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3.4.14) By Theorem 3.12, A(adj(A)) = det(A)In . Both sides are n × n matrices, so we can take the determinant of both to get det(A(adj(A))) = det(det(A)In ). The right hand side of this is a scalar matrix with det(A) in each of the n entries on the diagonal. Thus, its determinant is (det(A))n . We can apply Theorem 3.9 to the left side to get det(A(adj(A))) = det(A) det(adj(A)) = (det(A))n . If A is nonsingular, we can cancel a factor of det(A) from each side to get det(adj(A)) = (det(A))n−1 , as desired. If A is singular, then the right hand side of the desired equation is 0n−1 = 0. If adj(A) is also singular, then the desired equation is 0 = 0. If adj(A) is nonsingular, then Theorem 3.9 gives us A(adj(A)) = det(A)In = O. We can multiply both sides by (adj(A))−1 on the right to get A = A(adj(A))(adj(A))−1 = O(adj(A))−1 = O, so A would have to be the zero matrix. However, this means that adj(A) is also a zero matrix, and hence singular. 4.1.22) If u ∈ R2 , then let u = u + (−1)u = x y + (−1) x y = . We compute x y x y + −x −y = x−x y−y = 0 0 = 0. x Likewise, if u ∈ R3 , then let u = y . We compute z 0 x−x −x x x x u+(−1)u = y +(−1) y = y + −y = y − y = 0 = 0. 0 z−z −z z z z 4.2.2) a) No. Consider A = 1 1 1 0 and B = 1 0 1 1 . Then A and B are in V , but A + B is not. b) Yes. If A ∈ V , then kA multiplies the product of the entries of A by k 4 , so it is still zero. c) A = O, the zero matrix. d) Yes, as scalar multiplication by -1 gives a matrix in V as from part (b), and A + (−1)A = O. e) No, because it is not closed under addition. 4.2.7) 3 (no zero), 4 (no negative numbers), and b (can’t multiply by -1) fail to hold. As b does not hold, scalar multiplication is undefined, so it doesn’t entirely make sense to ask whether it has properties 5-8. 1 4.2.10) 4 and b fail to hold, as if x < 0, then we can’t multiply by -1. 4.2.14) That 0 is the only element in V tells us that u = 0, v = 0, and w = 0. Once we realize this, the equations of the various steps mostly assert that 0 = 0. a) u ⊕ v = 0 ⊕ 0 = 0 ∈ V 1) u ⊕ v = 0 ⊕ 0 = v ⊕ u 2) u ⊕ (v ⊕ w) = 0 ⊕ (0 ⊕ 0) = 0 ⊕ 0 = (0 ⊕ 0) ⊕ 0 = (u ⊕ v) ⊕ w 3) The zero element is 0, and we compute u + 0 = 0 ⊕ 0 = 0 = u and 0 + u = 0 ⊕ 0 = 0 = u. 4) As u = 0, we set −u = 0 and get u + −u = 0 ⊕ 0 = 0 and −u + u = 0 ⊕ 0 = 0. b) c · u = c · 0 = 0 ∈ V 5) c · (u ⊕ v) = c · (0 ⊕ 0) = c · 0 = 0 = 0 ⊕ 0 = (c · 0) ⊕ (c · 0) = (c · u) ⊕ (c · v) 6) (c + d) · u = (c + d) · 0 = 0 = 0 ⊕ 0 = (c · 0) ⊕ (d · 0) = (c · u) ⊕ (d · u) 7) c · (d · u) = c · (d · 0) = c · 0 = 0 = (cd) · 0 = (cd) · u 8) 1 · u = 1 · 0 = 0 = u 4.2.17) No. Let c = 1, u = 1, and v = 1 and consider property 5. We get 1 · (1 ⊕ 1) = (1 · 1) ⊕ (1 · 1), from which 1 · 1 = 2 ⊕ 2, or 2 = 4. 2