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# ELEM DIFF EQ & LIN A MATH 2090

LSU

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This 8 page Class Notes was uploaded by Madison Gottlieb Sr. on Tuesday October 13, 2015. The Class Notes belongs to MATH 2090 at Louisiana State University taught by Staff in Fall. Since its upload, it has received 17 views. For similar materials see /class/222637/math-2090-louisiana-state-university in Mathematics (M) at Louisiana State University.

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Date Created: 10/13/15

Solutions to some Math 20903 homework problems 41 19 The inversions of the permutation n n 7 1 n 7 2 2 1 are given by n n 71nn 7 2 n 1 71 i 1 n i 27 71 i 1 n i 37 a 71 i 1 1 3727371 2 1 Note that there are n 7 1 inversions in the first row 71 7 2 in the second rOW and nally 1 in the last row Therefore the total number of inversions equals to 710171 n71n721 2 Thus the sign attached to the term a1na2n71a3n72 39 39 39 an 1 is 41 W1W 42 21 The determinant of the coe icient matrix is given by 1 2 k 2 7k 1 7k612k3k274763k211k743k71k4 3 6 1 Note that the given system has an in nite number of solutions if and only if the above determinant is zero Which yields the values of k 13 and k 4 42 27 Bring out 71 from the first row and 12 from the second row to get 1 73 1 1 71 3 71 2 71 7 75 4 72 14 3 1 13 3 1 13 Then add the second row to the first row 71 3 71 3 1 13 4 72 14 4 72 14 3 1 13 3 1 13 Since the first and third rows are identical the determinant equals to 0 1 42 29 First note that detA is a polynomial of degree 3 Hence the equation detA 0 has three roots Then observe the following a Whne x 0 the third column is zero and so detA 0 b When at 71 the second and third columns are identical and so detA 0 c When at 2 the first and third columns are proportional and so detA 0 Therefore we conclude that all values of x for which detA 0 are 0 71 and 2 42 39 First note that the determinant is a polynomial in 442 of degree 3 Then observe the following a When we put at y the determinant is 0 because the first and second rows are identical This means that the determinant has a factor x 7 y b When we put at z the determinant is 0 because the first and third rows are identical This means that the determinant has a factor x 7 z c When we put y z the determinant is 0 because the second and third rows are identical This means that the determinant has a factor y 7 2 From the above facts we conclude that to 1 x 1 y2 cx7yz7zy7z 1 1 22 N168 where c is a constant To determine the constant a note that the determinant has a term 2122 On the other hand the right hand side of Equation 1 has a term 702422 Hence 0 71 Finally put 0 71 into Equation 1 to get 2 1 1 7y9672y72967yy722796 N168 N168 43 20 Use the first row cofactor exapnsion to get is 35 11 f1 7 1 71 7x 0 i1 710 0 1 i 71 0 1 716 7y 0 1 y 7y 71 1 7 z 7y 71 0 y 72 i1 0 72 1 0 72 1 i1 72 1 71 0 796727y796yy296727967y72 Macyzgtyyzzyz yzyz xyz2 On the other hand this problem can also be solved in an indirect way as follows First add the third and fourth columns to the second column to get y z 0 yz y z 716 1 71 7x 0 1 71 7x 0 1 71 7y 71 0 1 7y 0 0 1 yz 7y 0 1 72 1 i1 0 72 0 i1 0 2 1 0 Then in the last determinant add the second and third rows to the first row to obtain inc 1 71 ixiyiz 0 0 0 1 7y 0 l 7y 0 l fariyiz 71 0 7y2 72 71 0 72 71 0 Therefore we have 70 E i 51 96 1 1 7 71 0 1 iyz 7y 0 1 xyz2 y 72 71 72 71 0 52 11 The answer in the book is wrong The set R2 with the addition and scalar multiplication as defined in this problem is not a vector space First proof The set R2 is not closed under scalar multiplication For example which is not an element in R2 Second proof It is easy to see that the element satisfying Axiom 3 is given by 01 But then 170 azb 1 a707 which is impossible to be equal to 0 1 This means that 1 0 has no additive inverse for any 1 Thus Axiom 4 is not satisfied 57 9 We use the given four vectors in R4 to form the following matrix El 1 I NH OOJgtOOgtJgt NOWH What we need to find is a basis for the rowspace of this matrix Apply the row operations 1 multiply the first row by 72 and then add it to the second and fourth rows 2 multiply the first row by 71 and then add it to the third row This results the following matrix 1 4 l 3 0 0 l 71 0 0 71 l 0 0 0 0 l 4 l 3 0 0 l 71 0 0 0 0 i0 0 0 OJ Therefore the first and second row vectors form a basis for the row space and so the vectors 14 13 and 0 0 171 form a basis for the space spanned by the given four vectors 61 1900 First we need to write 1102 as a linear combination of 171 l l and 172 171 and so let 12 011 1 021 1 Comparing the components we see that 01 02 1 and 01 7 02 2 Solve for 01 and 02 to get 1 2 951 i 952 02 2 2 Then use the linearity of T and the given information on T171 and T172 to get 01 12 951102 2 1i2 2 Tlt1 2 7961 T171 T172 2 3 1 3952 T 21 7 2 In particular we have T4 72 7110 7171 61 21 Use the linearity of T to rewrite the given information as Apply 3 gtlt Eq 2 7 Eq 1 to get T171 8171 7 4432 Then apply Eq 1 72 gtlt Eq 2 to get N52 75431 3432 63 7 The kernel of T consists of all functions y such that Ty 0 namely y 7 y 0 This differential equation has auxilliary equation r2 7 l 0 which has two roots 1 l 71 Hence the differential equation y 7 y 0 has two linearly independent solutions 6 and e It follows that a basis for the kernel of T is given by 692 e 63 17 First we need to find what T17 equals to for any 17 E V Note that any 17 E V can be expressed as 17 01171 02172 03173 Use the linearity of T and the given information to get T01171 02172 03173 01T171 C2T172 03773 0121171 U72 02131 U72 03131 27472 201 C2 0201171 701 C2 20201172 Therefore the action of T on V is given by T01171 02172 03173 201 C2 0201171 701 C2 20201172 1 Now let us find KerT By Equation 1 we need to find all vectors 01171 02172 03173 in V such that 201 C2 0201171 701 C2 20201172 0 But 11711172 is a basis for W and so 201C20307 7017022030 Use the Gauss Jordan elirnination to get the solution 01 7315 02 5t 03 15 Thus the kernel of T consists of all vectors 17 of the form 43 7325431 51132 1433 t73171 5432 433 5 This means that the kernel of T is given by KerT 1573171 5172 173 l t E R Thus KerT is spanned by the vector 73171 5172 173 and so dimKerT 1 Next we find the range of T For this purpose we need to find out what vector 1 11171 M72 in W is the image of some vector 17 in V By Equation 1 we need to solve the equation 2010203a7 701 7 02 203 Use the Gauss Jordan elimination to solve this system of linear equations 0173tab 025t7a72b 0315 2 Note that for any a and b we can always find 01 02 03 as given by Equation 2 such that T01171 02172 03173 201 C2 0201171 01 C2 20201172 W71 57472 This means that for any vector u in W we can always find a vector 17 in V such that T17 117 Hence RangeT W and so dimRangeT 2 1 2 A 89 Extra problem Let A 3 4 Fmd e The characteristic polynomia is easily found to be p A2 75A72 Hence the eigenvalues are given by 5 V 33 5 7 v33 A2 2 2 We can derive the corresponding eigenvectors A1 a a U1 34M l 02 374ml Let S be the matrix S 171172 namely 34m 313 m Then we have the following identity 2 57M 0 2 5 9 0 71 5 AH Multiply S from the left and S 1 from the right to get 5ix 0 A s 2 1 0 57m 2 Therefore 6A is given by Now we can use row operations to find 5 1 fagE 1 8 33 2 33 51 V 3 Mm 7 1 saxE NE Finally put Equations 1 and 3 into Equation 2 and carry out matrix multiplication to get 73J 552 3 557 3 3 2e 2 ie 2 6A 7 2m m 35 5Fe 5 F 3x 552 73 55 x 2m 95 53 Let X1 and X2 be the Laplace transforms of 1 and 2 respectively Take the Laplace transform of the given differential equations with the initial conditions to get H V 8X1712X17X27 5X2X12X2 N From Equation 2 we solve for X2 1 X X 3 2 s 7 2 1 lt gt Put this X2 into Equation 1 1 3X1712X17 X1 572 which can be rewritten as 1 i 2 X X 1 8 1 s 7 2 1 7 or 2 i2 1 le 1 572 Thus we find X1 to be 2 s X 4 1 s i 22 1 gt 7 Use the First Shifting Theorem to take the inverse Laplace transform to get 1 621 cost Next use Equations 3 and 4 to find X2 1 X 2 37221 Again use the First Shifting Theorem to take the inverse Laplace transform to get 2 621 sint

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