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

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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 old-fashioned 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 two-fold 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 ...
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2.4-27 - Section 2.4 Problem 27 We are asked to show...

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