By Bernard Kolman, David R. Hill
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Extra resources for Instructor's Solutions Manual for Elementary Linear Algebra with Applications, 9th Edition
In Matlab the command null(A) produces an orthonormal basis for the null space of A. 31. Each set of two vectors is linearly independent since they are not scalar multiples of one another. In Matlab the reduced row echelon form command implies sets (a) and (b) are linearly independent while (c) is linearly dependent. 6, p. 242 2. (c). 4. (d). 1 1 0 0 1 0 0 1 0 0 c + c3 0 0 c1 + c4 + c2 + c3 + c4 = , then 1 = . The 0 0 1 1 0 1 1 1 0 0 c2 + c4 c2 + c3 + c4 0 0 first three entries imply c3 = −c1 = c4 = −c2 .
Since A = A−1 , we have det(A) = 1 det(A) =⇒ (det(A))2 = 1. Hence det(A) = ±1. (b) If AT = A−1 , then det(AT ) = det(A−1 ). But det(A) = det(AT ) and det(A−1 ) = 1 det(A) hence we have det(A) = 1 det(A) =⇒ (det(A))2 = 1 =⇒ det(A) = ±1. 16. 2, the only time we get terms which do not contain a zero factor is when the terms involved come from A and B alone. Each one of the column permutations of terms from A can be associated with every one of the column permutations of B. Hence by factoring we have det A O O B = (terms from A for any column permutation)|B| = |B| (terms from A for any column permutation) = (det B)(det A) = (det A)(det B).
Then u1 ⊕ (u ⊕ u1 ) = u1 ⊕ (u ⊕ u2 ) (u1 ⊕ u) ⊕ u1 = (u1 ⊕ u) ⊕ u2 0 ⊕ u1 = 0 ⊕ u2 u1 = u2 . 21. (b) c 0=c (0 ⊕ 0) = c 0⊕c (c) Let c u = 0. If c = 0, then so u = 0. 1 c 0 so c (c u) = 0 = 0. 1 c 0 = 0. Now 1 c (c u) = 1 c 22. Verify as for Exercise 9. Also, each continuous function is a real valued function. 23. v ⊕ (−v) = 0, so −(−v) = v. 24. If u ⊕ v = u ⊕ w, add −u to both sides. 25. If a u=b u, then (a − b) u = 0. 2. 3, p. 205 2. Yes. 4. No. 6. (a) and (c). 8. (a). 10. (c). 12. (a) Let a1 0 b1 A = 0 c1 0 d1 0 e1 be any vectors in W .
Instructor's Solutions Manual for Elementary Linear Algebra with Applications, 9th Edition by Bernard Kolman, David R. Hill