F?r welchen Wert des Parameters \(\lambda\) sind die Vektoren $$ \mathbf{a}_{1}=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right), \quad \mathbf{a}_{2}=\left(\begin{array}{r} -\lambda \\ 2 \\ 5 \end{array}\right) \text { und } \mathbf{a}_{3}=\left(\begin{array}{r} -2 \\ \lambda \\ 7 \end{array}\right) $$ linear abh?ngig?

The determinant of a matrix helps us determine whether sets of vectors are linearly dependent or independent. It is a special number that can be calculated from the elements of a square matrix. In our specific problem, we check if a 3x3 matrix formed by vectors \(\textbf{a}_1, \textbf{a}_2, \textbf{a}_3\) is linearly dependent by calculating its determinant. The determinant equals zero for the vectors to be linearly dependent, and the formula to calculate it involves expanding across rows or columns with minor determinants and co-factors.
Consider the matrix: \(\begin{pmatrix}1 & -\textbackslash{}lambda & -2 \ 1 & 2 & \textbackslash{}lambda \ 1 & 5 & 7 \textbackslash{}end{pmatrix}\). Expanding across the first row, we get the expression:
\[1 \cdot (2 \cdot 7 - 5 \cdot \textbackslash{}lambda) - (-\textbackslash{}lambda) \cdot (1 \cdot 7 - 1 \cdot \textbackslash{}lambda) + (-2) \cdot (1 \cdot 5 - 1 \cdot 2) \]
Simplifying this expression helps us to identify the specific values of \(\textbackslash{}lambda\) for which the vectors are linearly dependent.

Linear algebra deals with vectors, vector spaces, linear transformations, and systems of linear equations. It's a fundamental branch of mathematics that is vital in various applications in science and engineering. In our problem, we explore the concept of linear dependence through the context of vectors and matrices.
When we say vectors are linearly dependent, we mean that one vector in the set can be written as a linear combination of the others. Mathematically, for vectors \(\textbf{a}_1, \textbf{a}_2, \textbf{a}_3\) to be linearly dependent, there must exist scalars (coefficients) not all zero, such that \[c_1 \textbf{a}_1 + c_2 \textbf{a}_2 + c_3 \textbf{a}_3 = \textbf{0} \]
Setting up this equation in matrix form and finding when the determinant of the matrix is zero helps us determine the linear dependence condition.

Quadratic equations appear in the form \[ax^2 + bx + c = 0 \] and can be solved using the quadratic formula. In our exercise, after calculating the determinant, we ended up with a quadratic equation in \(\textbackslash{}lambda\).
Here’s the equation we need to solve: \(\textbackslash{}lambda^2 - 2\textbackslash{}lambda - 8 = 0\). The quadratic formula is given by: \[\lambda = \textbackslash{}frac{-b \pm \textbackslash{}sqrt{b^2 - 4ac}}{2a} \], where \{a, b, c\} are the coefficients from the equation.
Substituting the values, \{a = 1, b = -2, c = -8\}, we get:
\[\lambda = \textbackslash{}frac{2 \pm \textbackslash{}sqrt{4 + 32}}{2} = \textbackslash{}frac{2 \pm 6}{2} \]
This simplifies to \(\textbackslash{}lambda = 4\) or \(\textbackslash{}lambda = -2\). Therefore, these are the specific values for which our given vectors are linearly dependent.