Finding K For Vector (2, 1, K) In Linear Span Of S
In linear algebra, the concept of a linear span is fundamental. A linear span of a set of vectors is the set of all possible linear combinations of those vectors. In simpler terms, it's the space you can reach by stretching and adding the given vectors together. This article delves into the problem of determining whether a given vector belongs to the linear span of a set of vectors and, if so, finding the scalar values that define this linear combination. We will explore this through a specific example where we are given a set S of two vectors in three-dimensional space, S = {(1, 2, 3), (1, 0, -1)}, and we want to find the value of k for which the vector (2, 1, k) lies within the linear span of S. This involves understanding how to express a vector as a linear combination of other vectors, setting up a system of equations, and solving for the unknowns. The process combines algebraic manipulation with a conceptual grasp of vector spaces, making it a valuable exercise in linear algebra.
Let's consider the set S consisting of two vectors in three-dimensional space: S = {(1, 2, 3), (1, 0, -1)}. Our goal is to find the value of k for which the vector (2, 1, k) belongs to the linear span of S. This means we need to determine if there exist scalars (let's call them a and b) such that the linear combination of the vectors in S equals the vector (2, 1, k). Mathematically, this can be expressed as:
a(1, 2, 3) + b(1, 0, -1) = (2, 1, k)
This equation leads to a system of linear equations that we need to solve for a, b, and ultimately, k. The problem encapsulates key concepts in linear algebra, such as vector spaces, linear combinations, and the span of a set of vectors. It's a practical application of these concepts, requiring us to translate a theoretical definition into a concrete algebraic problem.
Setting up the Equations
To find the value of k, we first need to express the vector (2, 1, k) as a linear combination of the vectors in S. This involves finding scalars a and b such that:
a(1, 2, 3) + b(1, 0, -1) = (2, 1, k)
Expanding this equation, we get:
(a, 2a, 3a) + (b, 0, -b) = (2, 1, k)
Combining the vectors on the left side, we have:
(a + b, 2a, 3a - b) = (2, 1, k)
This vector equation gives us a system of three linear equations:
- a + b = 2
- 2a = 1
- 3a - b = k
This system of equations is the foundation for solving our problem. Each equation represents a component of the vectors, and by solving this system, we can find the values of a, b, and subsequently, k. The process of setting up these equations is crucial in many linear algebra problems, as it allows us to translate a vector problem into an algebraic one.
Solving for a and b
Now that we have the system of equations, our next step is to solve for the scalars a and b. We have the following equations:
- a + b = 2
- 2a = 1
- 3a - b = k
From the second equation, 2a = 1, we can easily solve for a:
a = 1/2
Now that we have the value of a, we can substitute it into the first equation, a + b = 2, to solve for b:
(1/2) + b = 2
b = 2 - (1/2)
b = 3/2
So, we have found the values a = 1/2 and b = 3/2. These values are crucial because they tell us how to combine the vectors in S to get the vector (2, 1, k). The process of solving for these scalars demonstrates the power of linear equations in solving vector problems.
Determining the Value of k
With the values of a and b determined, we can now find the value of k. We have the third equation from our system:
3a - b = k
Substituting the values we found for a and b (a = 1/2, b = 3/2) into this equation, we get:
3(1/2) - (3/2) = k
(3/2) - (3/2) = k
0 = k
Therefore, the value of k for which the vector (2, 1, k) belongs to the linear span of S is 0. This result means that the vector (2, 1, 0) can be expressed as a linear combination of the vectors (1, 2, 3) and (1, 0, -1). This completes the solution to our problem, showcasing how solving a system of linear equations can lead to the determination of a vector's components within a given span.
In summary, we have successfully determined the value of k for which the vector (2, 1, k) belongs to the linear span of the set S = {(1, 2, 3), (1, 0, -1)}. By setting up a system of linear equations based on the definition of a linear combination and solving for the scalars a and b, we found that k = 0. This exercise highlights the importance of understanding linear combinations and the span of vectors in linear algebra. It also demonstrates a practical application of solving systems of equations to address problems in vector spaces. The process involved translating a vector problem into an algebraic one, solving the algebraic problem, and then interpreting the results in the context of the original vector problem. This approach is fundamental in many areas of mathematics, physics, and engineering, making the understanding of these concepts crucial for anyone working in these fields. The problem we solved is a microcosm of the broader applications of linear algebra, showing how abstract concepts can be used to solve concrete problems.
