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Chapter 2: Problem 43

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. A system of linear equations having fewer equations than variables has no solution, a unique solution, or infinitely many solutions.

Short Answer

Expert verified
The statement is true. A system of linear equations with fewer equations than variables can have no solution or infinitely many solutions, because having more variables than equations indicates an underdetermined system, which lacks enough constraints to determine a unique solution. An example with no solution is when two parallel lines never intersect, and an example with infinitely many solutions is when any point lying on a specific line can be a valid solution.

Step by step solution

01

A system of linear equations can have no solution (inconsistent), a unique solution, or infinitely many solutions (dependent). The configuration between the number of equations and variables can play a role in determining the possible types of solutions. #Step 2: Analyze the case of fewer equations than variables#

When the number of equations is smaller than the number of variables, it is "underdetermined." Having more variables than equations means there are not enough constraints to determine a unique solution, which may lead to no solution or infinitely many solutions. #Step 3: Provide an illustration of each possible case#
02

Let's see some examples to illustrate each of the given possibilities in the statement: 1. No solution: Consider the following system of linear equations with 2 variables and 1 equation: \(x - y = 0\) \(x - y = 2\) This system has two parallel lines, and they never intersect. Therefore, it has no solution. 2. Infinitely many solutions: Consider the following system of linear equations with 2 variables and 1 equation: \(x - y = 0\) This system has infinitely many solutions, since any point that lies on the line \(x = y\) is a valid solution. Examples include \((1, 1), (-2, -2), (3, 3)\), and so on. #Step 4: Conclusion#

Based on our analysis and illustration, we can conclude that the statement is true. A system of linear equations with fewer equations than variables can either have no solution or infinitely many solutions. The configuration of having fewer equations than variables does not allow for the possibility of a unique solution.

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Most popular questions from this chapter

\(e\) $$ A=\left[\begin{array}{rr} 2 & 2 \\ -2 & -2 \end{array}\right] $$ Show that \(A^{2}=0\). Compare this with the equation \(a^{2}=0\), where \(a\) is a real number.

A university admissions committee anticipates an enrollment of 8000 students in its freshman class next year. To satisfy admission quotas, incoming students have been categorized according to their sex and place of residence. The number of students in each category is given by the matrix $$ \begin{array}{l} \text { In-state } \\ \text { A= Out-of-state } \\ \text { Foreign } \end{array}\left[\begin{array}{rr} 2700 & 3000 \\ 800 & 700 \\ 500 & 300 \end{array}\right] $$ By using data accumulated in previous years, the admissions committee has determined that these students will elect to enter the College of Letters and Science, the College of Fine Arts, the School of Business Administration, and the School of Engineering according to the percentages that appear in the following matrix: $$ B=\begin{array}{l} \text { Male } \\ \text { Female } \end{array}\left[\begin{array}{llll} 0.25 & 0.20 & 0.30 & 0.25 \\ 0.30 & 0.35 & 0.25 & 0.10 \end{array}\right] $$ Find the matrix \(A B\) that shows the number of in-state, outof-state, and foreign students expected to enter each discipline.

Solve the system of linear equations using the Gauss-Jordan elimination method. $$ \begin{array}{r} 2 x_{1}-x_{2}-x_{3}=0 \\ 3 x_{1}+2 x_{2}+x_{3}=7 \\ x_{1}+2 x_{2}+2 x_{3}=5 \end{array} $$

(a) write each system of equations as a matrix equation and (b) solve the system of equations by using the inverse of the coefficient matrix. $$ \begin{array}{l} \begin{array}{r} x_{1}+x_{2}+x_{3}+x_{4}=b_{1} \\ x_{1}-x_{2}-x_{3}+x_{4}=b_{2} \\ x_{2}+2 x_{3}+2 x_{4}=b_{3} \\ x_{1}+2 x_{2}+x_{3}-2 x_{4}=b_{4} \end{array}\\\ \text { where (i) } b_{1}=1, b_{2}=-1, b_{3}=4, b_{4}=0\\\ \text { and } \quad \text { (ii) } b_{1}=2, b_{2}=8, b_{3}=4, b_{4}=-1 \end{array} $$

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. A system of linear equations having more equations than variables has no solution, a unique solution, or infinitely many solutions.
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