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(a) Find a homogeneous linear system of two equations in three unknowns whose solution space consists of those vectors in \(R^{3}\) that are orthogonal to \(\mathbf{a}=(-3,2,-1)\) and \(\mathbf{b}=(0,-2,-2)\) (b) What kind of geometric object is the solution space? (c) Find a general solution of the system obtained in part (a), and confirm that Theorem 3.4 .3 holds.

Short Answer

Expert verified
(a) System: \([-3x_1 + 2x_2 - x_3 = 0, x_2 + x_3 = 0\]) \(b) A line in \( R^3 \) (c) General solution: \( t(-1, -1, 1) \), confirming the solution is a subspace.

Step by step solution

01

Identify Orthogonality Conditions

We need to find a homogeneous linear system where the solution vectors \( \mathbf{x} = (x_1, x_2, x_3) \) in \( R^3 \) are orthogonal to both \( \mathbf{a} = (-3, 2, -1) \) and \( \mathbf{b} = (0, -2, -2) \). To be orthogonal, the dot product of \( \mathbf{x} \) with each vector should be zero: \( \mathbf{a} \cdot \mathbf{x} = 0 \) and \( \mathbf{b} \cdot \mathbf{x} = 0 \).
02

Write Orthogonality Equations

The orthogonality condition for \( \mathbf{a} \) gives us the equation:\[ -3x_1 + 2x_2 - x_3 = 0. \]For \( \mathbf{b} \), the orthogonality condition gives us the equation:\[ -2x_2 - 2x_3 = 0. \]
03

Simplify the Equations

From the second equation, simplify the expression:\[ -2x_2 - 2x_3 = 0 \] which reduces to:\[ x_2 + x_3 = 0 \] or \[ x_2 = -x_3. \]
04

Describe the Homogeneous System

The homogeneous linear system consists of:1. \(-3x_1 + 2x_2 - x_3 = 0 \)2. \( x_2 + x_3 = 0 \).This system represents the vectors orthogonal to the given vectors \( \mathbf{a} \) and \( \mathbf{b} \).
05

Solve the System to Find the Solution Space

Substitute \( x_2 = -x_3 \) from the second equation into the first:\[ -3x_1 + 2(-x_3) - x_3 = 0 \]which simplifies to:\[ -3x_1 - 3x_3 = 0 \].Thus, \( x_1 = -x_3 \).Set \( x_3 = t \), a parameter, so:\[ x_1 = -t, \quad x_2 = -x_3 = -t, \quad x_3 = t \].The general solution is \( (x_1, x_2, x_3) = t(-1, -1, 1) \).
06

Determine the Geometric Object

The set of all solutions \( (x_1, x_2, x_3) = t(-1, -1, 1) \) with parameter \( t \) form a line through the origin in \( R^3 \).
07

Verify Theorem 3.4.3

Theorem 3.4.3 states that the solution space of a homogeneous linear system is a subspace of \( R^n \). Here, the line through the origin \( t(-1, -1, 1) \) is a subspace of \( R^3 \), confirming the theorem.

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Key Concepts

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

Orthogonality Conditions
Orthogonality is a mathematical condition where two vectors are perpendicular to each other. In this exercise, we are tasked with finding a homogeneous linear system where vectors in \(R^3\), denoted by \(\mathbf{x} = (x_1, x_2, x_3)\), must be orthogonal to the given vectors \(\mathbf{a} = (-3, 2, -1)\) and \(\mathbf{b} = (0, -2, -2)\). To determine orthogonality, we use the dot product rule: two vectors are orthogonal if their dot product is zero. This leads us to establish the two conditions:
  • \(\mathbf{a} \cdot \mathbf{x} = -3x_1 + 2x_2 - x_3 = 0\)
  • \(\mathbf{b} \cdot \mathbf{x} = -2x_2 - 2x_3 = 0\)
These equations express the need for the solution vector \(\mathbf{x}\) to be orthogonal to both \(\mathbf{a}\) and \(\mathbf{b}\), forming the basis of our next steps in defining the system.
Solution Space
The solution space of a homogeneous linear system refers to the set of all vectors that satisfy the given linear equations. After establishing the orthogonality conditions, we translate them into equations: 1. \(-3x_1 + 2x_2 - x_3 = 0\) and 2. \(x_2 + x_3 = 0\). These equations must be solved simultaneously to understand the vectors \(\mathbf{x} = (x_1, x_2, x_3)\) that satisfy both conditions. By substituting \(x_2 = -x_3\) from the second equation into the first, we simplify the system to a single parameter representation. This reduction helps identify the set of vectors forming the solution space. Solving these provides insight into how vectors can be manipulated in a space defined by linear equations and is crucial for deeper exploration.
General Solution
Finding the general solution of a linear system involves expressing the solutions in terms of one or more parameters. Here, the reduced form \(-3x_1 - 3x_3 = 0\) leads to \(x_1 = -x_3\). By setting \(x_3 = t\), where \(t\) is a parameter, we express the general solution as:
  • \(x_1 = -t\)
  • \(x_2 = -t\)
  • \(x_3 = t\)
This gives \( (x_1, x_2, x_3) = t(-1, -1, 1) \), which indicates that every solution is a scalar multiple of the vector \((-1, -1, 1)\). This general solution captures the entire set of solutions for the given linear system, providing a comprehensive understanding of the vectors that fit the criteria.
Geometric Object in R^3
In three-dimensional space \(R^3\), the solution space of a system of equations can describe various geometric objects, such as points, lines, or planes. In this scenario, the vector equation \( (x_1, x_2, x_3) = t(-1, -1, 1) \) describes a line passing through the origin of the coordinate system in \(R^3\).
  • The parameter \(t\) allows the vector to extend infinitely in both directions on the line.
  • The line only includes points where solutions are orthogonal to both \(\mathbf{a}\) and \(\mathbf{b}\), fulfilling the orthogonality condition.
Furthermore, according to Theorem 3.4.3, we have confirmed that the solution space must be a subspace in \(R^3\), which indeed a line through the origin is. Understanding these geometric implications helps visualize the solution space in a three-dimensional context, bridging algebraic solutions with geometric interpretations.

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