4 = 10 x 4 + 1 = 10 43217 + 1 = 43348 x 6 = g x 5 = 10 x 5 + 1 = 10 43348 + 1 = 43295 x 7 = g x 6 = 10 x 6 + 1 = 10 43295 + 1 = 43316 x 8 = g x 7 = 10 x 7 + 1 = 10 43316 + 1 = 43307 x 9 = g x 8 = 10 x 8 + 1 = 10 43307 + 1 = 43311 x 10 = g x 9 = 10 x 9 + 1 = 10 43311 + 1 = 43310 x 11 = g x 10 = 10 x 10 + 1 = 10 43310 + 1 = 43310 Hence the root of the equation is x 3 + x 2 100 = 0 is 43310
Find the real root of the equation cos x = 3x 1 correct to five decimal places using fixed point iteration method Let f x = cos x 3x + 1 = 0 f 0 = cos 0 3 0 + 1 = 2 +ve f 1 = cos 1 3 1 + 1 = 14597 +ve The root lies between 0 & 1 Since cos x 3x + 1 = 0 3x = cos x + 1 x = 1 3 1 + cos x x = g x = 1 3 1 + cos x Now, g x = 1 3 sin x = 1 sin x 3 g 0 = 1 3 sin 0 = 0 < 1 So that we can use the iteration method Let x 0 = 4 g 1 = 1 sin 1 = 0284 < 1 3 x 1 = g x 0 = 1 3 1 + cos x 0 = 1 3 x 2 = g x 1 = 1 3 1 + cos x 1 = 1 3 x 3 = g x 2 = 1 3 1 + cos x 2 = 1 3 1 + 06536 = 011545 1 + cos 011545 = 06644 1 + cos 06644 = 05957 x 4 = g x 3 = 1 3 1 + cos x 3 = 06092 x 5 = g x 4 = 1 3 1 + cos x 4 = 060669 x 6 = g x 5 = 1 3 1 + cos x 5 = 060717 x 7 = g x 6 = 1 3 1 + cos x 6 = 060708 x 8 = g x 7 = 1 3 1 + cos x 7 = 060710
x 9 = g x 8 = 1 3 1 + cos x 8 = 060710 Hence the root of the equation is cos x = 3x 1 is 060710 Triangularisation method or LU decomposition method 0r LU factorization method Solve the following system by triangularisation method x + y + z = 1, 4x + 3y z = 6, 3x + 5y + 3z = 4 The given system is equivalent to AX = B 1 1 1 4 3 1 3 5 3 x y z = 1 6 4 Let L = 1 0 0 l 21 1 0 l 31 l 32 1, U = u 11 u 12 u 13 0 u 22 u 23, Y = 0 0 u 33 y 1 y 2 y 3 Let LU = A 1 0 0 l 21 1 0 l 31 l 32 1 u 11 u 12 u 13 0 u 22 u 23 = 0 0 u 33 1 1 1 4 3 1 3 5 3 u 11 u 12 u 13 l 21 u 11 l 21 u 12 + u 22 l 21 u 13 + u 23 = l 31 u 11 l 31 u 12 + l 32 u 22 l 31 u 13 + l 32 u 23 + u 33 1 1 1 4 3 1 3 5 3 Equating the coefficients on both sides we get, u 11 = 1, u 12 = 1, u 13 = 1 l 21 u 11 = 4 l 21 1 = 4 l 21 = 4 l 21 u 12 + u 22 = 3 4 1 + u 22 = 3 u 22 = 1 l 21 u 13 + u 23 = 4 1 + u 23 = 1 u 23 = 1 4 = 5 u 23 = 5 l 31 u 11 = 3 l 31 1 = 3 l 31 = 3 l 31 u 12 + l 32 u 22 = 5 3 1 + l 32 1 = 5 l 32 = 3 5 l 32 = 2 l 31 u 13 + l 32 u 23 + u 33 = 3 3 1 + 2 5 + u 33 = 3 u 33 = 3 3 10 u 33 = 10
AX = B LUX = B LY = B were UX = Y 1 0 0 4 1 0 3 2 1 y 1 y 2 y 3 = 1 6 4 By forward substitution method we get y 1 = 1 4y 1 + y 2 = 6 4 1 + y 2 = 6 y 2 = 2 3y 1 2y 2 + y 3 = 4 3 1 2 2 + y 3 = 4 y 3 = 5 UX = Y 1 1 1 0 1 5 0 0 10 x y z = 1 2 5 By backward substitution method we get 10z = 5 z = 1 2 y 5z = 2 y 5 1 2 = 2 y = 5 2 2 y = 1 2 x + y + z = 1 x + 1 2 1 2 = 1 x = 1 Hence x = 1, y = 1 2, z = 1 2 Solve the following system by triangularisation method x + 5y + z = 14, 2x + y + 3z = 13, 3x + y + 4z = 17 The given system is equivalent to AX = B 1 5 1 2 1 3 3 1 4 x y z = 14 13 17 Let L = 1 0 0 l 21 1 0 l 31 l 32 1, U = u 11 u 12 u 13 0 u 22 u 23, Y = 0 0 u 33 y 1 y 2 y 3 Let LU = A
1 0 0 l 21 1 0 l 31 l 32 1 u 11 u 12 u 13 0 u 22 u 23 = 0 0 u 33 1 5 1 2 1 3 3 1 4 u 11 u 12 u 13 l 21 u 11 l 21 u 12 + u 22 l 21 u 13 + u 23 = l 31 u 11 l 31 u 12 + l 32 u 22 l 31 u 13 + l 32 u 23 + u 33 1 5 1 2 1 3 3 1 4 Equating the coefficients on both sides we get, u 11 = 1, u 12 = 5, u 13 = 1 l 21 u 11 = 2 l 21 1 = 2 l 21 = 2 l 21 u 12 + u 22 = 1 2 5 + u 22 = 1 u 22 = 9 l 21 u 13 + u 23 = 3 2 1 + u 23 = 3 u 23 = 1 l 31 u 11 = 3 l 31 1 = 3 l 31 = 3 l 31 u 12 + l 32 u 22 = 1 3 5 + l 32 9 = 1 l 32 = 14 9 l 31 u 13 + l 32 u 23 + u 33 = 4 3 1 + 14 9 1 + u 33 = 4 u 33 = 1 14 9 u 33 = 5 9 AX = B LUX = B LY = B were UX = Y 1 0 0 2 1 0 3 14 9 1 y 1 y 2 y 3 = 14 13 17 By forward substitution method we get y 1 = 14 2y 1 + y 2 = 13 2 14 + y 2 = 13 y 2 = 15 3y 1 + 14 9 y 2 + y 3 = 17 3 14 + 14 9 15 + y 3 = 17 y 3 = 5 3 UX = Y 1 5 1 0 9 1 0 0 5 9 x y z = 14 15 5 3 By backward substitution method we get
5 9 z = 5 3 z = 3 9y + z = 15 9y + 3 = 15 9y = 18 y = 2 x + 5y + z = 14 x + 5 2 + 3 = 14 x = 1 Hence x = 1, y = 2, z = 3