Question

Untersuchen Sie das lineare Gleichungssystem \(A_{k} \mathbf{x}=\mathbf{b}_{k}\) auf Lösbarkeit in Anhängigkeit der reellen Parameter \(\alpha, \beta\) und ermitteln Sie gegebenenfalls alle Lösungen für $$ A_{1}=\left(\begin{array}{lll} 1 & 3 & 1 \\ 2 & 3 & 1 \\ 3 & 3 & \alpha \end{array}\right), \mathbf{b}_{1}=\left(\begin{array}{l} 5 \\ 5 \\ 5 \end{array}\right) $$A_{2}=\left(\begin{array}{rrr} 1 & 1 & -1 \\ 2 & 3 & \beta \\ 1 & \beta & 3 \end{array}\right), \mathbf{b}_{2}=\left(\begin{array}{l} 1 \\ 3 \\ 2 \end{array}\right)

Short answer

A1 is solvable if \( \alpha \neq -1 \); infinitely many solutions when \( \alpha = -1 \). A2 is solvable for \( \beta \neq 1 \); otherwise check solution consistency.

Step-by-step solution

Analyze System A1 for Solvability

The matrix equation can be represented as: \( A_{1} \mathbf{x} = \mathbf{b}_{1} \). We begin by setting up the augmented matrix for \( A_{1} \): \[\left(\begin{array}{ccc|c}1 & 3 & 1 & 5 \2 & 3 & 1 & 5 \3 & 3 & \alpha & 5\end{array}\right)\]The solvability depends on the ability to reduce the matrix to row-echelon form without contradiction. We need to perform row operations to achieve this.

Perform Row Operations on A1

First, operate on \( R_2 \) and \( R_3 \) to eliminate the leading term under the first column.- Subtract \( 2 \times R_1 \) from \( R_2 \): Result is \( (0, -3, -1, -5) \).- Subtract \( 3 \times R_1 \) from \( R_3 \): Result is \( (0, -6, \alpha-3, -10) \).The updated matrix is:\[\left(\begin{array}{ccc|c}1 & 3 & 1 & 5 \0 & -3 & -1 & -5 \0 & -6 & \alpha-3 & -10\end{array}\right)\]Next, continue by making either second row or third row simple to proceed.

Complete Row Reduction for A1

- We perform \( R_3 = R_3 - 2 \times R_2 \), resulting in \( (0, 0, \alpha + 1, 0) \).- The system shows a pivot for every column except potentially the last depending on \( \alpha \).Analyze the condition when \( \alpha + 1 = 0 \). This produces no leading entry in the third column, leading to potential inconsistency.

Deduce Solution Conditions for A1

When \( \alpha = -1 \), the last row becomes \( (0, 0, 0, 0) \), indicating infinitely many solutions with a free variable associated with the third column.When \( \alpha eq -1 \), the system will have a unique solution due to the third pivot.

Analyze System A2 for Solvability

Set up the matrix for \( A_{2} \):\[\left(\begin{array}{ccc|c}1 & 1 & -1 & 1 \2 & 3 & \beta & 3 \1 & \beta & 3 & 2\end{array}\right)\]Start row operations to form a triangular matrix, allowing insights into solvability.

Perform Row Operations on A2

Use \( R_1 \) to eliminate the first column terms:- \( R_2 = R_2 - 2 \times R_1 \): Results in \( (0, 1, \beta + 2, 1) \).- \( R_3 = R_3 - R_1 \): Results in \( (0, \beta - 1, 4, 1) \).Setup as:\[\left(\begin{array}{ccc|c}1 & 1 & -1 & 1 \0 & 1 & \beta + 2 & 1 \0 & \beta - 1 & 4 & 1\end{array}\right)\]Proceed by making additional zeroes below the pivots.

Finalize Row Operations on A2

Use \( R_2 \) to simplify:- \( R_3 = R_3 - (\beta - 1) \times R_2 \).- Results in \( (0, 0, 4 - (\beta - 1)(\beta + 2), 1 - (\beta - 1)) \).Check conditions of the form \( 4x = b \) to determine special parameter cases.

Conclude System A2 Solvability and Solutions

For \( 4 - (\beta - 1)(\beta + 2) = 0 \), solve to check for - if \( \beta = 1 \). The system may be dependent on or contradictory.When \( \beta eq 1 \), consistent system with unique solution.

Key concepts

Linear Equations

Linear equations are mathematical statements expressing equality between two expressions with one or more variables. For instance, suppose we have variables \( x, y, \) and \( z \). An example of a linear equation could be \( 2x + 3y - z = 7 \). In the context of our exercise, the equations were formed into a system and expressed in the form \( A \mathbf{x} = \mathbf{b} \). Linear equations feature variable terms meant to be solved, revealing the values that make each equation true. These equations can manifest as different scenarios: they can have a unique solution, no solution, or infinitely many solutions.

When solving linear equations, we often deal with matrices, especially in systems of equations. This matrix approach helps simplify complex problems into more manageable forms, as demonstrated in the given exercise.

Augmented Matrix

An augmented matrix is a powerful tool used to solve systems of linear equations. It is created by appending the constant matrix, \( \mathbf{b} \), to the coefficient matrix \( A \). In this exercise, we built the augmented matrix from the given system equations. For example, for \( A_1 \), the augmented matrix was \( \begin{pmatrix}1 & 3 & 1 & | & 5 \ 2 & 3 & 1 & | & 5 \ 3 & 3 & \alpha & | & 5 \end{pmatrix} \). This matrix represents a compact form of the equation system without losing any information.

Using an augmented matrix allows for an efficient application of row operations to reveal important properties of the system, like dependency on parameters and solutions. By performing operations such as row swaps, scaling, or replacements, we simplify this matrix into a form that makes it easier to extract solutions or determine if a solution exists.

Row-Echelon Form

The concept of row-echelon form is essential when solving systems of linear equations using matrices. A matrix is said to be in row-echelon form when it meets the following criteria:

• All-zero rows, if any, are at the bottom of the matrix,
• The first non-zero entry in each row, known as a leading entry, appears to the right of the leading entry in the row above it,
• The leading entry in each non-zero row is 1.

To solve the system for \( A_1 \) in our exercise, we performed several row operations to transform the original matrix into this critical form. The row-echelon form facilitates easier backward substitution, allowing us to determine the parameter-dependent conditions like those dictated by \( \alpha \). It serves as a clear visual representation of the linear system's solvability.

Parameter Dependency

In systems of linear equations, parameter dependency refers to how the presence of variables like \( \alpha \) or \( \beta \) affects the existence and types of solutions. In our system, these parameters in the matrices \( A_1 \) and \( A_2 \) provided scenarios where solutions are contingent upon specific conditions.

For instance, the exercise showed that if \( \alpha = -1 \) for \( A_1 \), the system has infinitely many solutions, as the row-echelon form results in a row of all zeros. If \( \alpha eq -1 \), there is a unique solution. Similarly, for \( A_2 \), setting \( \beta = 1 \) can lead to potential conflicts or dependencies.

Understanding parameter dependency not only helps in identifying configurations where multiple solutions exist but also in recognizing when no or a unique solution is achievable.