Begründen Sie, warum die folgenden Vektoren linear abh?ngig sind: a) \(\quad \mathbf{a}=\left(\begin{array}{l}1 \\ 2 \\ 5\end{array}\right), \quad \mathbf{b}=\left(\begin{array}{l}2 \\ 0 \\ 2\end{array}\right) . \quad \mathbf{c}=\left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right)\) b) \(\quad \mathbf{a}=\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right), \quad \mathbf{b}=\left(\begin{array}{r}3 \\ -1 \\ 4\end{array}\right), \quad c=\left(\begin{array}{r}-3 \\ 0 \\ -3\end{array}\right)\) c) \(\quad \mathbf{a}_{1}=\left(\begin{array}{l}1 \\ 0 \\\ 1\end{array}\right), \quad \mathbf{a}_{2}=\left(\begin{array}{l}1 \\ 1 \\\ 0\end{array}\right), \quad \mathbf{a}_{3}=\left(\begin{array}{r}0 \\ -2 \\\ 2\end{array}\right)\) d) \(\quad a=\left(\begin{array}{l}1 \\ 2\end{array}\right), \quad b=\left(\begin{array}{l}3 \\ 1\end{array}\right), \quad c=\left(\begin{array}{l}2 \\ 0\end{array}\right), \quad d=\left(\begin{array}{l}-1 \\ -1\end{array}\right)\)

Step by step solution

01

Understanding Linear Dependence

Vectors are linearly dependent if one can be written as a linear combination of the others. This means that we can find scalars (not all zero) such that a combination of the vectors equals the zero vector.

02

Solve Part a

Check if vectors \(\mathbf{a} = \begin{pmatrix} 1 \ 2 \ 5 \end{pmatrix}, \mathbf{b} = \begin{pmatrix} 2 \ 0 \ 2 \end{pmatrix}, \mathbf{c} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\) are linearly dependent.Notice that \(\mathbf{c}\) is the zero vector. Therefore, it can always be written as 0 times any vector, making \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) linearly dependent by default.

03

Solve Part b

Check if vectors \(\mathbf{a} = \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}, \mathbf{b} = \begin{pmatrix} 3 \ -1 \ 4 \end{pmatrix}, \mathbf{c} = \begin{pmatrix} -3 \ 0 \ -3 \end{pmatrix} \) are linearly dependent.Notice that \(\mathbf{c}\) can be written as \(-3 \cdot \mathbf{a}\), hence making the vectors linearly dependent.

04

Solve Part c

Check if vectors \(\mathbf{a}_1 = \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}, \mathbf{a}_2 = \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix}, \mathbf{a}_3 = \begin{pmatrix} 0 \ -2 \ 2 \end{pmatrix} \) are linearly dependent.Set up the equation \(k_1 \mathbf{a}_1 + k_2 \mathbf{a}_2 + k_3 \mathbf{a}_3 = 0\): \(k_1 \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} + k_2 \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix} + k_3 \begin{pmatrix} 0 \ -2 \ 2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \).From the equations, solve for \(k_3\):\[k_1 + k_2 = 0, \ k_2 - 2k_3 = 0, \ k_1 + 2k_3 = 0.\]Find the solution \(k_1 = -2k_3, k_2 = 2k_3, k_3 = k_3\). Thus, since \(k_3 eq 0\), the vectors are linearly dependent.

05

Solve Part d

Check if vectors \(\mathbf{a} = \begin{pmatrix} 1 \ 2 \end{pmatrix}, \mathbf{b} = \begin{pmatrix} 3 \ 1 \end{pmatrix}, \mathbf{c} = \begin{pmatrix} 2 \ 0 \end{pmatrix}, \mathbf{d} = \begin{pmatrix} -1 \ -1 \end{pmatrix} \) are linearly dependent.Set up the equation \(k_1 \mathbf{a} + k_2 \mathbf{b} + k_3 \mathbf{c} + k_4 \mathbf{d} = 0\): \(k_1 \begin{pmatrix} 1 \ 2 \end{pmatrix} + k_2 \begin{pmatrix} 3 \ 1 \end{pmatrix} + k_3 \begin{pmatrix} 2 \ 0 \end{pmatrix} + k_4 \begin{pmatrix} -1 \ -1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \).From the equations, solve for \(k_4\):\[k_1 + 3k_2 + 2k_3 - k_4 = 0, \ 2k_1 + k_2 - k_4 = 0.\]Express \(k_1, k_2, k_3, k_4\) in terms of \(k_4\):\[k_1 = k_3 - k_4, \ k_2 = -2k_3 + 2k_4, \ k_4 eq 0.\]Hence, the vectors are linearly dependent.

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Key Concepts

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

linear algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear transformations. It is a foundational area that has applications in various fields including computer science, engineering, and physics. A key concept in linear algebra is understanding how vectors relate to one another and how they can be combined or manipulated. Another fundamental idea is the study of systems of linear equations, which leads to the concept of matrix algebra and determinants.

In our given exercise, we deal with concepts like vectors and their linear dependence, which is a cornerstone in linear algebra. Knowing how vectors can be combined and the conditions under which they can be set to be dependent or independent is crucial for deeper understanding and applications in various disciplines.

vector spaces

A vector space is a collection of vectors that can be added together and multiplied by scalars while still remaining within the collection. Vector spaces are fundamental structures in linear algebra.
They are defined by two main operations: vector addition and scalar multiplication. For instance, in our exercise, the given vectors are elements of a three-dimensional vector space.

Each vector in this space can be manipulated through these operations to understand their relationships more deeply. The zero vector, which is a vector where all components are zero, is always part of a vector space and serves as an identity element for vector addition. Recognizing the properties and structure of vector spaces helps in solving problems related to linear dependence and independence.

linear combinations

A linear combination of vectors involves adding together the vectors after multiplying them by scalars. This is a fundamental process in the study of vector spaces. A vector \(\textbf{d}\) can be represented as a linear combination of vectors \(\textbf{a}, \textbf{b}, \textbf{c}\), mathematically expressed as:
\[ k_1 \textbf{a} + k_2 \textbf{b} + k_3 \textbf{c} = \textbf{d} \]
where \(\textbf{a}, \textbf{b}, \textbf{c}\) are known vectors and \( k_1, k_2, k_3 \) are scalars.

If the vector \(\textbf{d}\) is zero and not all of the scalars are zero, the vectors are said to be linearly dependent. This is precisely what we examined in the exercise. By setting up equations based on these linear combinations, we verified how one vector can be expressed in terms of others, confirming their linear dependence.

dependence relations

Dependence relations among vectors are about determining when a set of vectors is linearly dependent. Vectors are considered linearly dependent if there exist scalars, not all zero, such that a linear combination of these vectors equals the zero vector.

For instance, in the exercise, we checked several sets of vectors for linear dependence by setting up equations like: \[ k_1 \textbf{a} + k_2 \textbf{b} + k_3 \textbf{c} = \textbf{0} \]
Here, finding non-zero scalars fulfilled the condition for linear dependence. Understanding dependence relations helps identify how different vectors in a space can be interrelated and if any vector can be constructed from others. This also simplifies problems like solving systems of equations and understanding the subspace relations within vector spaces.