HW10solutions

# HW10solutions - Section 6.3 2 The area of this triangle...

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

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

View Full Document

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Section 6.3 2. The area of this triangle equals a half of the area of the parallelogram spanned by 211 and 122. Therefore: 1 Areazé—ldeﬂvl v2)l=§}det(7 2) kid—501225. 12. Let A be a matrix with columns 111, v2, v3 and 174. We know that the volume of the 4-parallelepiped P spanned by 111 . . . v4 equals the absolute value of the determinant of A. Therefore we can write: ldetA l: VOlum€(P) Z llvill'll’v2ill'llvaill'Hmill S llvillllwll'llv3ll-Hv4ll- Equality holds when o HUQWH = Hog”, i.e. 712 is perpendicular to 211, 0 vanlf : H123“, i.e. U3 is perpendicular to 121 and v2, 0 Hm-“ = i.e. U4 is perpendicular to '01, oz and ’04 in other words the 4 columns of A are mutually orthogonal. By assumption the entries of the U28 are 0, 1 or —17 so the maximum length if each column is “Will 3 \/<ﬁ‘1)2 + ("1)2 + (“1)2 + ("1)2 = \/2 for all 2'. Equality holds when each column has length 27 i.e. all the entries of A are i1. We get: ldetA IS llvill ' llvzll ‘ llvall - Hull S WW =15- The maximal value of the determinant of a matrix of this kind is 16. As an example7 we construct a matrix with mutually orthogonal columns of length 2: 1 1 1 1 -1 —1 —1 ——1 1 ~—1 1 —1 Phat—AH 1 An explicit computation is required to Check that the determinant is indeed +16 (andnot — 16). 3 7 1 a: _ 1 2 , _ 1 22.8etA-(a a)—<4 11),?)—<3),andcall:c(x2)the unique solution of the system A\$ : b. Then: 1 7 det(ba2) (“(3 11> -10 x1: 2 — —— 2 clet(cz1 a2) dt 3 7 5 e 4 11 and 23. Similar to the previous problem. SetA 2 ( al a2 ) = < 36 23 >7 b 2 < (1) >7 and call x C :1 > the unique solution of the system Am 2 b. 2 Then: dt 1’3 wdet(ba2)_ e 0 7 M7 \$1-det(a.1 a2)mdet( 5 -3) 17 and For every 2’ and 3', we have denoted by AM the matrix obtained from A by deleting the ith row and the 3'” column. An easy computations shows that: o dew/11,1 =det<1 =1 ; det(A1,2):det :0 0 0 1 O 1 O dew/1173 : det 2 0 2 W2 d€t(A271) I < 0 1 > 2 0 1 1 1 0 2 1 —~ ~17de13(A2,3)—~ det < 2 O > Z 0 - det(A371 = det HO 1 1 HO OH 1 ; det(A) : —1 (by the Sarraus’ rule). H OH 0 (181-4143;; 2 det < ) . det(Ag)2 = det< > < 1 < > So we get: 1 1 0 ~2 —1 0 1 A‘1:——1 o —1 0 = 0 1 0 . “ —1 0 1 2 0 —1 26. Suppose that the entries of the matrix A are an integer numbers, and that det(A) : 1. Let us look at the inverse of A: 1 1 <--det(A1,1) 'd€t(A1’2) VLdedALB) )T A” 2 Z d t A ~—d€t<A2,1) +det(A2,2) ~d€t<A273) e ( > ~“ d6t<A31> ~det(A3,2) " (181114373) 3 " d€t<AL1> - det(A2,1) “ det(A371) = '- deli/113) d“ det<A23> — det(A372) ”' det(A1,3) — det(A273) —— det(A373) For every 2' and j, we have denoted by Am tha matrix obtained from A by deleting the ith row and the jth column. For instance <> <> 0 G21 a2,3 141,2 = 612,1 0 (12,3 2 < ’ G31 Cl C13; 0 03,3 7 3’3 and det(A172) Z a2,1a3,3 —- 02’3CL371. Because the entries of A are integers, it is clear that det(Ai,j) is always an integer. 80 A"1 has integer entries. 34. Because A'1 2 dab) adj(A), we can write: A . adj(A) : A. (det(A) A-l) 2 det(A) A - A“1 z det(A) I I A similar argument shows that adj(A) ‘ A z I. Section 7.1 3. If v is an eigenvector of A associated to the eigenvalue A, then (A+2])v: Av +2¢=Av+2v=()\+2)v. Au ’0 So u is an eigenvector of A + 2] associated to the eigenvalue A + 2. 6. Assume that v is an eigenvector of both A and B , so that 0 AU : AAU for some eigenvalue AA of A, and 0 Br) :2 A31) for some eigenvalue A3 of B. An easy computation shows that (AB)v 2 A(Bv) 2 A(>\Bv) = AB(AU) : (ABA/ﬂu. Therefore 1) is an eigenvector of AB with eigenvalue ABAA. ab Cd <22><a>=<2><¢<2>=<2> Notice that there are no conditions on b and d. 80 A z ( 5 b 8. We are looking for a matrix A :2 < > such that A61 2 561, i.e. 0 d 18. We notice that: 0 every non—zero vector in the plane V is an eigenvector with eigenvalue +1g because is unchanged by the reflection, and 0 every vector in Vi (2 the line through the origin perpendicular to V) is an eigenvector with eigenvalue -17 because is mapped into its opposite by the reflection. So7 if we want a basis of R3 consisting of eigenvectors, we should pick one vector in Vi and two linearly independent vectors in V. Of course, we have plenty of choices! 22. Because T01) = v, and v is non—zero, the vector v is an eigenvector of T of eigenvalue 1 (and so is any non—zero multiple of U). To show that there are no other eigenvectors, we prove that the equation T(:c) :2 A2: only has solutions of the form a: 2 cv, for some scalar c. The vectors 7; and w form a basis of R2, hence we can write: T(:c) : T(cv + dw) 2 cT(v) + dT(w) : co + d(v +10) 2 (c+d)v + dw. in order to have T(:c) :— Act : (Ac)v + (Ad)w, we must choose /\ 2 l and d z 0. So :0 2 cu. as expected. 36. The matrix A sould satisfy: oA<i>=5<i>=<1s>7and -A<:>=w<:>=<:2>- We can put the two equations together7 and write: 3 1 15 10 15 10 A(12>‘<5 20>¢>A"(5 20) 1 1510 2 —1 3 <i"4“§<5 20>(—1 3>@A"(—211>' m m a} ,8 w a 1 :«r» x : Mr Wm; 14% 213:3 Lg; 3;: {gag g: 2313 455w \aajgjgjﬁa‘é mg 5% VM 3” 1 “1mm ’2 5 2:“ E a“ M M ngffgéimﬁ :5 :; :3 3., :3 €19; Lﬁgﬁaaéwﬁég *{ééi’ gig; i ﬁx,” g§§ a 3“: “‘3 ‘ «4; gig") w"? : m “x ~¢ if} " ‘3 r giggi '1 3“ Wm x; @3143 rs» Sig”; 128:4 2% {36% z A \$593 gig?“ ﬁﬁgﬁ MW ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

HW10solutions - Section 6.3 2 The area of this triangle...

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

View Full Document
Ask a homework question - tutors are online