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

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

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