Page 390 - 35Linear Algebra

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n
This means we are going to have an idea of what A looks like for any n. Lets
look at the example of one of the matrices in the problem. Let

1 λ
A = .
0 1
n
Lets compute A for the first few n.

1 0
0
A =
0 1

1 λ
1
A =
0 1

1 2λ
2
A = A · A =
0 1

1 3λ
2
3
A = A · A = .
0 1
There is a pattern here which is that

1 nλ
n
A = ,
0 1
then we can think about the first few terms of the sequence
∞ n
X A 0 1 2 1 3
A
e = = A + A + A + A + . . . .
n! 2! 3!
n=0
Looking at the entries when we add this we get that the upper left-most entry
looks like this:
∞
1 1 X 1 1
1 + 1 + + + . . . = = e .
2 3! n!
n=0
Continue this process with each of the entries using what you know about Taylor
series expansions to find the sum of each entry.
2 × 2 Example
Lets go though and show how this 2×2 example satisfies all of these properties.
Lets look at

7 3
M =
11 5
We have a rule to compute the inverse

−1

a b 1 d −b
=
c d ad − bc −c a
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