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## MATRIX ALGEBRA & APPLS

by: Kennith Herman

16

0

6

# MATRIX ALGEBRA & APPLS MA 322

Kennith Herman
UK
GPA 3.54

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
6
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KARMA
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## Popular in Mathematics (M)

This 6 page Class Notes was uploaded by Kennith Herman on Friday October 23, 2015. The Class Notes belongs to MA 322 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 16 views. For similar materials see /class/228153/ma-322-university-of-kentucky in Mathematics (M) at University of Kentucky.

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Date Created: 10/23/15
MA322 Matrix Algebra Answers to assigned even problems July 26 2007 Section 11 10 3 5 6 3 14 211 24 a True See the blue shaded box on page 8 b False 1 1 and i i have the same number of rows but are not row equivalent c False By de nition inconsistent means no solution d True See de nition of equivalent on page 3 1 0 0 2 1 0 1 2 1 0 2 26 Many answers For example 0 1 0 1 R1R3 0 1 0 1 R2R3 0 1 1 1 0 0 1 0 0 0 1 0 0 0 1 0 Section 12 2 a Reduced echelon b echelon c not echelon d echelon 10 11 74 212 12 is free 13 77 14 11 79 7 713 12 2 613 314 13 is free 14 is free 15 0 16 a Unique solution b In nitly many solutions 20 a Inconsistent when h 9 and k f 6 b Unique solution when h f 9 c Many solutions when h 9 and k 6 24 1f the fth column is a pivot column then the reduced echelon form of the augmented matrix will have a row like 0 0 0 0 1 Since this is the augmented matrix the system is inconsistent See theorem 2 page 30 Many answers For example 11 12 13 4 211 212 213 5 Section 13 4 Just see me if you have questions about this one 6 7211 81 13 0 311 512 7 613 0 4 1 3 9 10 11 1 12 7 13 2 2 6 5 15 12 No b is not a linear combination of a1a2 a3 1 16 Many answers for example 1V1 1vz 0 20 The span is the 12plane Section 14 2 The product is not de ned because the number of columns of the rst matrix does not equal the number of rows of the vector 3 5 12x 45 1 18 The reduced echelon form of B has row of zeros ie there is a row without a pivot position Thus the columns of B do not span R4 see theorem 4 on page 4 32 Consider the matrix that has n columns vectors and m rows with n lt m There may be at most n pivot positions since there can be only one per column Since n lt m this implies that not all rows have a pivot position Therefore the columns of this matrix do not span all of Rm again by theorem 4 page 4 34 If the equation Ax b has a unique solution then the corresponding linear system does not have any free variables That means that there is a pivot in every column of A Since A is 3 X 3 this also implies that A has a pivot in every row Therefore the columns of A must span R3 by theorem 4 Section 154 6xzg 3 1 5 4 16x 3 13 3 10 134 The solution set is a line through p parallel to q 24 a False A nontrivial solution just means that not all entries are zero Some of the entries may be zero b True See example 2 page 51 c True lf 0 is a solution then is must be true that b 0 since A 0 0 d True See the paragraph after example 3 page 53 e False This is true only if Ax b is consistent Therefore in general it is false 26 Let p be a possible solution to Ax b By theorem 6 we know that the solution set of this equation is p vh where vh is any solution to the homogeneous equation Ax 0 Therefore if p is the only solution then vh must only be zero 30 a Since there are only two pivot positions there must be a free variable because the pivot positions correspond to basic variables ie nonfree variables So there must be nontrivial solutions b Since there are only two pivot positions there is not a pivot in every row Therefore there is not a solution for every b E R3 again by theorem 4 page 43 Section 17 2 Linearly independent 10 a No h b All h 16 Linearly dependent 18 Linearly dependent 38 True lf zlvl 12V Ing 0 had a nontrivial solution then so would the equation zlvl 12V Ing 0V4 0 However we know that this is not the case since V1V2V3V4 are linearly independent meaning the vector equation zlvl 12V Ing z4V4 0 has only the trivial solution Thus V1 V2 V3 are also linearly independent Section 4 x 3 which is the unique solution 8 5 rows and 4 columns 7 20 5 3 TO TCiV1 C2V2 vap since the v s span R 61TV1 02TV2 CpTVp since T is linear 010 CPO 0 32 Pick any vector With 12 f 0 and let 5 be any negative number Use this to check and see if Tcx CTX Section 19 4 1 1 i 1 1 W2 0 1 1 0 12 Look at the rotation matrix on page 84 Note that if we let 45 7r2 we get the matrix from number 4 So the transformation from 4 is the same as a rotation counterclockwise by 7r2 2 8i 1 4 18 0 0 0 1 Section 21 4 71 3 4A7513 78 2 76 74 1 3 75 15 SAM 740 35 30 40 10ABAC1 l 72 14 2 6 12 One example 1 3 20 The second column of AB is also all zeros because Abg A0 0 22 1f the columns of B are linearly dependent then there exists a nonzero vector x such that Bx 0 From this ABx ABX 0 Since x is nonzero the columns of AB must be linearly dependent 28 uTv vTu and uvT vuT by theorem 3 Section 22 8 Since AD I A lAD A II Which implies ID A 1 or D A l 24 1f the equation Ax b has as solution for each b in R then A has a pivot position in each row Since A is square the pivots must be on the diagonal of A so A is row equivalent to In Thus A is invertible 26 Take the formula for A 1 and actually verify that AA 1 A lA I 32 Not invertible Section 23 2 Not invertible because the determinant is zero Or you could note that the second column is 732 times the rst column 4 The columns of the matrix are linearly dependent because one of the columns is zero so the matrix is not invertible 22 Statment g of the lnvertible Matrix theorem is false for H so statement d is false too Section 31 2 2 4 20 10 6 16 2 20 The determinant of the rst is ad 7 be and the determinant of the second is Mad 7 be So scaling a row by k multiplies the determinant by k Section 32 6 A constant may be factored out of a row 12 114 16 21 24 linearly independent 34 detPAP 1 detP detA detP 1 detPdetP 1detA detPP 1detA detI detA detA Section 33 411 732zg 12 8 All real 8 11 723735 23 12 7275355193 7 3 7 71 3 7 12 adjA 0 0 5 SoA 1 0 0 5 2 71 74 2 71 74 18 Since the cofactors are just sums and products of elements of A all of the cofactors are integer Since the detA is 1 we get from the inverse formula that all of the entries of A 1 are integer Section 41 2 a Given in W and any scalar c the vector 5 is in W because CI cy 02zy which is 2 0 b Let u and v 8 Yes The zero polynomial is in the set If p and q are in the set then 10 4 0 100 q0 0 and cpxo 7 cltpltogtgt 7 0 22 Yes The zero matrix is in the set If B and C are in the set then FB C FB FC 0 and for scalar k FkB MFB 0 Section 42 8 W is not a vector space because the zero vector is not in W 12 To get the zero vector we need d 712 so that the third entry 2d 1 equals zero But then its impossible for the fouth entry to be zero So this is not a vector space 28 The two systems have the form AI v and AI 51 Since we are given that the rst system is consistant we know that v is in the column space of A Since colA is a vector space we know that 51 is in the column space too Thus AI 51 must also be consistant Section 43 4 This is a basis 6 This is not a basis since it does not span R3 but they are linearly independent 24 Let A be the matrix whose columns are the basis elements Since the columns of A are linearly independent we have a pivot in every column Since we have n vectors in R the matrix is square and so there is a pivot in every row as well Therefore the columns also span R Therefore we have a basis Section 44 Section 45 4 0 2 73 0 is a basis so the dimension is 2 0 71 l6 dimnullA0 dimcolA2 26 If dimV0 the statement is obvious because the zero vector space is the only subspace of the zero vector space Otherwise H has a basis Which contains n linearly independent vectors But these vectors are also linearly independent in V Since there are n of them and V is n dimensional these also form a basis for V Section 46 8 dimnullA2 colA cannot be R4 because the vectors in colA have 5 entries 10 l 12 2 20 No Since there are 2 free variables the null space of the coefficient matrix A is 2 dimensional Since there are 8 columns the rank of A is 6 dimensional and because there are 6 equations colA is a subspace of R5 Therefore colA R5 so Ax b is consistent for all b Section 47 2 7 5 a 4 3 10 b M 4 Section 51 4 Yes A 32 18 40 3 3 24 Many solutions As an example 2 has only 2 as an eigenvalue 26 We want to know the value of A if Ax Ax Where x f 0 Now A2 AAx AAx AAX A2x But A2x 0 since A2 0 so A2x 0 as well Since x f 0 it must be that A 0 Therefore the only eigenvalue of A is zero Section 52 6A2 7 11A 40 No real eigenvalues 107A3 14A 12 18h 6 20 detAT 7 AI detAT 7 AIT detA 7 AIT detA 7 AI Section 53 28 Since A has n linearly 39 A t 39 A is quot quot by the Tquot quot quot theorem So A can be Written as PDP l Therefore AT PDP UT 7 ltP1gtTDTltPTgt 7 PT 1DPT Therefore7 AT can be Written as QDQ 1 Where Q PT 17 so AT is diagonalizable Section 55 2 A3i21r2 A37i 21 10 A75i5iq 37r4T5 2 71 3 71 14141 op 3

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