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Chapter 7: Problem 28

Deal with the problem of solving \(\mathbf{A x}=\mathbf{b}\) when \(\operatorname{det} \mathbf{A}=0\) Suppose that det \(\mathbf{\Lambda}=0,\) and that \(\mathbf{x}=\mathbf{x}^{(0)}\) is a solution of \(\mathbf{A} \mathbf{x}=\mathbf{b} .\) Show that if \(\xi\) is a solution of \(\mathbf{A} \xi=\mathbf{0}\) and \(\alpha\) is any constant, then \(\mathbf{x}=\mathbf{x}^{(0)}+\alpha \xi\) is also a solution of \(\mathbf{A} \mathbf{x}=\mathbf{b} .\)

Short Answer

Expert verified
Short Answer: The proof relies on the given information that \(\mathbf{x}^{(0)}\) is a solution to the equation \(\mathbf{A x} = \mathbf{b}\) and \(\xi\) is a solution to the homogeneous system \(\mathbf{A} \xi =\mathbf{0}\). By expressing the vector \(\mathbf{x}\) as \(\mathbf{x} = \mathbf{x}^{(0)} + \alpha \xi\) and performing matrix operations, we arrive at the equation \(\mathbf{A} \mathbf{x} = \mathbf{b}\), proving that the sum of \(\mathbf{x}^{(0)}\) and any scalar multiple of \(\xi\) is also a solution to the original system of equations.

Step by step solution

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01

Define given information

We are given the following information: - \(\mathbf{A}\) is a matrix such that \(\operatorname{det} \mathbf{A} = 0\) - \(\mathbf{x}^{(0)}\) is a solution of the system of equations \(\mathbf{A x} = \mathbf{b}\) - \(\xi\) is a solution of the homogeneous system of equations \(\mathbf{A} \xi = \mathbf{0}\) - \(\alpha\) is any constant Our goal is to prove that the vector \(\mathbf{x} = \mathbf{x}^{(0)} + \alpha \xi\) is also a solution of the system of equations \(\mathbf{A x} = \mathbf{b}\).
02

Calculate the product \(\mathbf{A x}\)

Now we will find \(\mathbf{A} \mathbf{x}\) by substituting the expression of \(\mathbf{x}\) in the equation: $$\mathbf{A} \mathbf{x} = \mathbf{A} (\mathbf{x}^{(0)} + \alpha \xi)$$
03

Distribute the matrix \(\mathbf{A}\) over the sum

Now distribute the matrix \(\mathbf{A}\) over the sum inside the parenthesis: $$\mathbf{A} \mathbf{x} = \mathbf{A} \mathbf{x}^{(0)} + \mathbf{A} (\alpha \xi)$$
04

Utilize the given information and properties of matrices

Recall, it is given that \(\mathbf{A} \mathbf{x}^{(0)} = \mathbf{b}\) and \(\mathbf{A} \xi = \mathbf{0}\). We can rewrite the equation in step 3 utilizing this information: $$\mathbf{A} \mathbf{x} = \mathbf{b} + \alpha \mathbf{0}$$
05

Simplify the result

The scalar multiplication of \(\mathbf{0}\) with any constant \(\alpha\) will result in \(\mathbf{0}\): $$\mathbf{A} \mathbf{x} = \mathbf{b} + \mathbf{0}$$ Since the sum of \(\mathbf{b}\) and \(\mathbf{0}\) will be the same as \(\mathbf{b}\), our final equation is: $$\mathbf{A} \mathbf{x} = \mathbf{b}$$ This proves that \(\mathbf{x} = \mathbf{x}^{(0)} + \alpha \xi\) is also a solution of the given system of equations, as desired.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Homogeneous Equations
Homogeneous equations form a fundamental category within linear algebra. They are characterized by their simplicity: all of these equations equal zero. In mathematical terms, if we have a matrix \textbf{A} and a vector \textbf{x}, a homogeneous equation is given by \( \mathbf{A} \mathbf{x} = \mathbf{0} \).

When dealing with systems of linear equations, the notion of homogeneity provides a way to understand the structure and solution space of the system. If \( \mathbf{x}^{(0)} \) is a particular solution to the non-homogeneous equation \( \mathbf{A} \mathbf{x} = \mathbf{b} \) where \( \mathbf{b} \) is non-zero, any solution \( \xi \) to the associated homogeneous equation can be scaled by a constant \( \alpha \) and added to \( \mathbf{x}^{(0)} \) to yield another solution to the original equation. This principle is critical in understanding the linearity of these systems and how solutions can be 'built' from a particular solution and the homogeneous solution space.
Matrix Algebra
Matrix algebra involves operations such as addition, multiplication, and scalar multiplication that can be performed on matrices. Key to solving systems of equations using matrix algebra is the ability to manipulate matrices and vectors to reveal solutions.

In the exercise provided, the product \( \mathbf{A} \mathbf{x} \) is evaluated using matrix multiplication. It's important to understand that matrix multiplication is associative, allowing us to distribute \( \mathbf{A} \) across a sum within a product, as in \( \mathbf{A} (\mathbf{x}^{(0)} + \alpha \xi) = \mathbf{A} \mathbf{x}^{(0)} + \mathbf{A} (\alpha \xi) \). Furthermore, scalar multiplication interacts with matrix multiplication in such a way that \( \alpha \) can be factored in or out from the matrix product, a property used neatly in the solution to the exercise.
Determinants and Singularity
Determinants play a critical role in matrix theory, helping to determine whether a matrix is invertible or singular. The determinant of a square matrix \( \mathbf{A} \) is a scalar value that encodes certain properties of the matrix. When \( \operatorname{det} \mathbf{A} = 0 \)—indicating the matrix is singular—the matrix does not have an inverse, and this has significant implications for solving linear systems.

A singular matrix corresponds to a linear system that either has no solutions or an infinite number of solutions. In the provided exercise, the fact that \( \mathbf{A} \) is singular but a solution \( \mathbf{x}^{(0)} \) exists implies that there may be an infinite number of solutions. This foundational understanding facilitates our comprehension of why adding any scalar multiple of a solution to the homogeneous equation to a particular solution of the non-homogeneous equation yields another valid solution.
Superposition Principle
The superposition principle is a fundamental concept in linear systems which states that the sum of individual solutions to a linear equation still constitutes a valid solution. This concept arises directly from the linearity property of these equations, where adding or scaling solutions does not affect the validity of the results.

In our exercise scenario, this principle manifests in the conclusion that \( \mathbf{x} = \mathbf{x}^{(0)} + \alpha \xi \) is also a solution to \( \mathbf{A} \mathbf{x} = \mathbf{b} \). This demonstrates how, through superposition, we find a spectrum of solutions from a singular matrix equation. Notably, this principle is not confined to the realm of mathematics but is also applied in various fields of physics and engineering to analyze systems subjected to multiple forces or influences simultaneously.

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Most popular questions from this chapter

Consider the initial value problem $$ x^{\prime}=A x+g(t), \quad x(0)=x^{0} $$ (a) By referring to Problem \(15(c)\) in Section \(7.7,\) show that $$ x=\Phi(t) x^{0}+\int_{0}^{t} \Phi(t-s) g(s) d s $$ (b) Show also that $$ x=\exp (A t) x^{0}+\int_{0}^{t} \exp [\mathbf{A}(t-s)] \mathbf{g}(s) d s $$ Compare these results with those of Problem 27 in Section \(3.7 .\)

The coefficient matrix contains a parameter \(\alpha\). In each of these problems: (a) Determine the eigervalues in terms of \(\alpha\). (b) Find the critical value or values of \(\alpha\) where the qualitative nature of the phase portrait for the system changes. (c) Draw a phase portrait for a value of \(\alpha\) slightly below, and for another value slightly above, each crititical value. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\alpha} & {10} \\ {-1} & {-4}\end{array}\right) \mathbf{x} $$

Consider a \(2 \times 2\) system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\). If we assume that \(r_{1} \neq r_{2}\), the general solution is \(\mathbf{x}=c_{1} \xi^{(1)} e^{t_{1}^{\prime}}+c_{2} \xi^{(2)} e^{\prime 2},\) provided that \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent In this problem we establish the linear independence of \(\xi^{(1)}\) and \(\xi^{(2)}\) by assuming that they are linearly dependent, and then showing that this leads to a contradiction. $$ \begin{array}{l}{\text { (a) Note that } \xi \text { (i) satisfies the matrix equation }\left(\mathbf{A}-r_{1} \mathbf{I}\right) \xi^{(1)}=\mathbf{0} ; \text { similarly, note that }} \\ {\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(2)}=\mathbf{0}} \\ {\text { (b) Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(1)}=\left(r_{1}-r_{2}\right) \mathbf{\xi}^{(1)}} \\\ {\text { (c) Suppose that } \xi^{(1)} \text { and } \xi^{(2)} \text { are linearly dependent. Then } c_{1} \xi^{(1)}+c_{2} \xi^{(2)}=\mathbf{0} \text { and at least }}\end{array} $$ $$ \begin{array}{l}{\text { one of } c_{1} \text { and } c_{2} \text { is not zero; suppose that } c_{1} \neq 0 . \text { Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=\mathbf{0}} \\ {\text { and also show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=c_{1}\left(r_{1}-r_{2}\right) \boldsymbol{\xi}^{(1)} \text { . Hence } c_{1}=0, \text { which is }} \\\ {\text { a contradiction. Therefore } \xi^{(1)} \text { and } \boldsymbol{\xi}^{(2)} \text { are linearly independent. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Modify the argument of part (c) in case } c_{1} \text { is zero but } c_{2} \text { is not. }} \\ {\text { (e) Carry out a similar argument for the case in which the order } n \text { is equal to } 3 \text { ; note that }} \\ {\text { the procedure can be extended to cover an arbitrary value of } n .}\end{array} $$

Consider the system $$ x^{\prime}=A x=\left(\begin{array}{rrr}{1} & {1} & {1} \\ {2} & {1} & {-1} \\\ {-3} & {2} & {4}\end{array}\right) x $$ (a) Show that \(r=2\) is an eigenvalue of multiplicity 3 of the coefficient matrix \(\mathbf{A}\) and that there is only one corresponding cigenvector, namely, $$ \xi^{(1)}=\left(\begin{array}{r}{0} \\ {1} \\ {-1}\end{array}\right) $$ (b) Using the information in part (a), write down one solution \(\mathbf{x}^{(1)}(t)\) of the system (i). There is no other solution of the purely exponential form \(\mathbf{x}=\xi e^{y t}\). (c) To find a second solution assume that \(\mathbf{x}=\xi t e^{2 t}+\mathbf{\eta} e^{2 t} .\) Show that \(\xi\) and \(\mathbf{\eta}\) satisfy the equations $$ (\mathbf{A}-2 \mathbf{I}) \xi=\mathbf{0}, \quad(\mathbf{A}-2 \mathbf{I}) \mathbf{n}=\mathbf{\xi} $$ since \(\xi\) has already been found in part (a), solve the second equation for \(\eta\). Neglect the multiple of \(\xi^{(1)}\) that appears in \(\eta\), since it leads only to a multiple of the first solution \(\mathbf{x}^{(1)}\). Then write down a second solution \(\mathbf{x}^{(2)}(t)\) of the system (i). (d) To find a third solution assume that \(\mathbf{x}=\xi\left(t^{2} / 2\right) e^{2 t}+\mathbf{\eta} t e^{2 t}+\zeta e^{2 t} .\) Show that \(\xi, \eta,\) and \(\zeta\) satisfy the equations $$ (\mathbf{A}-2 \mathbf{l}) \xi=\mathbf{0}, \quad(\mathbf{\Lambda}-2 \mathbf{I}) \mathbf{\eta}=\mathbf{\xi}, \quad(\mathbf{A}-2 \mathbf{l}) \zeta=\mathbf{\eta} $$ The first two equations are the same as in part (c), so solve the third equation for \(\zeta,\) again neglecting the multiple of \(\xi^{(1)}\) that appears. Then write down a third solution \(\mathbf{x}^{(3)}(t)\) of the system (i). (e) Write down a fundamental matrix \(\boldsymbol{\Psi}(t)\) for the system (i). (f) Form a matrix \(\mathbf{T}\) with the cigenvector \(\xi^{(1)}\) in the first column, and the generalized eigenvectors \(\eta\) and \(\zeta\) in the second and third columns. Then find \(T^{-1}\) and form the product \(\mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}\). The matrix \(\mathbf{J}\) is the Jordan form of \(\mathbf{A}\).

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{cc}{-3} & {3 / 4} \\ {-5} & {1}\end{array}\right) $$
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