Solutions to Homework 7.
Math 110, Fall 2006.
Prob 3.1.3. To find the inverses of the elementary matrices, we must simply undo the corresponding
elementary operations:
(a)
001
010
100
,(b)
100
0 1/3 0
001
,(c)
100
010
201
.
Prob 3.1.5. If Eis an elementary matrix of type I corresponding to the swap or rows iand J, then E
has 1 in positions (i, j) and (j, i), zeros in positions (i, i) and (j, j), and the other entries exactly as in the
identity matrix of the appropriate size. But then Eis symmetric, i.e., its transpose is Eitself. Likewise, if
Eis of type II, then its transpose is itself. If Eis of type III and corresponds to adding row jmultiplied by
cto row i, then it looks like the identity matrix except for the entry cin position (i, j). Then its transpose
has cin position (j, i), which corresponds to adding row imultiplied by cto row j.
So, in all three cases Eis an elementary matrix if and only if so is Et.
Prob 3.2.2. (a) 2, (b) 3, (c) 2, (d) 1, (e) 3, (f) 3, (g) 1.
Prob 3.2.6.
(a) Tis invertible. Writing the matrix of Tin the standard basis for P2(IR) and inverting it, we get
T−1(ax2+bx +c) = −ax2−(4a+b)x−(10a+ 2b+c).
(b) Tis not invertible, since it has a nontrivial kernel, for example, T1 = 0.
(c) Tis invertible: the same approach as in (a) yields
T−1(a, b, c) = 1
6a−1
3b+1
2c, 1
2a−1
2c, −1
6a+1
3b+1
2c.
(d) Representing Twith respect to the standard bases for IR3and P2(IR) and inverting the obtained matrix,
we get
T−1(a+bx +cx2) = (c, (a−b)/2,(a+b−2c)/2).
(e) Tis invertible, and
T−1(a, b, c) = 1
2a−b+1
2cx2+−1
2a+1
2cx+b.
(f) Tis not invertible: for example, the matrix
A=−1−1
1 1
is in the kernel of T, since both tr(A) and tr(EA) are zero, and the other two traces are also zero since
tr(At) = tr(A) and tr(EA) = tr(AE) always.
Prob 3.2.8. Since cA = (cI)Aand since the matrix cI is invertible for c6= 0, we conclude that multipli-
cation by cI does not change the rank of A, i.e., rank(A) = rank(cA).
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