Math 43, Fall 2007
B. Dodson
Week 8:
Problem 3.2.32
Find the inverse of the elementary matrix
E
=
•
1
2
0
1
‚
.
Solution: Left multiplication by
E
does the 3rd elementary row
operation
R
1
→
R
1
+ 2
R
2
. (check)
To undo this row operation, we subtract 2
R
2
from the first row,
so
E

1
=
•
1

2
0
1
‚
. (why?)
Problem 3.2.40
Write the invertible matrix
A
=
•
2
4
1
1
‚
as a product of
elementary row matrices.
Solution. We reduce
A
to the identity matrix
I,
keeping
track of the row operations. We have
A
=
•
2
4
1
1
‚
→
•
1
1
2
4
‚
→
•
1
1
0
2
‚
→
•
1
1
0
1
‚
→
•
1
0
0
1
‚
=
I,
which in terms of elementary matrices looks like
E
4
E
3
E
2
E
1
A
=
I,
and gives
A
=
E

1
1
E

1
2
E

1
3
E

1
4
.
Tracing through the row operations
E
1
=
E

1
1
=
•
0
1
1
0
‚
;
E
2
=
•
1
0

2
1
‚
,
so
E

1
2
=
•
1
0
2
1
‚
,
E
3
=
•
1
0
0
1
2
‚
,
so
E

1
3
=
•
1
0
0
2
‚
; and
E
4
=
•
1

1
0
1
‚
,
so
E

1
4
=
•
1
1
0
1
‚
.
We get
A
=
•
0
1
1
0
‚ •
1
0
2
1
‚ •
1
0
0
2
‚ •
1
1
0
1
‚
.
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2
A collection
S
of vectors in
R
n
is a
Subspace
of
R
n
if
it satisfies the two subspace conditions, called
closure rules:
(1) for every
u, v
∈
S,
check
u
+
v
∈
S
(closure under vector +),
and (2) for every
u
∈
S,
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 Spring '08
 Dodson
 Linear Algebra, Multiplication, main fact, E3 E2 E1

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