Download 3.4.14) By Theorem 3.12, A(adj(A)) = det(A)I n. Both

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