Chapter 9: Problem 20
\(\mathbf{A}=\left[ \begin{array}{rr}{5} & {4} \\ {-1} & {0}\end{array}\right]\)
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 9: Problem 20
\(\mathbf{A}=\left[ \begin{array}{rr}{5} & {4} \\ {-1} & {0}\end{array}\right]\)
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
The inner product of two vectors is a generalization of the dot product, for vectors with complex entries. It is defined by $$ \begin{aligned} &(\mathbf{x}, \mathbf{y}):=\sum_{i=1}^{n} x \bar{y}_{i} &\text { where } \end{aligned} $$ \(\mathbf{x}=\operatorname{col}\left(x_{1}, x_{2}, \ldots, x_{n}\right), \quad \mathbf{y}=\operatorname{col}\left(y_{1}, y_{2}, \ldots, y_{n}\right) \quad\) are complex vectors and the overbar denotes complex conjugation. (a) Show that \((\mathbf{x}, \mathbf{y})=\mathbf{x}^{T} \overline{\mathbf{y}},\) where \(\overline{\mathbf{y}}=\operatorname{col}\left(\bar{y}_{1}, \bar{y}_{2}, \ldots, \bar{y}_{n}\right)\) (b) Prove that for any \(n \times 1\) vectors \(\mathbf{x}, \mathbf{y}, \mathbf{z}\) and any complex number \(\lambda\), we have $$ \begin{array}{l} (\mathbf{x}, \mathbf{y})=\overline{(\mathbf{y}, \mathbf{x})} \\ (\mathbf{x}, \mathbf{y}+\mathbf{z})=(\mathbf{x}, \mathbf{y})+(\mathbf{x}, \mathbf{z}) \\ (\lambda \mathbf{x}, \mathbf{y})=\lambda(\mathbf{x}, \mathbf{y}), \quad(\mathbf{x}, \lambda \mathbf{y})=\bar{\lambda}(\mathbf{x}, \mathbf{y}) \end{array} $$
\(\mathbf{A}=\left[ \begin{array}{rr}{1} & {1} \\ {-2} & {4}\end{array}\right]\)
Let \(\mathbf{A} :=\left[ \begin{array}{rr}{1} & {2} \\ {1} & {1}\end{array}\right], \quad \mathbf{B} :=\left[ \begin{array}{ll}{0} & {3} \\\ {1} & {2}\end{array}\right], \quad\) and \(\mathbf{C} :=\left[ \begin{array}{rr}{1} & {-4} \\ {1} & {1}\end{array}\right].\) Find: (a) AB. (b) (AB)C. (c) (A+B)C.
37\. To illustrate the connection between a higher-order equation and the equivalent first-order system, consider the equation $$\quad y^{\prime \prime \prime}(t)-6 y^{\prime \prime}(t)+11 y^{\prime}(t)-6 y(t)=0.$$ (a) Show that \(\left\\{e^{t}, e^{2 t}, e^{3 t}\right\\}\) is a fundamental solution set for \((11)\) . (b) Using the definition in Section 6.1 , compute the Wronskian of \(\left(e^{t}, e^{2 t}, e^{3 t}\right)\) . (c) Setting \(x_{1}=y, x_{2}=y^{\prime}, x_{3}=y^{\prime \prime},\) show that equa- tion \((11)\) is equivalent to the first-order system $$(12) \quad \mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}$$ where $$\mathbf{A} :=\left[ \begin{array}{rrr}{0} & {1} & {0} \\ {0} & {0} & {1} \\\ {6} & {-11} & {6}\end{array}\right].$$ (d) The substitution used in part (c) suggests that $$S :=\left\\{\left[ \begin{array}{c}{e^{t}} \\ {e^{t}} \\\ {e^{t}}\end{array}\right], \left[ \begin{array}{c}{e^{2 t}} \\ {2 e^{2 t}} \\\ {4 e^{2 t}}\end{array}\right], \left[ \begin{array}{c}{e^{3 t}} \\ {3 e^{3 t}} \\ {9 e^{3 t}}\end{array}\right]\right\\}$$ is a fundamental solution set for system \((12) .\) Verify that this is the case. (e) Compute the Wronskian of \(S .\) How does it compare with the Wronskian computed in part (b)?
Use the variation of parameters formula \((10)\) to derive a formula for a particular solution \(y_{p}\) to the scalar equation \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=g(t)\) in terms of two linearly independent solutions \(y_{1}(t), y_{2}(t)\) of the corresponding homogeneous equation. Show that your answer agrees with the formulas derived in Section \(4.6 .\) [Hint: First write the scalar equation in system form.]
What do you think about this solution?
We value your feedback to improve our textbook solutions.