Hassan Ouro-gneni
Vaz
MAT
343
ONLINE
B
Fall
2019
Assignment Section
1.5
Elementary
Matrices due 10/21/2019 at 11:59pm MST
1.
(1 point) Suppose that
A
is a 3
×
3 matrix. For each of
the question (a)-(d), find the elementary matrix
E
such that
EA
gives the matrix resulting from
A
after the given row operation
is performed. Then give
E
-
1
.
a.
R
2
←
1
5
R
2
E
=
,
E
-
1
=
b.
R
3
← -
3
R
3
E
=
,
E
-
1
=
c.
R
1
↔
R
3
E
=
,
E
-
1
=
d.
R
2
←
R
2
+
6
R
3
E
=
,
E
-
1
=
Solution:
a.
E
1
is obtained by multiplying by
1
2
the first row
of the identity matrix
I
=
1
0
0
1
. This gives
E
1
=
1
2
0
0
1
The inverse of
E
1
is obtained by multiplying the first row of the
identity by 2. This gives
E
-
1
1
=
2
0
0
1
.
b.
E
2
is obtained by multiplying by
-
6 the second row of
the identity matrix
I
=
1
0
0
1
. This gives
E
2
=
1
0
0
-
6
The inverse of
E
2
is obtained by dividing the first row of the
identity by
-
6. This gives
E
-
1
2
=
1
0
0
1
-
6
.
c.
E
3
is obtained by switching the first and second rows of
the identity matrix
I
=
1
0
0
1
. This gives
E
3
=
0
1
1
0
The inverse of
E
3
is obtained by switching the first and second
rows of the identity . This gives
E
-
1
3
=
0
1
1
0
.
Note that
E
3
=
E
-
1
3
. This is always the case for permutation
matrices, that is, matrices that are obtained by permuting the
rows of the identity matrix.
d.
E
4
is obtained by performing the row operation
r
2
←
r
2
+
7
r
1
on the identity matrix
I
=
1
0
0
1
. This gives
E
4
=
1
0
7
1
The inverse of
E
4
is obtained by performing the row op-
eration
r
2
←
r
2
-
7
r
1
on the identity matrix.
This gives
E
-
1
4
=
1
0
-
7
1
.
e.
E
5
is obtained by multiplying by
1
5
the second row of the
identity matrix
I
=
1
0
0
0
1
0
0
0
1
. This gives
E
5
=
1
0
0
0
1
5
0
0
0
1
The inverse of
E
5
is obtained by multiplying the second row of
the identity by 5. This gives
E
-
1
5
=
1
0
0
0
5
0
0
0
1
.
f.
E
6
is obtained by multiplying by
-
3 the third row of the
identity matrix
I
=
1
0
0
0
1
0
0
0
1
. This gives
E
6
=
1
0
0
0
1
0
0
0
-
3
The inverse of
E
6
is obtained by dividing the third row of the
identity by
-
3. This gives
E
-
1
6
=
1
0
0
0
1
0
0
0
1
-
3
.
g.
E
7
is obtained by switching the fist and the third row of the
identity matrix
I
=
1
0
0
0
1
0
0
0
1
. This gives
E
7
=
0
0
1
0
1
0
1
0
0
The inverse of
E
7
is obtained by switching the first and third
row of the identity . This gives
E
-
1
7
=
0
0
1
0
1
0
1
0
0
.
h.
E
8
is obtained by performing the row operation
r
2
←
r
2
+
6
r
3
on the identity matrix
I
=
1
0
0
0
1
0
0
0
1
.
This gives
E
8
=
1
0
0
0
1
6
0
0
1
The inverse of
E
8
is obtained by performing the row operation
r
2
←
r
2
-
6
r
3
the identity . This gives
E
-
1
8
=
1
0
0
0
1
-
6
0
0
1
.
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