Question

Untersuchen Sie die folgenden homogenen Gleichungssysteme und lösen Sie sie falls möglich. a) $$ \begin{gathered} x_{1}+x_{2}-x_{3}=0 & \text { b) } 2 x-3 y+z=0 \\ -x_{1}+3 x_{2}+x_{3}=0 & 4 x+4 y-z=0 \\ x_{2}+x_{3}=0 & x-\frac{3}{2} y+\frac{1}{2} z=0 \end{gathered} $$

Short answer

a) The solutions to the first system are (2s, -s, s) for every real number s. b) The solutions to the second system are (t, -6t, 0) for every real number t.

Step-by-step solution

Solving the first system

Consider the first system: \[ \begin{matrix} x_1+x_2-x_3 &=0 \ -x_1+3x_2+x_3 &=0 \ x_2+x_3 &=0 \end{matrix} \] Notice that the third equation is already simplified as much as possible, implying that \(x_2 = -x_3\). This value can be substituted into the first and second equations, resulting in two new equations: \[ \begin{matrix} x_1 - x_3 - x_3 &=0 \ -x_1 - 3x_3 + x_3 & =0 \end{matrix} \] which further simplify to: \[ \begin{matrix} x_1 - 2x_3 &=0 \ -x_1 - 2x_3 &=0 \end{matrix} \]

Further simplification of the first system

From the above two equations, it's clear that \(x_1 = 2x_3\). This system of equations has an infinite number of solutions. Any value of \(x_3\) would suffice. It's usual to let \(x_3 = s\), where \(s\) is a parameter, thus concluding that solutions are given by \( (x_1, x_2, x_3) = (2s, -s, s) \) with \( s \) in \(\mathbb{R}\).

Solving the second system

Now consider the second system: \[ \begin{matrix} 2x-3y+z &=0 \ 4x+4y-z &=0 \ x-\frac{3}{2}y+\frac{1}{2}z &=0 \end{matrix} \] The key here is to simplify the system by using operations such as multiplication or addition/subtraction of the equations. In this case, the third equation is \(1/2\) times the first equation, indicating this is a dependent system. This can be proven by multiplying the third equation by 2, yielding the first equation, i.e. \(2*(x-\frac{3}{2}y+\frac{1}{2}z) = 2x-3y+z\). Therefore, we have a dependent system with infinite solutions.

Further simplification of the second system

Now, if we add the first two equations, we get an equation \(6x+ y = 0\), which implies \(y = -6x\). Substituting this into the second equation, we find \(4x-4*6x = 0\), which simplifies to \(x = 0\). Thus, we find that \(y = -6x = 0\) and \(z = 3y = 0\). However, as mentioned earlier, this is a dependent system and will have infinite solutions. Thus, every value of \(x\) would suffice, and it is conventional to let \(x = t\), a parameter. Thus, the solutions are given by \( (x, y, z) = (t, -6t, 0) \) with \( t \) in \(\mathbb{R}\).

Key concepts

Linear Algebra

Linear Algebra is a branch of mathematics that deals with vectors, vector spaces, and linear transformations. It is fundamental in solving systems of equations and understanding the structure of mathematical models.

In Linear Algebra, systems of linear equations are crucial, usually expressed in the matrix form:

• Vectors: Central objects that can be added together and scaled.
• Matrices: Rectangular arrangements of numbers representing linear equations.
• Determinants: Special numbers calculated from matrices, helpful in assessing solutions.

These elements are used extensively for computations and theoretical analysis, often finding applications in different scientific disciplines. Understanding these concepts is key to mastering how systems of linear equations function.

System of Equations

A System of Equations is a set of two or more equations with the same set of unknowns. Solving these systems involves finding values for the unknowns that satisfy all equations simultaneously.

There are various methods to solve systems of equations, such as:

• Substitution: Solving one equation for a variable and then using that expression in other equations.
• Elimination: Simplifying the system by adding or subtracting equations to eliminate variables.
• Matrix Method: Using matrices to simplify and solve the system efficiently.

System of Equations can either be consistent (having one or more solutions) or inconsistent (having no solutions). Homogeneous equations, like in our exercise, always have at least one solution where all variables equal zero.

Infinite Solutions

When a System of Equations has infinite solutions, it means there are countless possibilities for the values of the variables that satisfy all the equations. This often occurs when the equations describe the same plane or line in space, essentially overlapping each other at every point.

For instance, in our exercise, both systems provided have an infinite number of solutions due to their homogeneous nature, sometimes indicating dependencies among the equations. You determine the set of solutions by introducing parameters to express the infinite possibilities.

Steps include:

• Identify the dependent equations.
• Substitute one variable with a parameter (e.g., set one variable as a parameter such as 's' or 't').
• Express other variables in terms of this parameter to capture all solutions.

An infinite solution means the system is balanced, with no unique point of intersection.

Dependent Systems

Dependent Systems are systems of equations where at least one of the equations can be derived from others by algebraic manipulation. This typically results in infinitely many solutions or sometimes no solution if not properly aligned.

Identifying factors include:

• Equations that are multiples of each other.
• Systems transforming into identity equations during simplification.
• Reduced row echelon form revealing rows of zeros, indicating overlap.

Recognizing a dependent system means observing that part of the information is redundant. Often, this arises from removing variables that produce zero upon simplification, leaving a parameter for flexibility in solutions. Their presence indicates a potential infinite solution scenario offering diverse solution sets.