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Chapter 1: Problem 6

a. Give the general solution of the equation \(x_{1}+5 x_{2}-2 x_{3}=0\) in \(\mathbb{R}^{3}\) (as a linear combination of two vectors, as in the text). b. Find a specific solution of the equation \(x_{1}+5 x_{2}-2 x_{3}=3\) in \(\mathbb{R}^{3}\); give the general solution. c. Give the general solution of the equation \(x_{1}+5 x_{2}-2 x_{3}+x_{4}=0\) in \(\mathbb{R}^{4}\). Now give the general solution of the equation \(x_{1}+5 x_{2}-2 x_{3}+x_{4}=3\).

Short Answer

Expert verified
The general solutions in R鲁 are: (a) \(\begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}=x_2\begin{bmatrix}-5 \\ 1 \\ 0 \end{bmatrix}+x_3\begin{bmatrix}2 \\ 0 \\ 1 \end{bmatrix}\), (b) \(\begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix}+x_2\begin{bmatrix}-5 \\ 1 \\ 0 \end{bmatrix}+x_3\begin{bmatrix}2 \\ 0 \\ 1 \end{bmatrix}\), and for R鈦�: (c) \(\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}=\begin{bmatrix} 3 \\ 0 \\ 0 \\ 0 \end{bmatrix}+x_2\begin{bmatrix}-5 \\ 1 \\ 0 \\ 0\end{bmatrix}+x_3\begin{bmatrix}2 \\ 0 \\ 1 \\ 0\end{bmatrix}+x_4\begin{bmatrix}-1 \\ 0 \\ 0 \\ 1\end{bmatrix}\).

Step by step solution

01

Part a: Finding general solution in R鲁

The given equation is: \(x_{1}+5x_{2}-2x_{3}=0\) To express this in terms of linear combinations of two vectors, let's consider the equation as a matrix equation: \[ \begin{bmatrix}1 & 5 & -2\end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 \] Let's solve this equation to find the general solution as a linear combination of two vectors. We'll do this by expressing \(x_1\) in terms of free variables \(x_2\) and \(x_3\). \(x_1 = -5x_2 + 2x_3\) Now we can write the general solution as: \(\begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}=x_2\begin{bmatrix}-5 \\ 1 \\ 0 \end{bmatrix}+x_3\begin{bmatrix}2 \\ 0 \\ 1 \end{bmatrix}\) So, the general solution in R鲁 is a linear combination of the vectors \(\begin{bmatrix}-5 \\ 1 \\ 0 \end{bmatrix}\) and \(\begin{bmatrix}2 \\ 0 \\ 1 \end{bmatrix}\).
02

Part b: Finding specific and general solutions in R鲁

The given equation is: \(x_{1}+5x_{2}-2x_{3}=3\) Using the general solution found in part (a), we can add a specific solution to get the final general solution to this equation. Now we need to find one specific solution, so we can set one variable, say \(x_3\), equal to 0, to find a combination of \(x_1\) and \(x_2\). Thus, we have the equation: \(x_{1}+5x_{2}-2\cdot 0 =3\) Setting \(x_2 = 0\): \(x_1 = 3\) Now the specific solution is \(\begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix}\). The general solution can be expressed as a linear combination of the previous general solution found in part (a), plus the specific solution: \(\begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix}+x_2\begin{bmatrix}-5 \\ 1 \\ 0 \end{bmatrix}+x_3\begin{bmatrix}2 \\ 0 \\ 1 \end{bmatrix}\)
03

Part c: Finding general solutions in R鈦�

For this part, we have the equations: 1. \(x_{1}+5x_{2}-2x_{3}+x_{4}=0\) 2. \(x_{1}+5x_{2}-2x_{3}+x_{4}=3\) First, we'll solve equation (1) for the general solution in R鈦�. We can express this equation as a matrix equation as well: \[ \begin{bmatrix}1 & 5 & -2 & 1\end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = 0 \] We notice that \(x_2\), \(x_3\), and \(x_4\) are free variables, and we can write \(x_1\) in terms of these free variables. \(x_1 = -5x_2 + 2x_3 - x_4\) The general solution can then be written as a linear combination of these vectors: \(\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}=x_2\begin{bmatrix}-5 \\ 1 \\ 0 \\ 0\end{bmatrix}+x_3\begin{bmatrix}2 \\ 0 \\ 1 \\ 0\end{bmatrix}+x_4\begin{bmatrix}-1 \\ 0 \\ 0 \\ 1\end{bmatrix}\) Now we need to find the general solution for equation (2) in R鈦�. For this, we can simply add a specific solution to the general solution found for equation (1). Setting \(x_2 = 0, x_3 = 0, x_4 = 0\), we can find a specific solution for this equation: \(x_1 = 3\) The specific solution is \(\begin{bmatrix} 3 \\ 0 \\ 0 \\ 0 \end{bmatrix}\). Therefore, the general solution for equation (2) in R鈦� is: \(\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}=\begin{bmatrix} 3 \\ 0 \\ 0 \\ 0 \end{bmatrix}+x_2\begin{bmatrix}-5 \\ 1 \\ 0 \\ 0\end{bmatrix}+x_3\begin{bmatrix}2 \\ 0 \\ 1 \\ 0\end{bmatrix}+x_4\begin{bmatrix}-1 \\ 0 \\ 0 \\ 1\end{bmatrix}\)

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Key Concepts

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

Systems of Linear Equations
Systems of linear equations involve multiple linear equations that are solved together. Each equation relates linear combinations of variables, and the goal is to find values for these variables that satisfy all the equations simultaneously. In the given exercise, the equations provided each involve three or four variables, indicating a system that exists in three-dimensional or four-dimensional space.

The solutions to these systems are often expressed as points, lines, or planes in space, depending on how many solutions exist. There can be a single solution, infinitely many solutions, or no solution at all. The nature of these solutions is determined by the arrangement and the coefficients in the equations.

Understanding systems of linear equations is crucial because they are foundational in algebra. They form the basis for more complex operations in linear algebra, such as transformations and matrix analysis, which are useful in fields like engineering, computer science, and economics.
General Solution
The general solution of a system of linear equations provides a description of all possible solutions. It often involves expressing one or more variables as a function of free variables. These free variables can take any value, giving the solution its range of possibilities.

For example, in part (a) of the exercise, the general solution in \( \mathbb{R}^{3}\) is a linear combination of two vectors: \( \begin{bmatrix}x_1 \ x_2 \ x_3 \end{bmatrix}=x_2\begin{bmatrix}-5 \ 1 \ 0 \end{bmatrix}+x_3\begin{bmatrix}2 \ 0 \ 1 \end{bmatrix}\).

This means that the general solution can be described with two parameters (the free variables \(x_2\) and \(x_3\)), allowing an infinite number of solutions that form a plane in the three-dimensional space. For part (c) of the exercise, the addition of a fourth dimension means even more freedom in the solution, often leading to solutions described as hyperplanes, extending infinitely in four-dimensional space.
Vector Representation
Vector representation is fundamental in describing solutions to systems of linear equations. Each solution can often be expressed as a linear combination of vectors. In the context of the given exercise, we take the solutions that are determined by free variables and express these as weighted sums (linear combinations) of basis vectors.

These basis vectors guide us along the directions in which the solution expands. In \(\mathbb{R}^{3}\) for example, the vectors \(\begin{bmatrix}-5 \ 1 \ 0 \end{bmatrix}\) and \(\begin{bmatrix}2 \ 0 \ 1 \end{bmatrix}\) span the plane of solutions, illustrating how changes in the free variables traverse the solution space. This allows for visualization and deeper understanding of the solution set as a geometric object, enriching our comprehension beyond mere numeric values.

It is through vector representation that abstract algebraic concepts are given shape, supporting applications such as computer graphics and physics simulations, where the position or state of objects is frequently described in vector form.
Matrix Equation
A matrix equation succinctly represents a system of linear equations, making complex calculations easier. Instead of writing individual equations, we can express the entire system with matrices, such as \(A\mathbf{x} = \mathbf{b}\), where \(A\) is the coefficient matrix, \(\mathbf{x}\) is the vector of variables, and \(\mathbf{b}\) is the result vector.

In the original exercise, the matrix form simplifies solving by allowing us to use linear algebra techniques such as row reduction, scaling matrices, or employing software for larger systems. This is advantageous because operations on matrices follow consistent rules, which simplifies manipulation, especially for computer algorithms.

The matrix representation is fundamental for algorithms in numerical methods and optimizations, forming the basis of many existence and uniqueness theorems in linear algebra, like the rank theorem or Cramer's rule. Understanding this formalism is essential for advanced topics in mathematics and real-world applications.

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Most popular questions from this chapter

Find a normal vector to the given hyperplane and use it to find the distance from the origin to the hyperplane. a. \(\mathbf{x}=(-1,2)+t(3,2)\) b. The plane in \(\mathbb{R}^{3}\) given by the equation \(2 x_{1}+x_{2}-x_{3}=5\) "c. The plane passing through \((1,2,2)\) and orthogonal to the line \(\mathbf{x}=(3,1,-1)+\) \(t(-1,1,-1)\) d. The plane passing through \((2,-1,1)\) and orthogonal to the line \(\mathbf{x}=(3,1,1)+\) \(t(-1,2,1)\) *e. The plane spanned by \((1,1,4)\) and \((2,1,0)\) and passing through \((1,1,2)\) f. The plane spanned by \((1,1,1)\) and \((2,1,0)\) and passing through \((3,0,2)\) g. The hyperplane in \(\mathbb{R}^{4}\) spanned by \((1,-1,1,-1),(1,1,-1,-1)\), and \((1,-1,-1,1)\) and passing through \((2,1,0,1)\)

(A computer or calculator may be helpful in solving this problem.) Find numbers \(a, b\), \(c, d, e\), and \(f\) so that the five points \((0,2),(-3,0),(1,5),(1,1)\), and \((-1,1)\) all lie on the conic $$ a x^{2}+b x y+c y^{2}+d x+e y+f=0 . $$ Show, moreover, that \(a, b, c, d, e\), and \(f\) are uniquely determined up to a common factor.

Verify algebraically that the following properties of vector arithmetic hold. (Do so for \(n=2\) if the general case is too intimidating.) Give the geometric interpretation of each property. a. For all \(\mathbf{x}, \mathbf{y} \in \mathbb{R}^{n}, \mathbf{x}+\mathbf{y}=\mathbf{y}+\mathbf{x}\). b. For all \(\mathbf{x}, \mathbf{y}, \mathbf{z} \in \mathbb{R}^{n},(\mathbf{x}+\mathbf{y})+\mathbf{z}=\mathbf{x}+(\mathbf{y}+\mathbf{z})\). c. \(\mathbf{0}+\mathbf{x}=\mathbf{x}\) for all \(\mathbf{x} \in \mathbb{R}^{n}\). d. For each \(\mathbf{x} \in \mathbb{R}^{n}\), there is a vector \(-\mathbf{x}\) so that \(\mathbf{x}+(-\mathbf{x})=\mathbf{0}\). e. For all \(c, d \in \mathbb{R}\) and \(\mathbf{x} \in \mathbb{R}^{n}, c(d \mathbf{x})=(c d) \mathbf{x}\). f. For all \(c \in \mathbb{R}\) and \(\mathbf{x}, \mathbf{y} \in \mathbb{R}^{n}, c(\mathbf{x}+\mathbf{y})=c \mathbf{x}+c \mathbf{y}\). g. For all \(c, d \in \mathbb{R}\) and \(\mathbf{x} \in \mathbb{R}^{n},(c+d) \mathbf{x}=c \mathbf{x}+d \mathbf{x}\). h. For all \(\mathbf{x} \in \mathbb{R}^{n}, 1 \mathbf{x}=\mathbf{x}\).

(from Henry Burchard Fine's A College Algebra, 1905) Two points move at constant rates along the circumference of a circle whose length is \(150 \mathrm{ft}\). When they move in opposite senses they meet every 5 seconds; when they move in the same sense they are together every 25 seconds. What are their rates?

Find the general solution of each of the following equations (presented, as in the text, as a combination of an appropriate number of vectors). a. \(x_{1}-2 x_{2}+3 x_{3}=4\left(\right.\) in \(\left.\mathbb{R}^{3}\right)\) "d. \(x_{1}-2 x_{2}+3 x_{3}=4\left(\right.\) in \(\left.\mathbb{R}^{4}\right)\) b. \(x_{1}+x_{2}-x_{3}+2 x_{4}=0\left(\right.\) in \(\left.\mathbb{R}^{4}\right)\) e. \(x_{2}+x_{3}-3 x_{4}=2\left(\right.\) in \(\left.\mathbb{R}^{4}\right)\) \({ }^{*} \mathrm{c} . x_{1}+x_{2}-x_{3}+2 x_{4}=5\) (in \(\mathbb{R}^{4}\) )
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