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"" }}{PARA 270 "" 0 "" {TEXT 260 0 "" }}{PARA 270 "" 0 "" {TEXT 260 16 "By Russell Blyth" } {TEXT 260 0 "" }}{PARA 270 "" 0 "" {TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 29 "restart: with(LinearAlgebra):" }{MPLTEXT 1 261 0 "" }}}{SECT 1 {PARA 272 "" 0 "" {TEXT 262 43 "Compute rank usi ng row and column reduction" }{TEXT 262 0 "" }}{EXCHG {PARA 270 "" 0 " " {TEXT 260 45 "Define a matrix A (A was generated randomly)." }{TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 94 "A := Matrix( [[4, -1, -4, -1], [-1, -4, -2, 3], [4, 4, -3, -3], [4, -1, 0, 5], [1, \+ 5, -2, 3]]);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 65 "We also generate the 5x5 and 4x4 identity matrices for later u se." }{TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 24 " I5 := IdentityMatrix(5);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 7 "\nI4 := " }{MPLTEXT 1 261 15 "IdentityMatrix(" }{MPLTEXT 1 261 3 "4);" } {MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 135 "It's li kely that the randomly generated matrix A has rank 4 (why?); we reduce it using elementary row and column operations to the form" }{TEXT 260 0 "" }}{PARA 273 "" 0 "" {TEXT 263 1 " " }{XPPEDIT 2 0 "Matrix([[I r, O1], [O2, O3]])" "6#-I'MatrixG6$I*protectedGF&I(_syslibG6\"6#7$7$I# IrGF(I#O1GF(7$I#O2GF(I#O3GF(" }{TEXT 263 2 " ," }{TEXT 263 0 "" }} {PARA 270 "" 0 "" {TEXT 260 295 "where r is the rank of the matrix A \+ and O1, O2, O3 are matrices with all entries equal to zero, as a chec k. We keep track of the elementary operations we use, in the form of e lementary matrices, so we can confirm that the row and column reductio n can be accomplished via matrix multiplication." }{TEXT 260 0 "" }}} {EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 28 "DD := RowOperation(A,1,1 /4);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 30 "\nE1 := RowOperation(I5 ,1,1/4);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 47 "Next eliminate the entries below the leading 1." }{TEXT 260 0 "" } }}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 31 "DD := RowOperation(DD, [2,1],1);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 32 "\nE2 := RowOperati on(I5,[2,1],1);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 32 "DD := RowOperation(DD,[3,1],-4);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 33 "\nE3 := RowOperation(I5,[3,1],-4);" } {MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 32 "D D := RowOperation(DD,[4,1],-4);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 33 "\nE4 := RowOperation(I5,[4,1],-4);" }{MPLTEXT 1 261 0 "" }}} {EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 32 "DD := RowOperation(DD,[5 ,1],-1);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 33 "\nE5 := RowOperatio n(I5,[5,1],-1);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 113 "Next we use elementary column operations to eliminate t he entries in the first row to the right of the leading 1." }{TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 36 "DD := Column Operation(DD,[2,1],1/4);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 37 "\nG 1 := ColumnOperation(I4,[2,1],1/4);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 34 "DD := ColumnOperation(DD,[3,1], 1);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 35 "\nG2 := ColumnOperation( I4,[3,1],1);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 36 "DD := ColumnOperation(DD,[4,1],1/4);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 37 "\nG3 := ColumnOperation(I4,[4,1],1/4);" } {MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 163 "We have cleared out the entries below and to the right of the leading 1. We n ow start over with the lower right submatrix, ignoring the first row a nd column of DD." }{TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 31 "DD := RowOperation(DD,2,-4/17);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 32 "\nE6 := RowOperation(I5,2,-4/17);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 32 "DD := RowOpe ration(DD,[3,2],-5);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 33 "\nE7 := RowOperation(I5,[3,2],-5);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 35 "DD := RowOperation(DD,[5,2],-21/4);" } {MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 36 "\nE8 := RowOperation(I5,[5,2], -21/4);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 39 "DD := ColumnOperation(DD,[3,2],-12/17);" }{MPLTEXT 1 261 0 " " }{MPLTEXT 1 261 40 "\nG4 := ColumnOperation(I4,[3,2],-12/17);" } {MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 38 "D D := ColumnOperation(DD,[4,2],11/17);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 39 "\nG5 := ColumnOperation(I4,[4,2],11/17);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 274 "" 0 "" {TEXT 264 47 "Now we move on to the low er right 3 x 2 matrix." }{TEXT 264 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 32 "DD := RowOperation(DD,3,-17/43);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 33 "\nE9 := RowOperation(I5,3,-17/43);" } {MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 32 "D D := RowOperation(DD,[4,3],-4);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 34 "\nE10 := RowOperation(I5,[4,3],-4);" }{MPLTEXT 1 261 0 "" }}} {EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 35 "DD := RowOperation(DD,[5 ,3],80/17);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 37 "\nE11 := RowOper ation(I5,[5,3],80/17);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 38 "DD := ColumnOperation(DD,[4,3],21/43);" } {MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 39 "\nG6 := ColumnOperation(I4,[4, 3],21/43);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 49 "Now we are at the final lower right 2 x 1 matrix." }{TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 32 "DD := RowOperati on(DD,4,43/342);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 34 "\nE12 := Ro wOperation(I5,4,43/342);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> \+ " 0 "" {MPLTEXT 1 261 37 "DD := RowOperation(DD,[5,4],-187/43);" } {MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 39 "\nE13 := RowOperation(I5,[5,4] ,-187/43);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 153 "The matrix DD is now in the form specified above; note O2 and O3 are not present (they are 0 x 0 matrices, if you prefer), and O2 i s a 4 x 1 zero matrix." }{TEXT 260 0 "" }}{PARA 270 "" 0 "" {TEXT 260 0 "" }}{PARA 270 "" 0 "" {TEXT 260 74 "Compute the products of the ele mentary row and column matrices separately." }{TEXT 260 0 "" }}} {EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 48 "B := E13.E12.E11.E10.E9. E8.E7.E6.E5.E4.E3.E2.E1;" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> \+ " 0 "" {MPLTEXT 1 261 23 "C := G1.G2.G3.G4.G5.G6;" }{MPLTEXT 1 261 0 " " }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 112 "Check that left multipica tion by B and right multiplication by C accomplish the row and column \+ reductions above." }{TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 6 "B.A.C;" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 20 "The final matrix DD!" }{TEXT 260 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 182 "Exercise: Reduce the following matrix to the form above. Keep track of the elementary matrices involved and compute the matrix product corresponding to the row and column reduct ions." }{TEXT 260 0 "" }}{PARA 270 "" 0 "" {TEXT 260 118 "(Suggestion: use different names for matrices than those used above. The variable \+ D is protected, and cannot be used.)" }{TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 51 "AA := Matrix([[1,2,1,3,3],[-1,1,2,1,3 ],[1,0,1,1,1]," }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 33 "\n \+ [1,3,4,5,7]]);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 0 "" }}} }{SECT 1 {PARA 272 "" 0 "" {TEXT 262 8 "Inverses" }{TEXT 262 0 "" }} {EXCHG {PARA 270 "" 0 "" {TEXT 260 68 "Let's investigate the algorithm for finding inverses via an example." }{TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 44 "A:= RandomMatrix(3,3,generator= rand(-3..3));" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 139 "Find the inverse of the matrix A. The equation Y = AX compute s Y given X. We need to compute X given Y. Define general Y, and solve for X: " }{TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 16 "Y := ;" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 15 "\nA ugA := ;" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 37 "\nRedA := Redu cedRowEchelonForm(AugA);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 48 "Let's extract the expression for X in terms of Y" } {TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 23 "Xvec : = Column(RedA,4);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 35 "y1vec := subs(y1=1,y2=0,y3=0,Xvec);" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 36 "\ny2vec := subs(y1=0,y2=1,y3=0,Xvec);" } {MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 36 "\ny3vec := subs(y1=0,y2=0,y3=1 ,Xvec);" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 207 "Thus Xvec = y1*y1vec + y2*y2vec + y3*y3vec. If we make a matrix w ith y1vec, y2vec, y3vec as the columns, then y1, y2 and y3 are the coe fficients of the linear combination giving X. We write this as BY, whe re" }{TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 25 "B := ;" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 6 "Check:" }{TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 " " {MPLTEXT 1 261 4 "B.Y;" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 270 "" 0 "" {TEXT 260 57 "This is Xvec. Thus X = BY. Hence B is the inverse of A!" }{TEXT 260 0 "" }}{PARA 270 "" 0 "" {TEXT 260 0 "" }}{PARA 270 "" 0 "" {TEXT 260 29 "Check the products AB and BA:" }{TEXT 260 0 "" }}}{EXCHG {PARA 271 "> " 0 "" {MPLTEXT 1 261 4 "A.B;" }{MPLTEXT 1 261 0 "" }{MPLTEXT 1 261 5 "\nB.A;" }{MPLTEXT 1 261 0 "" }}}{EXCHG {PARA 275 "" 0 "" {TEXT 263 61 "Note the first column of B arises from solving AX = [1, 0, 0]" }}}}{PARA 276 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 15 10 1804 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }