1.6 - (.4 Dr. Doom Lin Alg MATH185O Section/leifPage 1 of 5...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (.4 Dr. Doom Lin Alg MATH185O Section/leifPage 1 of 5 Chapter 1 - Linear Equations and Matrices Section 1.6 - Further Results on Systems of Equations and Invertibility We’ve seen the following Thm. before in an earlier section, and now we are in a position to actually prove it. Theorem 1.6.1. Every linear system has no solution, exactly one solution, or infinitely many solutions. [37.001597 A E cle is consiéreut , AM) Mammal. SAY iT HRS MOLE “raw: A, soun'm’. ( new snow we LMU€ «ab—MAW sot-0W) . Lfi's 5A7 )_<\ ,)§,_ ALF. ain't-nor suLOTiaM . 1.5 A5. : E = AEL. LeT 3—<°= 55"!" A()£°)= A(§|—§1)= ANSI-AWE?—t b‘b‘g' pow LET \46 M. er Aw sum. , Cansioen. ALE‘+‘L>_<_=)= Aw Mag, = W49: mg =(o. So gel-kicf. is A: sow-nth“, 8. since [4 maxim-7 , we eisz 00— mm soLuTI‘e-k. We now have another method for solving certain linear systems using in- verses. Theorem 1.6.2. If A is an invertible n X 71 matrix, and b is any n X 1 col- umn matrix, then the linear system Ax = b has exactly one solution, namely, x = A‘1b. Proof: A£= '2 5" AAA 3 = Ark L" I! =A-\l9 4:) 25 = A’lb. .— To sum) Soumén Ls umhue. SM Lang; Au semi-(Sass. So Ayah :Arz ,i - _— ’ Ax‘zA-¥16=) AA¥\:A\A¥J‘;‘° 154:3}42", Xvi)“ , _ LL Dr. Doom Lin Alg MATH185O SectionjflfPage 2 of 5 Example: 171 -- 2362 -- 3363 = 5 2:61 -- 5362 -- 3363 = 3 $1 -- -- 8363 = 17 In matrix form we can write this as Ax = b where 1 2 3 :61 5 A = 2 5 3 , X = :62 , b = 3 1 0 8 :63 17 —40 16 9 We get 14—1 = 13 —5 —3 5 —2 —1 -. -—Ro u. a 5‘ ' 1‘: A 19: ['5 -S *3 3 1'" Linear Systems with a Common Coefficient Matrix If one has to solve a sequence of systems that share the same coefficient matrix (so the column of constants is the only thing that changes): Ax=b1,Ax=b2,...,Ax=bk, then provided A is invert' le we can solve the syste s using matrix inversion, getting corresponding lutions x = A_1b1,x = A_1b2,. . . ,x = A_1bk L_____1 L‘J I.“ But a more efficient method that works even if A is singular, is using Gauss- Jordan Elimination on the augmented matrix lA|b1|b2|~~~|bk~l IL Dr. Doom Lin Alg MATH185O Seetionfl Page 3 of 5 Example: Three systems AX 2 b1, AX 2 b2, AX 2 b3 1 2 3 2 0 1 WhGI‘eA: 2 5 3 ,andb1= 3 7132: 3 ,b3— 1 1 0 8 5 2 —2 _ \ L 3 'L O I [Almlm [a g 443‘» 4 m H, fl L o b 5 7. 'L W A: a _S .3 2- «7.1 ‘ Z 3 1 ° ' g -1 q o ‘ =2» -\ a -\ an AM 23'. 9-3‘£\ O '2 g 3 7. '3 \ v . — "l Sou: \s LsA L,‘ ’11 " '1 L 'L ‘5 7. o \ ———>[o \—3 —\\3‘—1-} W- AZ’bz a 23223221 0 o —| \ fl, -‘§ SOUL) \s §=At1911\l,z,8 -% ‘ L 1: L ° , cuau'rafit .110 1-5 M. 23:44) 0 o l ’\ '2 5 s,ng L; ll :K‘Lp; Rpm-“.2 ‘ 3 o \ 3 o a o \ o —L( -7,\ N (2122359., 0 o u —\ -2 S . — l o 0 l3 LL L”. (iv—Lil; i o 1 o --(L —ZI [H] o o I “l ~31. s {fU 9? LC. Dr. Doom Section% Page 4 of 5 Lin Alg MATH1850 Theorem 1.6.3. Let A be a square matrix. (a) If B is a square matrix satisfying BA 2 I then B = Ail. (b) If B is a square matrix satisfying AB = I then B = A4. Proof, Ca.) AN, Sqom Monica as {AH-E cm 2- 9am , sum A is \A’chfm, Confirm Ax =0 ,wam sun 1050i». save is um‘qfi. 3A1 33. 2161 A01 mu sourdous 00'- A§=g, S, =9 Pas-=qu :2 EAg, =Q>Ag4 =3 I¥‘-=:\: E1 -.-7 5:)“ _ $0 TUNA Sou» 1‘ timid?- my erJAéu, TH“, A" $615. 55““ "W BzA' M = IbhA")=(LsA)A"= IA“= a This is the theorem that keeps on growing: Theorem 1.6.4. Equivalent Statements If A is an n X n matrix, then the following statements are equivalent, that is, they are all true or all false. (a) A is invertible. (b) Ax = 0 has only the trivial solution. (c) The reduced row echelon form of A is [7,. (d) A is expressible as a product of elementary matrices. (e) Ax = b is consistent for every n X 1 matrix b. ( ) Ax = b has exactly one solution for every n X 1 matrix b. Sum» Proof: (4») => ((3 => (a) => 0») (A) '9 (:2) “(6 (s A» em.qu Tun. — ‘ (£3 -: (g) '16 obvious C 1: 37am \MS subm'mis {was sym u. mm) (c) =7Cq.) leaunes A MT qt- mu, ASSUME A§=b (s Cofi¢\s:\'ekK Fol. Aw H5 th—buv— 1M Faun mm 4751-2143 Ame («ISIS—W : 5 ° 0 a 7.. ' _ . A! 1 ’ Q. J ' ‘- 3 a " . _ a o o ' LANA. we SOLU'nonfS )5. , )5, , ,5“ Macc—rwew consuls“. Ahm- w = [MAMA-u \m: LIV—Z mum C. & AC=1. so m mm m. C=A". so f A is (ABEL-Elena , -OO LC Dr. Doom Lin Alg MATH185O SectionM Page 5 of 5 Theorem 1.6.5. Let A and B be square matrices of the same size. If AB is invertible, then so are A and B. We’ve actually seen the converse of this theorem in an earlier section (Thm.1.4.6.). We shall be better equipped to give a proof of it in a later chapter. A Fundamental Problem: Let A be a fixed n X 71 matrix. Find all m X 1 matrices b such that the system of equations Ax = b consistent. Example: Determining consistency of a linear system: a: + 2y — 2,2 2 b1 —33 — y —— 3,2 2 b2 —33 —— 4,2 2 b3 233 23'1Q2 svsTEM Ks canémem’ H‘F ...
View Full Document

This note was uploaded on 11/10/2009 for the course MATH 1850 taught by Professor Mihaibeligan during the Spring '09 term at UOIT.

Page1 / 7

1.6 - (.4 Dr. Doom Lin Alg MATH185O Section/leifPage 1 of 5...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online