2.4-27 - Section 2.4, Problem 27 We are asked to show...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Section 2.4, Problem 27 We are asked to show whether the transformation y1 x + x2 =1 y2 x1 · 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 1·0 0·1 and there are y for which no solution exists x + x2 0 =1 x1 · x2 1 ⇒ x2 = −x1 ⇒ −x2 = 1 (impossible). 1 To determine exactly which y give no solution, one solution, or many solutions, we can solve the system the old-fashioned way. The equations are x1 + x2 = y1 x1 · x2 = y2 . Solving for x2 and substituting, x1 · (y1 − x1 ) = y2 ⇒ x2 − y1 x1 + y2 = 0. 1 y1 ± √ √ 2 y1 −4y2 , 2 y1 y 2 −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 two-fold when y1 > 4y2 . We can plot these regions on the y1 − 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

Page1 / 2

2.4-27 - Section 2.4, Problem 27 We are asked to show...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online