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Linear Algebra With Applications
Found in: Page 97
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

Which of the (nonlinear) transformations from R2 to R2to in Exercises 25 through 27 are invertible? Find the inverse if it exists.26. role="math" localid="1660365558437" [y1y2]=[x2x13+x2]

The inverse of the given transformation is x1=y2-y13x2=y1

See the step by step solution

Step by Step Solution

Step 1: Check whether the transformation is invertible or not

The given transformation is y1=x2y2=x13+x2.

We can rewrite the transformation as f;R2R2.such that f(x1,x2)=(x2,x13+x2)

Then the given transformation is invertible if and only if f is invertible.

(a) Suppose f(x1,x2)=f(z1,z2)for some (x1,x2),f(z1,z2)R2.Then x13=z13and .So x1z1are zero or nonzero together.

Now (x1-z1)(x12+x1z1+z12)=x13-z13=0 (1)

Now x12+x1z1+z12=0x1z1=(x1+z1)2and -3x1z1=(x1-z1)2. So x1,z1=0. If x1z1are nonzero, then x12+x1z1+z12=0and hence from (1) x1=z1.If x1,z1are zero, then x1=z1.Thus x1=z1and x13=z13. Sofis one-one.

(b) Let (z1,z2)R2. Then there exists x1Rsuch that x13=z2-z1. Therefore f(x1,z1)=(z1,x13+z1)=z1,z2. Hence fis onto.

Step 2: Find the inverse of the given transformation 

So,fis invertible and therefore the given transformation is invertible. Now f-1(z1,z2)=z2-z13,z1 Therefore inverse of the given transformation is x1=y2-y13x2=y1

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