Chapter 6: Problem 1
Prove: If \(\mathbf{F}\) is invertible, then \(\mathbf{F}^{-1}\) is unique.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 6: Problem 1
Prove: If \(\mathbf{F}\) is invertible, then \(\mathbf{F}^{-1}\) is unique.
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Find an affine transformation \(\mathbf{A}\) that so well approximates the branch \(\mathbf{G}\) of \(\mathbf{F}^{-1}\) defined near \(\mathbf{U}_{0}=\mathbf{F}\left(\mathbf{X}_{0}\right)\) that $$\lim _{\mathbf{U} \rightarrow \mathbf{U}_{0}} \frac{\mathbf{G}(\mathbf{U})-\mathbf{A}(\mathbf{U})}{\left|\mathbf{U}-\mathbf{U}_{0}\right|}=\mathbf{0}$$ (a) \(\left[\begin{array}{l}u \\ v\end{array}\right]=\mathbf{F}(x, y)=\left[\begin{array}{l}x^{4} y^{5}-4 x \\ x^{3} y^{2}-3 y\end{array}\right], \quad \mathbf{X}_{0}=(1,-1)\) (b) \(\left[\begin{array}{l}u \\ v\end{array}\right]=\mathbf{F}(x, y)=\left[\begin{array}{c}x^{2} y+x y \\ 2 x y+x y^{2}\end{array}\right], \quad \mathbf{X}_{0}=(1,1)\) (c) \(\left[\begin{array}{l}u \\ v \\ w\end{array}\right]=\mathbf{F}(x, y, z)=\left[\begin{array}{c}2 x^{2} y+x^{3}+z \\ x^{3}+y z \\\ x+y+z\end{array}\right], \quad \mathbf{X}=(0,1,1)\) (d) \(\left[\begin{array}{l}u \\ v \\ w\end{array}\right]=\mathbf{F}(x, y, z)=\left[\begin{array}{c}x \cos y \cos z \\ x \sin y \cos z \\ x \sin z\end{array}\right], \quad \mathbf{X}_{0}=(1, \pi / 2, \pi)\)
Prove: If \(\mathbf{U}=\mathbf{F}(\mathbf{X})\) and \(\mathbf{X}=\mathbf{G}(\mathbf{U})\) are inverse functions, then $$\frac{\partial\left(u_{1}, u_{2}, \ldots, u_{n}\right)}{\partial\left(x_{1}, x_{2}, \ldots, x_{n}\right.} \frac{\partial\left(x_{1}, x_{2}, \ldots, x_{n}\right)}{\partial\left(u_{1}, u_{2}, \ldots, u_{n}\right)}=1$$ Where should the Jacobians be evaluated?
Find \(c \mathbf{A}\). (a) \(c=4, \mathbf{A}=\left[\begin{array}{cccc}2 & 2 & 4 & 6 \\ 0 & 0 & 1 & 3 \\\ 3 & 4 & 7 & 11\end{array}\right]\) (b) \(c=-2, \mathbf{A}=\left[\begin{array}{rrr}1 & 3 & 0 \\ 0 & 1 & 2 \\ 1 & -1 & 3\end{array}\right]\)
Let \(u=u(x, y)\) be determined near (1,1) by $$ x^{2} y u+2 x y^{2} u^{3}-3 x^{3} y^{3} u^{5}=0, \quad u(1,1)=1 $$ Find \(u_{x}(1,1)\) and \(u_{y}(1,1)\)
Find the matrix of \(\mathbf{L}\). (a) \(\mathbf{L}(\mathbf{X})=\left[\begin{array}{l}3 x+4 y+6 z \\ 2 x-47+2 z \\\ 7 x+2 y+3 z\end{array}\right]\) (b) \(\mathbf{L}(\mathbf{X})=\left[\begin{array}{l}2 x_{1}+4 x_{2} \\ 3 x_{1}-2 x_{2} \\ 7 x_{1}-4 x_{2} \\ 6 x_{1}+x_{2}\end{array}\right]\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.