Page 391 - 35Linear Algebra

G.6 Matrices 391

So this means that

1 5 −3
M −1 =
35 − 33 −11 7
Lets check that M −1 M = I = MM −1 .

1 5 −3 7 3 1 2 0
−1
M M = = = I
35 − 33 −11 7 11 5 2 0 2
You can compute MM −1 , this should work the other way too.
Now lets think about products of matrices

1 3 1 0
Let A = and B =
1 5 2 1
Notice that M = AB. We have a rule which says that (AB) −1 = B −1 A −1 .
Lets check to see if this works

1 5 −3 −1 1 0
−1
A = and B =
2 −1 1 −2 1
and

1 0 5 −3 1 2 0
−1
−1
B A = =
−2 1 −1 1 2 0 2
Hint for Review Problem 3
Firstnote that (b) implies (a) is the easy direction: just think about what it
means for M to be non-singular and for a linear function to be well-defined.
Therefore we assume that M is singular which implies that there exists a non-
zero vector X 0 such that MX 0 = 0. Now assume there exists some vector X V
such that MX V = V , and look at what happens to X V + c · X 0 for any c in your
field. Lastly don’t forget to address what happens if X V does not exist.

Hint for Review Question 4

In the text, only inverses for square matrices were discussed, but there is a
notion of left and right inverses for matrices that are not square. It helps
to look at an example with bits to see why. To start with we look at vector
spaces

2
3
Z = {(x, y, z)|x, y, z = 0, 1} and Z = {(x, y)|x, y = 0, 1} .
2 2
These have 8 and 4 vectors, respectively, that can be depicted as corners of
a cube or square:

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