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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 sot0W) .
Lﬁ's 5A7 )_<\ ,)§,_ ALF. ain'tnor suLOTiaM . 1.5 A5. : E = AEL.
LeT 3—<°= 55"!" A()£°)= A(§—§1)= ANSIAWE?—t b‘b‘g' pow LET \46 M. er Aw sum. , Cansioen. ALE‘+‘L>_<_=)= Aw Mag, = W49: mg =(o. So gelkicf. is A: sownth“, 8. since [4 maxim7 , we eisz 00— mm soLuTI‘ek. 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 lAb1b2~~~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'. 93‘£\ 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 15 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 {AHE 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 47512143 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 = [MAMAu \m:
LIV—Z
mum C. & AC=1.
so m mm m. C=A". so f A is (ABELElena , 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 ...
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This note was uploaded on 11/10/2009 for the course MATH 1850 taught by Professor Mihaibeligan during the Spring '09 term at UOIT.
 Spring '09
 MihaiBeligan
 Linear Algebra, Algebra

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