{VERSION 6 0 "IBM INTEL NT" "6.0" } {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 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{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 "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 12 "Homework #11" }}{PARA 0 "" 0 "" {TEXT 257 20 "Section 5.3. Ex. 6a)" }{TEXT -1 157 " If the v ectors were on the same line they would be all multiples of the same v ector. Since this is not the case, the vectors do not belong to the sa me line." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 258 6 "Ex. 8)" } {TEXT -1 195 " We can see that v1+v2-v3=0 so the three vectors are lin early dependent. From that equation we can solve for v3, v2 and v1 in \+ terms of the other vectors. For example v3=v1+v2, v2=v3-v1, v1=v3-v2. " }}{PARA 0 "" 0 "" {TEXT 259 5 "Ex 9)" }{TEXT -1 213 ". The vectors f orm a lineary dependent set iff the matrix formed with their component s has less than three leading ones. This happens if the determinant of the matrix if zero. So we will solve for the determinant:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "with(LinearAlgebra):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "A:=Matrix([[lambda,-1/2,-1/2],[-1/2 ,lambda, -1/2],[-1/2,-1/2,lambda]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "d:=Determinant(A);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "solve(d=0,lambda);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "As it can be seen, the solutions to the determinant of the matr ix equal zero are 1, and -1/2." }}{PARA 0 "" 0 "" {TEXT 260 19 "Sectio n 5.4 Ex. 4b)" }{TEXT -1 263 " Notice that the sum of the second and t hird polynomials equals the first polynomial, therefore the three vect ors are not linearly independent. Since the dimension of P2 is 3, we n eed three vectors for a basis, therefore the given vectors cannot be a basis in P2." }}{PARA 0 "" 0 "" {TEXT 261 7 "Ex. 7b)" }{TEXT -1 108 " We need to solve the vector equation: c1u1+c2u2=w. This leads to a li near system whose augmented matris is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "A1:=Matrix([[2,3,1],[-4,8,1]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ReducedRowEchelonForm(A1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "The values in the third column in the RREF abov e are the coeeficients c1 and c2 respectively." }}{PARA 0 "" 0 "" {TEXT 262 7 "Ex. 9a)" }{TEXT -1 90 " As above, we need to solve the ve ctor equation: c1v1+c2v2+c3v3=v, whihc leads to a system" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "A2:=Matrix([[1,2,3,2],[0,2,3,-1],[0 ,0,3,3]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ReducedRowEch elonForm(A2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "The solution is \+ seen in the fourth column in the RREF above." }}{PARA 0 "" 0 "" {TEXT 264 7 "Ex. 14)" }{TEXT -1 50 ". We write the matrix of the system and \+ reduce it:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "A3:=Matrix([[ 1,-4,3,-1],[2,-8,6,-2]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ReducedRowEchelonForm(A3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 245 "The solution set is given by: (4u-3s+t,u,s,t)=u(4,1,0,0)+s(-3,0,1 ,0)+t(1,0,0,1). So the dimension of the solution space (which is the n ullspace of the matrix) is three, and the basis vectors appear above. \+ They can be obtained with Maple as well:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "NullSpace(A3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 263 9 "Ex. 18a) " }{TEXT -1 298 "Look at the given equation as a system in three unknowns with only one equation. We have to find the nullspace \+ of the matrix again. With z=3t, y=3s we get x=2s-5t, so the solution i s: (2s-5t,3s,3t)=s(2,3,0)+t(-5,0,3), and again the two vectors are a b asis in the solution space. The dimension is 2." }}{PARA 0 "" 0 "" {TEXT 265 4 "18c)" }{TEXT -1 108 " We have a line, and if we write it \+ in vector form: (2t, -t, 4t)=t(2,-1,4), we see that a basis is (2,-1,4 )." }}{PARA 0 "" 0 "" {TEXT 266 19 "Section 5.5 Ex. 3c)" }{TEXT -1 276 " Let us write the augmented matrix. The vector b is in the column space of the matrix if and only if the augmented matrix does not have a leading one in the last column, which ie equivalent to saying that \+ rank(A4)=rank(A4|b) (A4 is the name of the matrix in the Maple output) ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "A4:=Matrix([[1,-1,1,5] ,[9,3,1,1],[1,1,1,-1]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ReducedRowEchelonForm(A4);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "As you can see rank(A4)=3 so b is in the column space of the given matri x." }}{PARA 0 "" 0 "" {TEXT 267 8 "Ex. 6c)." }{TEXT -1 53 " We let Map le do everything (we trust it, don't we?) " }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 45 "A5:=Matrix([[1,4,5,2],[2,1,3,0],[-1,3,2,2]]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ReducedRowEchelonForm(A5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "NullSpace(A5);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 268 9 "Ex. 11b) " }{TEXT -1 147 "We form \+ a matrix in which the given vectors are rows. Then we will find a basi s for the Row space of the amatrix by getting the RREF of the matrix. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "A6:=Matrix([[-1,1,-2,0] ,[3,3,6,0],[9,0,0,3]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 " ReducedRowEchelonForm(A6);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 225 "Si nce we get three leading ones, the three vectors are linearly independ ent, so a basis the space they span can be formed with the original ve ctors, but the rows in the above RREF can be chosen as well. Here is M aple at work:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "RowSpace(A 6);" }}}}{MARK "28" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }