6. For any m x m matrix A, Gauss Elimination without pivoting consists of: Algorithm 2...

  

Transcribed Image Text:

6. For any m x m matrix A, Gauss Elimination without pivoting consists of: Algorithm 2 Gauss Elimination Without Pivoting in words for k=1 to m-1 Add suitable multiples of row k to the rows beneath to introduce zeroes below the main diagonal in column k. end (a) Show that each iteration of the above algorithm can be effected by left- multiplying A by a matrix Lx=1-le; where l is the vector of multipliers for the kth column of A (the first k entries of t are 0) and e, is a vector in C with one in the kth position and zeroes elsewhere. Give a simple formula for the non-zero entries of (b) Show that for each k, L¹ = 1 + le (c) Show that the matrix L = L₁ L₂... L- is just 1+le; ++lmem. (d) Explain briefly why the result in (c) means that L is lower triangular. (e) Explain briefly why A can be factored into A - LU after the algorithm has completed. (See over for the rest of Q.6.) (f) When partial pivoting is applied, we have Lm-1Pm-1Lm-2Pm-2... L2P₂L1 P₁A=U where each P; swaps row j with one of the rows j+1,..., m (if necessary) to make the absolute value of the "pivot" A, as large as possible. Defining: show that TT₁ = Pm-1Pm-2... P Bened Lm-1 Pm-1 Lm-2Pm-2... L₂P₂L₁ P₁ = L^_L_2... L₂L ₁. Hint: first show that I₁ = P₁1 and then that L., LT-2L+1 PL (g) Show that the LU factorisation A = LU (without pivoting) is now replaced by PALU (with pivoting). Here P = ₁ and L-LL...L₁. (h) Finally, show that the matrix L is lower triangular as it case. in the no-pivoting 2 ot 7 6. For any m x m matrix A, Gauss Elimination without pivoting consists of: Algorithm 2 Gauss Elimination Without Pivoting in words for k=1 to m-1 Add suitable multiples of row k to the rows beneath to introduce zeroes below the main diagonal in column k. end (a) Show that each iteration of the above algorithm can be effected by left- multiplying A by a matrix Lx=1-le; where l is the vector of multipliers for the kth column of A (the first k entries of t are 0) and e, is a vector in C with one in the kth position and zeroes elsewhere. Give a simple formula for the non-zero entries of (b) Show that for each k, L¹ = 1 + le (c) Show that the matrix L = L₁ L₂... L- is just 1+le; ++lmem. (d) Explain briefly why the result in (c) means that L is lower triangular. (e) Explain briefly why A can be factored into A - LU after the algorithm has completed. (See over for the rest of Q.6.) (f) When partial pivoting is applied, we have Lm-1Pm-1Lm-2Pm-2... L2P₂L1 P₁A=U where each P; swaps row j with one of the rows j+1,..., m (if necessary) to make the absolute value of the "pivot" A, as large as possible. Defining: show that TT₁ = Pm-1Pm-2... P Bened Lm-1 Pm-1 Lm-2Pm-2... L₂P₂L₁ P₁ = L^_L_2... L₂L ₁. Hint: first show that I₁ = P₁1 and then that L., LT-2L+1 PL (g) Show that the LU factorisation A = LU (without pivoting) is now replaced by PALU (with pivoting). Here P = ₁ and L-LL...L₁. (h) Finally, show that the matrix L is lower triangular as it case. in the no-pivoting 2 ot 7

Answer rating: 100% (QA)

for any mxm Gauss of Algorthim 2 Elimination pivoting for k1 mal So that each matrin A above lett m... View the full answer