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Chapter 1: Problem 6

F?r welche reellen Werte des Parameters \(\lambda\) besitzt das homogene lineare Gleichungssystem nicht-triviale L?sungen? a) \(\left(\begin{array}{rr}-\lambda & 1 \\ 1 & -\lambda\end{array}\right)\left(\begin{array}{l}x \\\ y\end{array}\right)=\left(\begin{array}{l}0 \\ 0\end{array}\right)\) b) \(\left(\begin{array}{ccrr}(2+\lambda) & 0 & 0 & 0 \\ 0 & (2-\lambda) & 0 & 0 \\ 1 & -2 & -\lambda & -1 \\ 2 & -4 & 1 & -\lambda\end{array}\right)\left(\begin{array}{l}x_{1} \\ x_{2} \\ x_{3} \\\ x_{4}\end{array}\right)=\left(\begin{array}{l}0 \\ 0 \\ 0 \\\ 0\end{array}\right)\)

Short Answer

Expert verified
Part a: \(\textbackslash lambda = \textbackslash pm 1\). Part b: \(\textbackslash lambda = \textbackslash 2, -2\).

Step by step solution

01

- Understand the problem

Determine for which real values of parameter \(\textbackslash lambda\) the homogeneous linear systems have non-trivial solutions.
02

- Non-trivial solution criteria

Non-trivial solutions exist if the determinant of the coefficient matrix is zero.
03

Part a) Set up the determinant

Determine the determinant of the matrix \(\textbackslash begin\textbraceleft array\textbracright\textbraceleft rr\textbracright -\textbackslash lambda & 1 \textbackslash\textbackslash 1 & -\textbackslash lambda \textbackslash end\textbraceleft array\textbracright\)
04

Step 3a - Calculate the determinant for part a)

Calculate the determinant: \[ \textbackslash begin\textbraceleft vmatrix\textbracright -\textbackslash lambda & 1 \textbackslash\textbackslash 1 & -\textbackslash lambda \textbackslash end\textbraceleft vmatrix\textbracright = (\textbackslash -\textbackslash lambda) \textbackslash cdot (\textbackslash -\textbackslash lambda) - 1 \textbackslash cdot 1 = \textbackslash lambda^2 - 1 \ \textbackslash lambda^2 - 1 = 0 \ \textbackslash lambda^2 = 1 \ \textbackslash lambda = \textbackslash \textbackslash pm 1 \]
05

- Part b) Set up the determinant

Determine the determinant of the matrix: \(\textbackslash begin\textbraceleft array\textbracright \textbackslash ( 2 + \textbackslash lambda & 0 & 0 & 0 \textbackslash \textbackslash 0 & 2 - \textbackslash lambda & 0 & 0 \textbackslash \textbackslash 1 & -2 & \textbackslash -\textbackslash lambda & -1 \textbackslash \textbackslash 2 & -4 & 1 & -\textbackslash lambda \textbackslash end\textbraceleft array\textbracright \)
06

- Calculate the determinant for part b)

Use cofactor expansion along the first row: \[ \textbackslash begin\textbraceleft vmatrix\textbracright 2 + \textbackslash lambda & 0 & 0 & 0 \textbackslash\textbackslash 0 & 2 - \textbackslash lambda & 0 & 0 \textbackslash\textbackslash 1 & -2 & -\textbackslash lambda & -1 \textbackslash\textbackslash 2 & -4 & 1 & -\textbackslash lambda \textbackslash end\textbraceleft vmatrix\textbracright = (2 + \textbackslash lambda) \textbackslash cdot \textbackslash begin\textbraceleft vmatrix\textbracright 2 - \textbackslash lambda & 0 & 0 \textbackslash \textbackslash -2 & -\textbackslash lambda & -1 \textbackslash \textbackslash -4 & 1 & -\textbackslash lambda \textbackslash end\textbraceleft vmatrix\textbracright \] Continue expanding the 3x3 determinant: \[ = (2 + \textbackslash lambda) \textbackslash cdot \textbackslash left\textbraceleft (2 - \textbackslash lambda) \textbackslash cdot \textbackslash left\textbraceleft - \textbackslash lambda \textbackslash cdot (-\textbackslash lambda) - (-1) \textbackslash cdot (-4) \textbackslash right\textbraceleft \textbackslash right\textbraceleft = (2 + \textbackslash lambda) \textbackslash cdot \textbackslash left\textbraceleft (2 - \textbackslash lambda) (\textbackslash lambda^2 - 4) \textbackslash right\textbraceleft = 0 \] Solve the resulting polynomial equation.
07

- Polynomial solutions for part b)

Solve the polynomial \[ (2+\textbackslash lambda)(2-\textbackslash lambda)(\textbackslash lambda^2-4)=0 \] The solutions are \[ \textbackslash lambda=2, \textbackslash lambda=-2 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinant of a Matrix
In linear algebra, the determinant of a matrix is a special number that can be computed from its elements. The determinant provides important properties of the matrix, such as whether it is invertible or not. For a 2x2 matrix, the determinant is calculated as follows: \[ \text{det} \begin{pmatrix} a & b \ c & d \end{pmatrix} = ad - bc \]
For a matrix to have a non-trivial solution (any solution other than all variables being zero), its determinant must be zero. This is because only when the determinant is zero, the system of equations has infinitely many solutions or no unique solutions.
In our first example: \[ \begin{pmatrix} -\text{λ} & 1 \ 1 & -\text{λ} \end{pmatrix} \] The determinant of this matrix is calculated by: \[ (-\text{λ})(-\text{λ}) - (1)(1) = \text{λ}^2 - 1 = 0\] Therefore, \[ \text{λ} = 1 \text{ or } \text{λ} = -1 \]
Det sets \text{λ} to these values so that the determinant is zero, indicating that the system has non-trivial solutions.
Non-Trivial Solutions
Non-trivial solutions are solutions to a system of equations where not all the variables are zero. In the context of homogeneous linear systems, this means finding values of the variables that satisfy the equations without all being zero simultaneously. To find these solutions, we look for when the determinant of the system's coefficient matrix is zero.
This is because a zero determinant implies that the matrix is singular, meaning its rows or columns are linearly dependent. Consequently, the system of equations doesn't have a unique solution but instead has either infinitely many solutions or no non-zero solutions. Let's consider the second matrix from the exercise: \[ \begin{pmatrix} 2 + \text{λ} & 0 & 0 & 0 \ 0 & 2 - \text{λ} & 0 & 0 \ 1 & -2 & -\text{λ} & -1 \ 2 & -4 & 1 & -\text{λ} \end{pmatrix} \] After calculating and setting the determinant of this 4x4 matrix to zero, we solve for \text{λ} which gives us \[ \text{λ} = 2 \text{ and} \text{λ} = -2 \] Hence, these values make the system have non-trivial solutions.
Linear Algebra
Linear algebra is a branch of mathematics that deals with vectors, vector spaces (or linear spaces), linear transformations, and systems of linear equations. It is foundational for fields like computer science, physics, engineering, and more.
In this particular example, we deal with homogeneous linear systems. Such systems can be expressed in matrix form as: \[ A\textbf{x} = \textbf{0} \] where \textbf{x} is a vector of variables, and \textbf{0} is the zero vector. The goal is to find solutions to this equation.
When dealing with homogeneous systems, the key points to remember are:
  • If the determinant of the coefficient matrix A is not zero, the system has only the trivial solution (\textbf{x} = \textbf{0}).
  • If the determinant of A is zero, the system may have infinitely many solutions, which are the non-trivial solutions.
This exercise showcases the application of finding determinants and solving for parameters that provide non-trivial solutions, a key aspect of linear algebra that has widespread significance in various scientific and engineering disciplines.

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Most popular questions from this chapter

Zeigen Sie, daB die folgenden Matrizen unitär sind: $$ \begin{aligned} &\mathbf{A}=\frac{1}{\sqrt{12}}\left(\begin{array}{ccc} 2 & 1-\sqrt{3} j & -1-\sqrt{3} j \\ -2 j & \sqrt{3}-j & \sqrt{3}+j \\ -2 & 2 & -2 \end{array}\right) \\ &B=\frac{1}{\sqrt{3}}\left(\begin{array}{cc} 1+j & 1 \\ j & -1-j \end{array}\right), \quad C=\left(\begin{array}{ll} j & 0 \\ 0 & j \end{array}\right) \end{aligned} $$

Berechnen Sie die Determinanten der folgenden 2-reihigen Matrizen: a) \(A=\left(\begin{array}{rr}2 & 3 \\ 4 & -5\end{array}\right)\) b) \(\quad B=\left(\begin{array}{ll}a & a \\ b & b\end{array}\right)\) c) \(C=\left(\begin{array}{ll}3 & 11 \\ x & 2 x\end{array}\right)\)

Berechnen Sie mit den \((2,3)\)-Matrizen $$ \begin{aligned} &\mathbf{A}=\left(\begin{array}{ccc} 3+j & 2-j & 2 \\ 2 & 1+j & 1 \end{array}\right), \quad B=\left(\begin{array}{lll} 1 & 2+2 j & 1 \\ 0 & 4-3 j & 1 \end{array}\right) \\ &\text { und } \quad C=\left(\begin{array}{ccc} 1-j & 2 & 3-j \\ 1+j & 5 & 1+j \end{array}\right) \end{aligned} $$ die folgenden Ausdr?cke: a) \(\mathbf{A}+\mathbf{B}+\mathbf{C}\) b) \(3 \mathrm{~A}-4(\mathrm{~B}-2 \mathbf{C})\) c) \(\quad 2 \mathrm{~A}^{\mathbf{T}}-3(\mathbf{B}-\mathbf{C})^{\mathbf{T}}\)

Lösen Sie die folgenden nicht-quadratischen linearen Gleichungssysteme mit Hilfe elementarer Umformungen in den Zeilen der erweiterten Koeffizientenmatrix (Gau\betascher Algorithmus): \(2 x_{1}-3 x_{2}=11\) a) \(-5 x_{1}+x_{2}=-8\) b) \(\left(\begin{array}{rrrr}1 & 1 & 2 & 0 \\ -3 & 2 & 0 & 1 \\ 8 & -2 & -2 & 2\end{array}\right)\left(\begin{array}{c}x_{1} \\ x_{2} \\ x_{3} \\\ x_{4}\end{array}\right)=\left(\begin{array}{l}1 \\ 5 \\ 0\end{array}\right)\) \(x_{1}-5 x_{2}=16\) \(3 x_{2}-5 x_{3}+x_{4}=0\) c) $$ \begin{aligned} &-x_{1}-3 x_{2}-x_{4}=-5 \\ &-2 x_{1}+x_{2}+2 x_{3}+2 x_{4}=2 \\ &-3 x_{1}+4 x_{2}+2 x_{3}+2 x_{4}=8 \end{aligned} $$

Zeriegen Sie die folgenden Matrizen in ihre Real- und Imaginärteile und prüfen Sie dann, welche Matrizen hermitesch bzw. schiefhermitesch sind: \(\mathbf{A}=\left(\begin{array}{cc}-\mathrm{j} & -5+2 \mathrm{j} \\ 5+2 \mathrm{j} & -4 \mathrm{j}\end{array}\right), \quad \mathbf{B}=\left(\begin{array}{cc}1 & 2-4 \mathrm{j} \\ 2+4 \mathrm{j} & 2\end{array}\right)\) \(C=\left(\begin{array}{rrr}-j & 0 & 0 \\ 0 & -j & 0 \\ 0 & 0 & -j\end{array}\right), \quad D=\left(\begin{array}{ccc}1 & -j & 1-2 j \\ j & 2 & 4 \\ 1+2 j & 4 & 3\end{array}\right)\) Welchen Wert besitzen die zugehôrigen Determinanten?
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