Introduction to Linear algebra-Strang-Solutions-Manual_ver13

2 t3 2t 3 a aa 0 0 det at a d kakkb kkc k2 at a d 4

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Unformatted text preview: A ¤ 0 (b) The determinant is linear in its first column so x1 ja1 a2 a3 jCx2 ja2 a2 a3 jCx3 ja3 a2 a3 j. The last two determinants are zero because of repeated columns, leaving x1 ja1 a2 a3 j which is x1 det A. 5 If the first column in A is also the right side b then det A D det B1 . Both B2 and B3 are singular since a column is repeated. Therefore x1 D jB1 j=jAj D 1 and x2 D x3 D 0. 2 3 2 3 2 1 0 321 3 1 An invertible symmetric matrix 6 7 1 (b) 4 2 4 2 5. 6 (a) 4 0 05 3 has a symmetric inverse. 4 7 123 0 1 3 7 If all cofactors D 0 then A 1 would be the zero matrix if it existed; cannot exist. (And   11 the cofactor formula gives det A D 0.) A D has no zero cofactors but it is not 11 invertible. " # " # 6 3 0 300 This is .det A/I and det A D 3. T 3 1 1 and AC D 0 3 0 . The 1; 3 cofactor of A is 0. 8CD 6 2 1 003 Multiplying by 4 or 100: no change. and also det A 1 D 1. Now A is the inverse of C , so A can be found from the cofactor matrix for C . 9 If we know the cofactors and det A D 1, then C T D A 1 T 10 Take the determinant of AC T D .det A/I . The left side gives det AC T D .det A/.det C / while the right side gives .det A/n . Divide by det A to reach det C D .det A/n 1 . 11 The cofactors of A are integers. Division by det A D ˙1 gives integer entries in A 12 Both det A and det A 1 1 are integers since the matrices contain only integers. But det A 1= det A so det A must be 1 or 1. " # " # 013 1 2 1 1 3 6 2 and A 1 D C T . 13 A D 1 0 1 has cofactor matrix C D 5 210 1 3 1 . 1 14 (a) Lower triangular L has cofactors C21 D C31 D C32 D 0 (b) C12 D C21 ; C31 D C13 ; C32 D C23 make S 1 symmetric. (c) Orthogonal Q has cofactor matrix C D .det Q/.Q 1 /T D ˙Q also orthogonal. Note det Q D 1 or 1. 15 For n D 5, C contains 25 cofactors and each 4 by 4 cofactor has 24 terms. Each term needs 3 multiplications: total 1800 multiplications vs.125 for Gauss-Jordan. ˇ ˇ 16 (a) Area ˇ 3 2 ˇ D 10 (b) and (c) Area 10=2 D 5, these triangles are half of the 14 parallelogram in (a). ˇ ˇ ˇ ˇ ˇ3 1 1ˇ ˇi j kˇ C ˇ 1 3 1 ˇD 20. Area of faces D ˇ 3 1 1 ˇ D 2i 2j p 8k 17 Volume D ˇ Dˇ ˇ ˇ length of cross product length D 6 2 113 131 ˇ ˇ ˇ ˇ ˇ2 1 1ˇ ˇ ˇ 1ˇ 1ˇ 2 1 1ˇ ˇD5 18 (a) Area 2 ˇ 3 4 1 ˇ (b) 5 C new triangle area 2 ˇ 0 5 1 ˇ D 5 C 7 D 12. ˇ 19 ˇ 2 2 051 ˇ ˇ 1ˇ D 4 D ˇ2 3 101 ˇ 2 ˇ because the transpose has the same determinant. See #22. 13 D Solutions to Exercises 57 p 1 C 1 C 1 C 1 D 2. The volume det H is 24 D 16. (H=2 has orthonormal columns. Then det.H=2/ D 1 leads again to det H D 16:) The maximum volume L1 L2 L3 L4 is reached when the edges are orthogonal in R4 . p With entries 1 and 1 all lengths are 4 D 2. The maximum determinant is 24 D 16, p achieved in Problem 20. For a 3 by 3 matrix, det A D . 3/3 can’t be achieved by ˙1. This question is still waiting for a solution! An 18:06 student showed me how to transform the parallelogram for A to the parallelogram for AT , without changing its area. (Edges slide along themselves, so no change in baselength or height or area.) 2...
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