Math
2940
Solutions, Fall
2011
Section
2
.
5
2
)
0
0
1
0
1
0
1
0
0
looks like the identity, except the first and third rows (or columns) are
switched. So switching them again should be the identity. You can check
0
0
1
0
1
0
1
0
0

1
=
0
0
1
0
1
0
1
0
0
.
P
=
0
1
0
0
0
1
1
0
0
looks like the identity, except the rows (or columns) are permuted
backwards (first to third, third to second, second to first).
So doing this two more times
should be the identity. You can check
0
1
0
0
0
1
1
0
0

1
=
0
1
0
0
0
1
1
0
0
2
=
0
0
1
1
0
0
0
1
0
.
3
) We have 10
x
+ 20
y
= 1 and 20
x
+ 50
y
= 0.
Adding

2 times the first equation to
the second gives 10
y
=

2 so
y
=

1
/
5 and then
x
= 1
/
2 so the first column of
A

1
is
1
/
2

1
/
5
.
Solving We have 10
t
+ 20
z
= 0 and 20
t
+ 50
z
= 1. Adding

2 times the first equation
to the second gives 10
z
= 1 so
z
= 1
/
10 and then
t
=

1
/
5 so the first column of
A

1
is

1
/
5
1
/
10
.
We have
A

1
=
1
/
2

1
/
5

1
/
5
1
/
10
.
5
)

1
0
1
0
1
0
0
0
1
2
=
1
0
0
0
1
0
0
0
1
.
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 '05
 HUI
 Math, Linear Algebra, Algebra, Invertible matrix, Identity element, Identity matrix, ﬁrst equation

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