Question

(911 (1) (a) Recall that a square matrix A has an LU decomposition if we can write it as the product A = LU of a lower triangular matrix and an upper triangular matrix. Show that the matrix 0 1 21 A= 3 4 5 (6 7 9] does not have an LU decomposition 0 0 Uji U12 U13 O 1 2 Il 21 l22 0 0 U22 U23 = 3 4 5 (131 132 133 0 0 U33 6 7 9 Hint: use that the upper-left entry of A is zero and consider the determinants of L, U, and A. (b) To get around the problem that not all matrices have LU decompositions, the idea of “pivoting” is used. A square matrix P is a permutation matrix if it has exactly one entry of 1 in each row and each column and entries of 0 elsewhere. For a square matrix A, an LU decomposition with partial pivoting is a decomposition of the form LU=PA where L is lower triangular, U is upper triangular, and P is a permutation matrix. Using the permutation matrix 0 0 1] P=100 0 1 0 find an LU decomposition with partial pivoting for the matrix A given in part (a) above and with the further property that the lower triangular matrix L has ones on its diagonal. [6+4 = 10 marks]

Answer 1

Page: 1 O I (a) A- 1 2 3 4 5 6 7 9 Iwe Show A does not have Lu decomposition. as det A = 01/27 - 30)+2 (21-24) 3-6-3 . det li da 2 d33 lo lai dao lgi dar da Ull uiz 43 det - un 933 O U22 U23 a33 So if LU - det (Lu) - det (A) det (L) . det (U) - det (A) - = (111d 22 lgg). (4,1 422 423) det (2) to & det (ulto li to, uito ㅋ
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Page : 2 lll up to A but du ull 0 1 27 Luc - 3 45 6 7 9 با ۱ LU- ДЧ, о but we show la kito A ha A does not have Lu decomp. I Soi pa c (6 o 6 2 PA :10 6 7 9 - 6 7 1 2 O 3 4 S
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Page 3 Now we find Lu decomposition of PA 6 7 91 PA 2 R₂ R₂ - ( 1 ) Ri (131 E” 3 y 5 n 6 7 9 I 2 R3 – Rz - ( 42 ) R2 (: 13, 22 1 2 12 12 7 9 - o / L is just made up of the multipliers we used in Gaussian elimination with l's on the diagonal. 1 o / / 1 a 6 7 9 6 7 1 O 1 2 2 3 4 5 / 1 -1/2 L
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