{VERSION 4 0 "APPLE_PPC_MAC" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "" -1 256 "" 0 1 93 41 59 0 0 1 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 257 "" 0 1 93 41 59 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 0 1 93 41 59 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 0 1 93 41 59 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 261 "" 0 1 93 41 59 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE " " -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 16 "LU Decomposition" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT 259 35 "Part 1. The basic LU decomposition" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "with(linalg) :\nA:=matrix([[1,-3,1],[2,-4,2],[2,2,-3]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 185 "Recall that elementary matrices are created by applying a single row operation to an identity matrix. Complete step s 1-4 in the space below, adding text and command lines as necessary. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "I3:=diag(1,1,1):\nE1:=addrow(I3,1,2,-2);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 162 "T he following command computes an LU decomposition of the specified mat rix and assigns names to the lower and upper triangular factors. To s ee both factors, use " }{TEXT 263 5 "print" }{TEXT -1 4 " or " }{TEXT 262 5 "evalm" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "LUdecomp(A,L='L2',U='U2'): print(L2); pri nt(U2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 22 "Part 2. \+ More examples" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "B:=matrix([[4,2,2,-2],[2,0,5,1],[-2,-3,5,4]]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "Answer steps 1-3 here, ad ding lines as necessary." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "Us e the following matrix for step 4." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "C:=matrix([[1,2,3],[-2,-1,-5],[-1,4 ,-3],[2,1,1]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 50 "Part 3. Solving Equatio ns using LU decompositions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "A:=matrix([[1,-3,1],[2,-4,2],[2,2,-3]]);\nb :=[-10,4,1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 134 "Note tha t multiplying Row 1 by -2 and adding it to Row 2 requires 8 flops (4 m ultiplications and 4 additions). Answer steps 1-3 here." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "Use the following vector in step 4." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "c:=[1,4,0]; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 257 33 "Part 4. A modified decomposition" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "Enter R a nd use it for steps 1-4." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "R:=matrix([[0,3,-2],[-1,3,0],[2,3,-6],[-2,0,4 ]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "For step 5, enter \+ S and use the given commands to find U. Then explain how to find L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "S :=matrix([[1,1,-1,0],[-1,-1,1,-1],[-1,0,1,-1],[1,0,-1,0],[-1,-1,1,1]]) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "addrow(S,1,2,1): addro w(%,1,3,1): addrow(%,1,4,-1): addrow(%,1,5,1);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 15 "swaprow(%,2,3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "addrow(%,2,4,1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "addrow(%,3,4,-1): addrow(%,3,5,1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 16 "Part 5. Summary" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "Answer the summary questi ons here." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }