This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Section 2.4, Problem 27
We are asked to show whether the transformation
y1
x + x2
=1
y2
x1 Â· x2
y1
for which the solution is not unique
y2 is invertible. As we saw in section, there are y = 1
1+0
0+1
=
=
0
1Â·0
0Â·1
and there are y for which no solution exists
x + x2
0
=1
x1 Â· x2
1 â‡’ x2 = âˆ’x1 â‡’ âˆ’x2 = 1 (impossible).
1 To determine exactly which y give no solution, one solution, or many solutions, we can solve
the system the oldfashioned way. The equations are
x1 + x2 = y1
x1 Â· x2 = y2 .
Solving for x2 and substituting,
x1 Â· (y1 âˆ’ x1 ) = y2 â‡’ x2 âˆ’ y1 x1 + y2 = 0.
1
y1 Â± âˆš âˆš 2
y1 âˆ’4y2
,
2 y1 y 2 âˆ’4y2 1
so x2 =
. Looking under
This quadratic equation has solution x1 =
2
2
2
the radical, we see that this solution does not exist when y1 < 4y2 , is unique when y1 = 4y2 ,
2
and is twofold when y1 > 4y2 . We can plot these regions on the y1 âˆ’ y2 plane: 1 4 No Solution U ni
qu
e So lu tio n 2 0 2 Two Solutions 4
4 2 0 2 Figure 1: Solvability Regions 2 4 ...
View
Full
Document
 Spring '11
 EIns

Click to edit the document details