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Chapter 14: Problem 3

Ist ein reelles lineares Gleichungssystem \((\boldsymbol{A} \mid \boldsymbol{b})\) mit \(n\) Unbekannten und \(n\) Gleichungen f眉r ein \(b\) eindeutig l枚sbar, so ist es dies auch f眉r jedes \(b\) - stimmt das?

Short Answer

Expert verified
Answer: The statement is true if and only if the determinant of the coefficient matrix is non-zero. If the determinant is non-zero, the system will have a unique solution regardless of the 'b' vector.

Step by step solution

01

Understand the problem

The problem is asking whether a system of n linear equations with n variables, represented as matrix equation (A | b), is always uniquely solvable for any given vector 'b' if it's solvable for one 'b'. 'A' represents the coefficient matrix, and 'b' represents the solution vector. It's important to note that this is a square system; the number of equations is equal to the number of unknowns.
02

Calculate the determinant

The determinant of a square matrix A is a special number that can be calculated from the matrix. If the determinant of the matrix A (det(A) or |A|) is not zero, then the system of equations has a unique solution. If the determinant is zero, this means that the system of equations does not have a unique solution or no solution at all.
03

Apply Cramer's rule

Cramer's rule is a mathematical formula used to solve systems of linear equations with as many equations as there are variables. According to this rule, if the determinant of the coefficients matrix 'A' is non-zero, then the system is uniquely solvable.
04

Evaluate the statement

If the determinant of matrix 'A' is non-zero, the system will have a unique solution regardless of the 'b' vector. This means that if the system is solvable for a specific 'b' (with non-zero determinant), it is also solvable for any 'b'. If the determinant is zero, it would depend on the 'b' vector whether solutions exist and if they are unique. So, in conclusion, the statement is only necessarily true if the determinant of matrix 'A' is non-zero.
05

Formulate the conclusion

So, the original statement: "A real linear system of equations with n unknowns and n equations is uniquely solvable for one 'b', so it is also for any 'b'" is true if and only if the determinant of the coefficient matrix is non-zero.

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Most popular questions from this chapter

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: $$ \begin{aligned} -2 x_{1}+3 x_{2} &=2 m \\ x_{1}-5 x_{2} &=-11 \end{aligned} $$

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

Im Ursprung \(\mathbf{0}=(0,0,0)\) des \(\mathbb{R}^{3}\) laufen die drei St盲be eines Stabwerks zusammen, die von den Punkten $$ a=(-2,1,-5), \quad b=(2,-2,-4), \quad c=(1,2,-3) $$ ausgehen.

Berechnen Sie die L枚sungsmenge des komplexen linearen Gleichungssystems: $$ \begin{aligned} 2 x_{1} &+\mathrm{i} x_{3} &=\mathrm{i} \\ x_{1}-3 x_{2}-\mathrm{i} x_{3} &=2 \mathrm{i} \\ \mathrm{i} x_{1}+x_{2}+x_{3} &=1+\mathrm{i} \end{aligned} $$

Aus zwei Gold-Silber-Legierungen, in denen sich die Metallmassen wie \(2: 3\) bzw. wie \(3: 7\) verhalten, sind \(8 \mathrm{~kg}\). einer neuen Legierung mit dem Verh盲ltnis \(5: 11\) herzustellen. Wie viel Kilogramm der Legierungen sind dabei zu verwenden?
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