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Chapter 14: Problem 1

Haben (reelle) lineare Gleichungssysteme mit zwei verschiedenen L枚sungen stets unendlich viele L枚sungen?

Short Answer

Expert verified
Answer: No, it is not true. A real linear system cannot have exactly two different solutions. It can either have no solutions, one unique solution, or infinitely many solutions.

Step by step solution

01

Understanding the properties of linear systems

In a real linear system of equations, each equation represents a straight line in the cartesian coordinate system. The solutions of the linear system are the points where the lines intersect.
02

Analyzing the cases of intersection

There are three possible cases for the intersections of the lines: 1. The lines are parallel and distinct, meaning they never intersect. In this case, there is no solution to the linear system. 2. The lines are coincident, meaning they are, in fact, the same line. In this case, there are infinitely many solutions to the linear system, as every point on the line is a solution. 3. The lines intersect at a single point. In this case, there is one unique solution to the linear system.
03

Evaluate the possibility of two different solutions

According to the cases analyzed in Step 2, a linear system of equations can have no solution, one unique solution, or infinitely many solutions. There is no case in which a linear system can have exactly two different solutions.
04

Conclusion

Based on the analysis of linear systems of equations, a real linear system with two different solutions does not exist. Therefore, it is incorrect to say that real linear systems with two different solutions always have infinitely many solutions.

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Most popular questions from this chapter

Bestimmen Sie die L枚sungsmengen \(L\) der folgenden linearen Gleichungssysteme und untersuchen Sie deren geometrische Interpretationen $$ \begin{array}{rr} 2 x_{1}+3 x_{2}=5 & 2 x_{1}-x_{2}+2 x_{3}=1 \\ x_{1}+x_{2}=2 & x_{1}-2 x_{2}+3 x_{3}=1 \\ 3 x_{1}+x_{2}=1, & 6 x_{1}+3 x_{2}-2 x_{3}=1 \\ x_{1}-5 x_{2}+7 x_{3}=2 \end{array} $$

F眉r welche \(a \in \mathbb{R}\) hat das System $$ \begin{aligned} (a+1) x_{1}+\left(-a^{2}+6 a-9\right) x_{2}+(a-2) x_{3} &=1 \\ \left(a^{2}-2 a-3\right) x_{1}+\left(a^{2}-6 a+9\right) x_{2}+\quad 3 x_{3} &=a-3 \\ (a+1) x_{1}+\left(-a^{2}+6 a-9\right) x_{2}+(a+1) x_{3} &=1 \end{aligned} $$ keine, genau eine bzw. mehr als eine L枚sung? F眉r \(a=0\) und \(a=2\) berechne man alle L枚sungen.

Es sind reelle Zahlen \(a, b, c, d, r, s\) vorgegeben. Begr眉nden Sie, dass das lineare Gleichungssystem $$ \begin{aligned} &a x_{1}+b x_{2}=r \\ &c x_{1}+d x_{2}=s \end{aligned} $$ im Fall \(a d-b c \neq 0\) eindeutig l枚sbar ist und geben Sie die eindeutig bestimmte L枚sung an. Bestimmen Sie zus盲tzlich f眉r \(m \in \mathbb{R}\) die L枚sungsmenge des folgenden linearen Gleichungssystems: $$ \begin{aligned} -2 x_{1}+3 x_{2} &=2 m \\ x_{1}-5 x_{2} &=-11 \end{aligned} $$

Gibt es ein (reelles) lineares Gleichungssystem mit weniger Gleichungen als Unbekannten, welches eindeutig l枚sbar ist?

Beim freien Fall eines K枚rpers vermutet man, dass die Strecke \(s\) in Meter m, welche der fallende K枚rper zur眉cklegt, von der Fallzeit \(t\) in Sekunden s, der Erdbeschleunigung \(g\) in \(\mathrm{m} / \mathrm{s}^{2}\) und des Gewichts \(M\) des fallenden K枚rpers in \(\mathrm{kg}\) abh盲ngt. Ermitteln Sie durch Dimensionsanalyse aus dieser Vermutung eine Formel f眉r die Fallstrecke.
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