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Expert-verified Found in: Page 97 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Which of the (nonlinear) transformations from ${{\mathbit{R}}}^{{\mathbf{2}}}{\mathbf{}}{\mathbf{to}}{\mathbf{}}{{\mathbit{R}}}^{{\mathbf{2}}}$to in Exercises 25 through 27 are invertible? Find the inverse if it exists.26. role="math" localid="1660365558437" $\mathbf{\left[}\begin{array}{c}{\mathbf{y}}_{\mathbf{1}}\\ {\mathbf{y}}_{\mathbf{2}}\end{array}\mathbf{\right]}{\mathbf{=}}\mathbf{\left[}\begin{array}{c}{\mathbf{x}}_{\mathbf{2}}\\ {\mathbf{x}}_{\mathbf{1}}^{\mathbf{3}}\mathbf{+}{\mathbf{x}}_{\mathbf{2}}\end{array}\mathbf{\right]}$

The inverse of the given transformation is $\left|\begin{array}{c}{x}_{1}=\sqrt{{y}_{2}-{y}_{1}}\\ {x}_{2}={y}_{1}\end{array}\right|$

See the step by step solution

## Step 1: Check whether the transformation is invertible or not

The given transformation is $\left|\begin{array}{c}{y}_{1}={x}_{2}\\ {y}_{2}={x}_{1}^{3}+{x}_{2}\end{array}\right|$.

We can rewrite the transformation as $f;{R}^{2}\to {R}^{2}$.such that $f\left({x}_{1},{x}_{2}\right)=\left({x}_{2},{x}_{1}^{3}+{x}_{2}\right)$

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

(a) Suppose $f\left({x}_{1},{x}_{2}\right)=f\left({z}_{1},{z}_{2}\right)$for some $\left({x}_{1},{x}_{2}\right),f\left({z}_{1},{z}_{2}\right)\in {R}^{2}$.Then ${x}_{1}^{3}={z}_{1}^{3}$and .So ${x}_{1}{z}_{1}$are zero or nonzero together.

Now $\left({x}_{1}-{z}_{1}\right)\left({x}_{1}^{2}+{x}_{1}{z}_{1}+{z}_{1}^{2}\right)={x}_{1}^{3}-{z}_{1}^{3}=0$ (1)

Now ${x}_{1}^{2}+{x}_{1}{z}_{1}+{z}_{1}^{2}=0{x}_{1}{z}_{1}={\left({x}_{1}+{z}_{1}\right)}^{2}$and $-3{x}_{1}{z}_{1}={\left({x}_{1}-{z}_{1}\right)}^{2}$. So ${x}_{1},{z}_{1}=0$. If ${x}_{1}{z}_{1}$are nonzero, then ${{x}_{1}}^{2}+{x}_{1}{z}_{1}+{{z}_{1}}^{2}=0$and hence from (1) ${x}_{1}={z}_{1}$.If ${x}_{1},{z}_{1}$are zero, then ${x}_{1}={z}_{1}$.Thus ${x}_{1}={z}_{1}$and ${{x}_{1}}^{3}={{z}_{1}}^{3}$. Sofis one-one.

(b) Let $\left({z}_{1},{z}_{2}\right)\in {R}^{2}$. Then there exists ${x}_{1}\in R$such that ${{x}_{1}}^{3}={z}_{2}-{z}_{1}$. Therefore $f\left({x}_{1},{z}_{1}\right)=\left({z}_{1},{{x}_{1}}^{3}+{z}_{1}\right)=\left({z}_{1},{z}_{2}\right)$. Hence fis onto.

## Step 2: Find the inverse of the given transformation

So,fis invertible and therefore the given transformation is invertible. Now ${f}^{-1}\left({z}_{1},{z}_{2}\right)=\left(\sqrt{{z}_{2}-{z}_{1}},{z}_{1}\right)$ Therefore inverse of the given transformation is $\left|\begin{array}{c}{x}_{1}=\sqrt{{y}_{2}-{y}_{1}}\\ {x}_{2}={y}_{1}\end{array}\right|$ ### Want to see more solutions like these? 