nn nn n n n n n n n n a a a a a a a a a a a a a a a a a k1after first iteration

# Nn nn n n n n n n n n a a a a a a a a a a a a a a a a

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. . . . . . . . . . . . . . . . 0 . . 0 . . nn nn n n n n n n n n a a a a a a a a a a a a a a a a a k=1(after first iteration) i1 ij ij 1j 11 a a = a - a × a Where i = 2, n ; j = 1, n+1 1 3 1 3 3 33 1 2 2 23 22 1 1 1 13 12 11 . . 0 0 . . . . . . . . . . . . . . . . 0 0 . . 0 . . nn nn n n n n n n n a a a a a a a a a a a a a a a k=2 (after second iteration)  i2 ij ij 2j 22 a a = a - a × a Where i = 3, n ; j = 2, n+1
Forward Elimination: Algorithm June 21, 2016 ME262 Numerical Analysis Sessional 12 where i = 2, n ; j=1, n+1 k= 1 ik ij ij kj kk a a = a - a × a where k = 1, n-1 i = k+1, n ; j=k, n+1 i1 ij ij 1j 11 a a = a - a × a  i2 ij ij 2j 22 a a = a - a × a      i3 ij ij 3j 33 a a = a - a × a k=1,2, 3 … n- 1 k= 2 where i = 3, n ; j=2, n+1 k= 3 where i = 4, n ; j=3, n+1
Forward Elimination: Code June 21, 2016 ME262 Numerical Analysis Sessional 13 where k = 1, n-1 (iteration index) i = k+1, n (row index) j = k, n+1 (column index) for k=1:n-1 for i=k+1:n factor=a ik / a kk ; for j=k:n+1 a ij =a ij -factor × a kj ; end end end ik ij ij kj kk a a = a - a × a
Backward Substitution June 21, 2016 ME262 Numerical Analysis Sessional 14 3 33 2233 2 22 1122133 1 11 11121311 222322 3333
Backward Substitution June 21, 2016 ME262 Numerical Analysis Sessional 15 Perform back substitution to get {X} {X} by simple division Substitute this into (n-1) th equation Solve for x n-1 Repeat the process to solve for x n-2 , x n-3 , …. x 2 , x 1 ) ( ) ( , ) ( , 2 n 1 n n 2 n n 1 n 1 n 2 n 1 n 1 n b x a x a ) ( ) ( 1 n nn 1 n n n a b x
Backward Substitution June 21, 2016 ME262 Numerical Analysis Sessional 16 j n i j ij in ii i x a a a x 1 1 1 11 12 13 14 1 1 1 22 23 24 2 2 1 33 34 3 3 1 44 4 4 1 1 1 1 . 0 . 0 0 . 0 0 0 . . . . . . . . 0 0 0 0 . n n n n n n n n n n nn nn a a a a a a a a a a a a a a a a a a a a       
Problem June 21, 2016 ME262 Numerical Analysis Sessional 17 By using gauss elimination method find the augmented upper triangular matrix of the following equations: 0 2 2 2 3 2 2 7 3 4 6 5 6 6 4 2 4 3 2 1 4 2 1 4 3 2 1 x x x x x x x x x x x x x 0

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