John Q. Student
MATH 211 (Siefken): Homework #2
Due May 26, 2011
Question #1
(a) Find the inverse of
A
=
3
1
2
6
3
5
1
1
3
by Gauss–Jordan elimination.
To find
A

1
we apply GaussJordan elimination to [
A

I
].
3
1
2
1
0
0
6
3
5
0
1
0
1
1
3
0
0
1
R
2
7→
R
2

2
R
1
→
3
1
2
1
0
0
0
1
1

2
1
0
1
1
3
0
0
1
R
3
7→
R
3

1
/
3
R
1
→
3
1
2
1
0
0
0
1
1

2
1
0
0
2
/
3
7
/
3

1
/
3
0
1
R
3
7→
R
3

2
/
3
R
2
→
3
1
2
1
0
0
0
1
1

2
1
0
0
0
5
/
3
1

2
/
3
1
R
2
7→
R
2

3
/
5
R
3
→
3
1
2
1
0
0
0
1
0

13
/
5
7
/
5

3
/
5
0
0
5
/
3
1

2
/
3
1
R
1
7→
R
1

6
/
5
R
3
→
3
1
0

1
/
5
4
/
5

6
/
5
0
1
0

13
/
5
7
/
5

3
/
5
0
0
5
/
3
1

2
/
3
1
R
1
7→
R
1

R
2
→
3
0
0
12
/
5

3
/
5

3
/
5
0
1
0

13
/
5
7
/
5

3
/
5
0
0
5
/
3
1

2
/
3
1
R
1
7→
R
1
/
3
→
1
0
0
4
/
5

1
/
5

1
/
5
0
1
0

13
/
5
7
/
5

3
/
5
0
0
5
/
3
1

2
/
3
1
R
3
7→
3
/
5
R
3
→
1
0
0
4
/
5

1
/
5

1
/
5
0
1
0

13
/
5
7
/
5

3
/
5
0
0
1
3
/
5

2
/
5
3
/
5
and so we see
A

1
=
1
5
4

1

1

13
7

3
3

2
3
.
(b) Solve the following system of equations
3
x
1
y
2
z
=

1
6
x
3
y
5
z
=
2
x
+
y
+3
z
=
3
.
We notice that
A
is th coefficient matrix for the system of equations given. That is, the system of equations
is equivalent to the matrix equation
A
x
=

1
2
3
. Since we have already computed
A

1
, multiplying both
sides by
A

1
gives
x
y
z
=
1
5
4

1

1

13
7

3
3

2
3

1
2
3
=

9
/
5
18
/
5
2
/
5
.
Question #2
For each of the following statements, if they are true,
prove them
, otherwise provide a specific
counter
example
.
(a) If
A
and
B
are both 2
×
2 matrices which are not diagonal (and so neither is the zero matrix), then
AB
=
BA
implies
A
=
B
.
Question #2 continued on next page. . .
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John Q. Student
MATH 211 (Siefken): Homework #2
Due May 26, 2011
Consider
A
=
3
1
1
3
and
B
=
3

1

1
3
. We have
AB
=
BA
=
8
0
0
8
,
but
A
6
=
B
.
(b) If
c
∈
R
with
c
6
= 0 and
A
is invertible, then (
cA
)

1
=
1
c
A

1
.
1
c
A

1
is well defined since
A
is invertible and
c
6
= 0. Multiplying out, we see (
cA
)
1
c
A

1
=
c
c
AA

1
=
I
and
1
c
A

1
A
=
c
c
A

1
A
=
I
, and so since
1
c
A

1
satisfies the properties of (
cA
)

1
, we conclude (
cA
)

1
=
1
c
A

1
.
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 Summer '10
 SOSPEDRAALFONSO
 Linear Algebra, Algebra, GaussJordan Elimination, Matrices, Gauss–Jordan Elimination, John Q. Student

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