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Adding the terms on the left and adding the terms on the right, we cancel out
y
and ﬁnd
3
x
=
u
+
v
, or
x
= (
u
+
v
)
/
3
.
From the ﬁrst equation above we have
y
=
x

u
, and so
y
= (
u
+
v
)
/
3

u
= (
v

2
u
)
/
3
.
Thus, with
±
x
y
²
=
1
3
±
u
+
v
v

2
u
²
,
we have
f
³±
x
y
²´
=
±
u
v
²
. Since
±
u
v
²
is an arbitrary vector in
IR
2
, this shows that
f
is onto. Since, as we have seen in solving for
x
and
y
, this is the
only
vector with
f
³±
x
y
²´
=
±
u
v
²
,
f
is one to one. Since it is both one to one and onto, it is invertible. In
fact, we have already found the formula for the inverse:
f

1
³±
u
v
²´
=
1
3
±
u
+
v
v

2
u
²
.
The key thing to note is how the solution of all three parts revolves around solving the
equation (
*
). That is the point of the question.
1.7
Consider the function from
IR
2
to
IR
2
deﬁned by
f
µh
x
y
i¶
=
h
x
2

y
2
x

2
i
.
(a)
Find all solutions of
f
µh
x
y
i¶
=
h
0
0
i
.
(b)
Is
f
invertible? If so ﬁnd a formula for the inverse. If not, explain why not.
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.
 Fall '08
 Gladue
 Calculus

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