{VERSION 5 0 "APPLE_PPC_MAC" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 2 0 0 0 0 0 0 0 1 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 6 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 4 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 3" -1 5 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Warning" -1 7 1 {CSTYLE "" -1 -1 "Co urier" 1 10 0 0 255 1 2 2 2 2 2 1 1 1 3 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Out put" -1 12 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Exercise" -1 256 1 {CSTYLE " " -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 -12 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 4 "" 0 "" {TEXT -1 27 "2.3 Applications to System s" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "In-c lass demo by Russell Blyth." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "restart: with(linalg): with(plots): with(plottools):" }}{PARA 7 " " 1 "" {TEXT -1 80 "Warning, the protected names norm and trace have b een redefined and unprotected\n" }}{PARA 7 "" 1 "" {TEXT -1 50 "Warnin g, the name changecoords has been redefined\n" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 17 "Some computations" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "We start with a random 4 x 6 matrix." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "A := randmatrix(4,6,entries=rand(-5..5)); rank( A);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG-%'matrixG6#7&7(\"\"%!\" \"!\"%F+F+F,7(!\"#\"\"$F*F*!\"$F07(F*F+\"\"!\"\"&\"\"\"F37(F.F/!\"&F6F 4F3" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"%" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 62 "Does the vector (3,7,-5,3,2,-10) belong to the row spac e of A?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "We need to solve the following system:" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 30 "B:=vector(6,[3,7,-5,3,2,-10]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"BG-%'vectorG6#7(\"\"$\"\"(!\"&F)\"\"#!#5" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "k1*row(A,1)+k2*row(A,2)+k3*r ow(A,3)+k4*row(A,4)=evalm(B);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#/,**& %#k1G\"\"\"-%'vectorG6#7(\"\"%!\"\"!\"%F-F-F.F'F'*&%#k2GF'-F)6#7(!\"# \"\"$F,F,!\"$F6F'F'*&%#k3GF'-F)6#7(F,F-\"\"!\"\"&F'F=F'F'*&%#k4GF'-F)6 #7(F4F5!\"&FCF'F=F'F'-F)6#7(F5\"\"(FCF5\"\"#!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "We get the augmented matrix as follows:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "AAug := augment(transpose(A),B);" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%AAugG-%'matrixG6#7(7'\"\"%!\"#F*F+ \"\"$7'!\"\"F,F.F,\"\"(7'!\"%F*\"\"!!\"&F37'F.F*\"\"&F3F,7'F.!\"$\"\" \"F8\"\"#7'F1F7F5F5!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Put in \+ reduced echelon form:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "AA ugGJ := gaussjord(AAug);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'AAugGJG -%'matrixG6#7(7'\"\"\"\"\"!F+F+F+7'F+F*F+F+F+7'F+F+F*F+F+7'F+F+F+F*F+7 'F+F+F+F+F*7'F+F+F+F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "We see that the system is inconsistent, so B does not lie in the row space o f A." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 95 "N ow let's find a basis for the row space of A - find the reduced row ec helon form for A itself:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "AGJ := gaussjord(A);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$AGJG-%'mat rixG6#7&7(\"\"\"\"\"!F+F+#!$R#\"#!*#!$$G\"#O7(F+F*F+F+#!#)*\"#X#!#\"* \"#=7(F+F+F*F+#!\"(\"\"$#!#V\"\"'7(F+F+F+F*#\"#<\"\"*#\"$8\"F8" }}} {EXCHG {PARA 5 "" 0 "" {TEXT -1 0 "" }{TEXT 256 102 "Explicitly check \+ that each original row of A is the appropriate linear combination of t he rows of AGJ:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "row(A,1) ;\nevalm(A[1,1]*row(AGJ,1)+A[1,2]*row(AGJ,2)+A[1,3]*row(AGJ,3)+A[1,4]* row(AGJ,4));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7(\"\"%!\" \"!\"%F(F(F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7(\"\"%!\" \"!\"%F(F(F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "row(A,2);\n evalm(A[2,1]*row(AGJ,1)+A[2,2]*row(AGJ,2)+A[2,3]*row(AGJ,3)+A[2,4]*row (AGJ,4));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7(!\"#\"\"$\" \"%F)!\"$F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7(!\"#\"\"$ \"\"%F)!\"$F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "row(A,3); \nevalm(A[3,1]*row(AGJ,1)+A[3,2]*row(AGJ,2)+A[3,3]*row(AGJ,3)+A[3,4]*r ow(AGJ,4));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7(\"\"%!\" \"\"\"!\"\"&\"\"\"F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7( \"\"%!\"\"\"\"!\"\"&\"\"\"F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "row(A,4);\nevalm(A[4,1]*row(AGJ,1)+A[4,2]*row(AGJ,2)+A[4,3]*row( AGJ,3)+A[4,4]*row(AGJ,4));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vecto rG6#7(!\"#\"\"$!\"&F)\"\"\"\"\"&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-% 'vectorG6#7(!\"#\"\"$!\"&F)\"\"\"\"\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "Check again to see why B is not in the row space, using \+ the coefficients that would make the first four entries correct:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "evalm(B);\nevalm(B[1]*row(AG J,1)+B[2]*row(AGJ,2)+B[3]*row(AGJ,3)+B[4]*row(AGJ,4));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7(\"\"$\"\"(!\"&F'\"\"#!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7(\"\"$\"\"(!\"&F'#!$H&\"#!*#!$b \"\"#O" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Now let's find a basis \+ for the column space of A" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "At := transpose(A);\nAtGJ := gaussjord(At);\nrank(At);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#AtG-%'matrixG6#7(7&\"\"%!\"#F*F+7&!\"\"\" \"$F-F.7&!\"%F*\"\"!!\"&7&F-F*\"\"&F27&F-!\"$\"\"\"F77&F0F6F4F4" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%%AtGJG-%'matrixG6#7(7&\"\"\"\"\"!F+F +7&F+F*F+F+7&F+F+F*F+7&F+F+F+F*7&F+F+F+F+F/" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 104 "Now look at the null space of A. Actually, we get a better basis if we use the reduced row echelon form:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "evalm(AGJ);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'matrixG6#7&7(\" \"\"\"\"!F)F)#!$R#\"#!*#!$$G\"#O7(F)F(F)F)#!#)*\"#X#!#\"*\"#=7(F)F)F(F )#!\"(\"\"$#!#V\"\"'7(F)F)F)F(#\"#<\"\"*#\"$8\"F6" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 53 "BasisNS1 := nullspace(A);\nBasisNS2 := nulls pace(AGJ);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%)BasisNS1G<$-%'vectorG 6#7(\"\"\"#\"$V$\"#i\"\"!#\"#]\"#J#\"$X'F-#!$0\"F1-F'6#7(F.#!$l'\"$C\" F*#!##)F1#!%:9F;#\"$R#F-" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)BasisNS 2G<$-%'vectorG6#7(#\"$R#\"#!*#\"#)*\"#X#\"\"(\"\"$#!#<\"\"*\"\"\"\"\"! -F'6#7(#\"$$G\"#O#\"#\"*\"#=#\"#V\"\"'#!$8\"F@F7F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 304 "In either case the null space has dimension two, that is, nullity(A)=2. \+ The second basis clearly shows the basis elements corresponding to the nonpivot columns of AGJ. It is also clear that these two vectors are independent. Check that any linear combination of the basis vectors l ies in the null space:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "x := c1*BasisNS2[1]+c2*BasisNS2[2];\nevalm(A &* x);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"xG,&*&%#c1G\"\"\"-%'vectorG6#7(#\"$$G\"#O#\"#\"* \"#=#\"#V\"\"'#!$8\"F2\"\"!F(F(F(*&%#c2GF(-F*6#7(#\"$R#\"#!*#\"#)*\"#X #\"\"(\"\"$#!#<\"\"*F(F8F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vec torG6#7&\"\"!F'F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "Fi nally, note that rank(A) + nullity(A) = 4 + 2 = 6 = # columns of A." } }}}}{MARK "1 0 0" 0 }{VIEWOPTS 1 1 0 3 2 1804 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }