Math 43, Fall 2007
B. Dodson
Week 9:
Monday: Finish Suggested Hw8, start on Hw9 material
1.
Determinants
2.
Properties of Dets
3. Eigenvalues and Eigenvectors (2-by-2 in 4.1;
then n-by-n in 4.3)
——–
We compute det
2
1
5
4
2
3
9
5
1
using the
(first) row expansion (by minors):
det
(
A
) = 2
fl
fl
fl
fl
2
3
5
1
fl
fl
fl
fl
-
1
fl
fl
fl
fl
4
3
9
1
fl
fl
fl
fl
+ 5
fl
fl
fl
fl
4
2
9
5
fl
fl
fl
fl
= 2(2
-
15)
-
(4
-
27) + 5(20
-
18)
= 2(
-
13)
-
(
-
23) + 5(2) =
-
26 + 33 = 7
.
———–
Problem
Reduce
A
=
2
1
3
5
3
0
1
2
4
1
4
3
5
2
5
3
to an upper triagular
matrix and use the reduction to find
det
(
A
)
.
Solution:
A
→
-
1
1
2
3
3
0
1
2
4
1
4
3
5
2
5
3
(
r
1
→
r
1
-
r
2
)
→
-
1
1
2
3
0
3
7
11
0
5
12
15
0
7
15
18
(
r
2
→
r
2
+ 3
r
1
, r
3
→
r
3
+ 4
r
1
, r
4
→
r
4
+ 5
r
1
)
2
→
-
1
1
2
3
0
3
7
11
0
5
12
15
0
1
1
-
4
(
r
4
→
r
4
-
2
r
2
)
——–
→
-
1
1
2
3
0
1
1
-
4
0
5
12
15
0
3
7
11
(
r
2
↔
r
4
)
→
-
1
1
2
3
0
1
1
-
4
0
0
7
35
0
0
4
23
(
r
3
→
r
3
-
5
r
2
, r
4
→
r
4
-
3
r
2
)
→
-
1
1
2
3
0
1
1
-
4
0
0
-
1
-
11
0
0
4
23
(
r
3
→
r
3
-
2
r
4
)
→
-
1
1
2
3
0
1
1
-
4
0
0
-
1
-
11
0
0
0
-
21
=
U
(
r
4
→
r
4
+ 4
r
3
).
Now we have det
U
= (
-
1)(1)(
-
1)(
-
21) =
-
21
,
and det
A
= (
-
1) det
U
= 21
,
since all of the 3rd
elementary operations change the determinant by a
factor of (1) – so no change - there are no 2nd
EROs and exactly one 1st ERO, with each 1st ERO
changing the determinant by a factor of (
-
1)
.
Notice that, for the problem of finding the value of
the determinant, we do not need a reduced row echelon
or 1’s on the diagonal, and stop at a triangular form.
3
3. Eigenvalues and Eigenvectors
—————
A vector
v
= 0 in
R
n
is an
eigenvector with eigenvalue
λ
of
an
n
-by-
n
matrix
A
if
Av
=
λv.
We re-write the vector equation as (
A
-
λI
n
)
v
= 0
,
which is a homogeneous system with coef
matrix (
A
-
λI
)
,
and we want
λ
so that the system has a non-trivial solution.
