Winter 2006 Homework 3 Page: 1
Applied Linear Algebra MATH 102
◦pg. 126 # 2
3 2
7 4 −1
=1
4·3−2·74 -2
-7 3 =-2 1
7/2 -3/2 .
◦pg. 126 # 10
a) False: The product matrix is invertible, but the product of inverses
should be in the reverse order. See Theorem 6 (b).
b) True, by Theorem 6 (b).
c) True, by Theorem 4.
d) True, by Theorem 7.
e) False, The last past of Theorem 7 is misstated here. The elemen-
tary row operations that reduce Ato Inreduce Into A−1.
◦pg. 127 # 24
If the equation Ax=bhas a solution for each bin Rn, then Ahas
a pivot position in each row by Theorem 4 in Section 1.4. Since Ais
square, the pivots must be on the diagonal of A. It follows that Ais
row equivalent to In. By Theorem 7, Ais invertible.
Remark 1. I gave a longer answer in section because I was re-deriving
Theorem 4 of Section 1.4. You do not need to include those details for
this particular question. However, for your benefit please find them
below.
Proof of Theorem 4 of Section 1.4: The statements (a), (b), and (c)
are all equivalent by the definition of linear combination and spanning,
as given in this section. The difficulty with this Theorem is proving
that (a) is equivalent to (d), namely
For A, an m×nmatrix, if for each b∈Rm, the equation Ax=bhas
a solution, then matrix Ahas a pivot position in every row.
We prove this by noting that if Ax=eihas a solution, where eiis the
unit vector which is 1 in the ith coordinate, and zero elsewhere, then
row-reducing the augmented matrix [A|b] will give us [A′|λ·b] where
λ∈R. This reduced system must have a solution, and thus A′, hence
Amust have a pivot in row i. We thus obtain that there is a pivot in
every row of A.
Also see Example 3 in Section 1.4 for more details.
