Introduction to Linear algebra-Strang-Solutions-Manual_ver13

A i d 2 7 c 10 d 0 when d 2 or d 5 those are

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Unformatted text preview: to ABC (use     00 01 it twice) (c) False: det.4A/ is 4 det A (d) False: A D ,B D , 01 10   0 1 AB BA D is invertible. 1 0 n 4 Exchange rows 1 and 3 to show jJ3 j D 1. Exchange rows 1 and 4, then 2 and 3 to show jJ4 j D 1. 5 jJ5 j D 1, jJ6 j D 1, jJ7 j D 1. Determinants 1; 1; 1; 1 repeat so jJ101 j D 1. 6 To prove Rule 6, multiply the zero row by t D 2. The determinant is multiplied by 2 (Rule 3) but the matrix is the same. So 2 det.A/ D det.A/ and det.A/ D 0. 7 det.Q/ D 1 for rotation and det.Q/ D 1 for reflection .1 2 sin2  2 cos2  D 1/. 8 QT Q D I ) jQj2 D 1 ) jQj D ˙1; Qn stays orthogonal so det can’t blow up. 9 det A D 1 from two row exchanges . det B D 2 (subtract rows 1 and 2 from row 3, then columns 1 and 2 from column 3). det C D 0 (equal rows) even though C D A C B ! 10 If the entries in every row add to zero, then .1; 1; : : : ; 1/ is in the nullspace: singular A has det D 0. (The columns add to the zero column so they are linearly dependent.) If every row adds to one, then rows of A I add to zero (not necessarily det A D 1). 11 CD D D C ) det CD D . 1/n det DC and not det DC . If n is even we can have an invertible CD . 12 det.A 1 1 . ad b c / divides twice by ad ad b c b c (once for each row). This gives .ad b c/2 D 13 Pivots 1; 1; 1 give determinant D 1; pivots 1; 2 ; 3=2 give determinant D 3. 14 det.A/ D 36 and the 4 by 4 second difference matrix has det D 5. 15 The first determinant is 0, the second is 1 2 t 2 C t 4 D .1 t 2 /2 . . Solutions to Exercises 52 16 A singular rank one matrix has determinant D 0. The skew-symmetric K also det K D 0 (see #17). 17 Any 3 by 3 skew-symmetric K has det.K T / D det. K / D . 1/3 det.K/. This is det.K/. But always det.K T / D det.K/. So we must have det.K/ D 0 for 3 by 3. ˇ ˇ ˇ ˇ ˇ ˇ 1 a a2 ˇ ˇ1 ˇ ˇ a a2 ˇ ˇ ˇ ˇ ˇ b a b 2 a2 ˇ ˇ 18 ˇ 1 b b 2 ˇ D ˇ 0 b a b 2 a2 ˇ D ˇ ˇ ˇ ˇ ˇ ˇ c a c 2 a2 ˇ (to reach 2 by 2, ˇ 1 c c2 ˇ ˇ 0 c a c 2 a2 ˇ eliminate a and a2 in row ˇ by column ˇoperations). Factor out b a and c 1 ˇ 1 bCa ˇ ˇ the 2 by 2: .b a/.c a/ ˇ ˇ 1 c C a ˇ D .b a/.c a/.c b /. / D 1, 6 and det.U / D 36. 2 by 2 matrix: det.U / D ad; det.U / D a d . If ad ¤ 0 then det.U 1 / D 1=ad .   a Lc b Ld det reduces to .ad b c/.1 L`/. The determinant changes if you c `a d `b do two row operations at once. Rules 5 and 3 give Rule 2. (Since Rules 4 and 3 give 5, they also give Rule 2.) 1 det.A/ D 3; det.A 1 / D 3 ; det.A I / D 2 4 C 3. The numbers  D 1 and  D 3 give det.A I / D 0. Note to instructor: If you discuss this exercise, you can explain that this is the reason determinants come before eigenvalues. Identify  D 1 and  D 3 as the eigenvalues of A.     3 1 18 7 1 1 2 2 1 has det 10 . det.A/ D 10, A D , det.A / D 100, A D 10 14 11 2 4 det.A I / D 2 7 C 10 D 0 when  D 2 or  D 5; those are eigenvalues. Here A D LU with det.L/ D 1 and det.U / D 6 product of pivots, so also det.A/ D 6. det.U 1 L 1 / D 1 D 1= det.A/ and det.U 1 L 1 A/ is det I D 1. 6 When the i , j entry is ij , row 2 D 2 time...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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