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In the world of vectors, a linear combination involves creating a new vector by adding scaled versions of other vectors. Imagine you have a collection of vectors, and you wish to see if you can express another vector as a combination of these. By using linear combinations, you determine the coefficients (or scalars) needed with these vectors to reconstruct any vector in that set. For the exercise given, we need to express the vector \( \mathbf{w} = \begin{bmatrix} 1 \ 3 \ 4 \end{bmatrix} \) as a linear combination of the vectors in \( \mathcal{B} \).

• The scalars \( a \) and \( b \) are used to show that \( a\begin{bmatrix} 3 \ 1 \ 4 \end{bmatrix} + b\begin{bmatrix} 5 \ 1 \ 6 \end{bmatrix} = \begin{bmatrix} 1 \ 3 \ 4 \end{bmatrix} \).
• Finding appropriate values of \( a \) and \( b \) confirms whether \( \mathbf{w} \) can be obtained from the given set of vectors.

Understanding this concept is crucial as it helps us understand how vectors relate in multi-dimensional spaces, ensuring that the vector \( \mathbf{w} \) is indeed expressible using vectors from \( \mathcal{B} \).
Linear combinations are at the heart of vector spaces and are a foundational concept in linear algebra.

A spanning set is essentially a collection of vectors that can be used to generate a whole vector space through linear combinations. When a vector \( \mathbf{w} \) is said to be within the 'span' of set \( \mathcal{B} \), it means that \( \mathbf{w} \) can be expressed as a linear combination of vectors in \( \mathcal{B} \). In simpler terms, you can create \( \mathbf{w} \) using these given vectors.

• The spanning set or span of vectors \( \mathcal{B} \) includes all possible linear combinations of the vectors it contains.
• For this problem, showing that \( \mathbf{w} \) is in the span of \( \mathcal{B} \) means finding the scalars \( a \) and \( b \) such that \( a\begin{bmatrix} 3 \ 1 \ 4 \end{bmatrix} + b\begin{bmatrix} 5 \ 1 \ 6 \end{bmatrix} = \begin{bmatrix} 1 \ 3 \ 4 \end{bmatrix} \).

Spanning sets help us determine the reach or extent of vector combinations, providing a clear picture of what vectors can be derived from others in a vector space. Understanding this explains the significance of each vector's potential within the set.

A system of equations is a collection of equations that are solved together to find common variable values. In the context of vectors, determining a linear combination requires solving a system of equations. The step-by-step solution presents these equations derived from expressing \( \mathbf{w} \) as a linear combination of vectors in \( \mathcal{B} \).

• Each derived equation corresponds to the individual components of the vectors being combined to form \( \mathbf{w} \).
• The system involves:
  • \( 3a + 5b = 1 \)
  • \( a + b = 3 \)
  • \( 4a + 6b = 4 \)

Solving such a system applies algebraic techniques to find values of \( a \) and \( b \). These values confirm the vector \( \mathbf{w} \)'s expression in terms of the basis vectors. Solving these precisely gives insight into how vectors interact and can be reconstructed within a span. Clear knowledge of solving systems of equations is vital here, reinforcing the connectivity and eventual simplification leading to the coordinate vector.

A basis is essentially the building blocks of a vector space. It’s a subset of vectors that are linearly independent and span the vector space. For the exercise given, \( \mathcal{B} \) is considered as a basis for a subspace including \( \mathbf{w} \), provided \( \mathbf{w} \) can be expressed using vectors in \( \mathcal{B} \).

• A basis must meet two key conditions:
  • The vectors must span the vector space or subspace.
  • The vectors in it should be linearly independent (they can't be expressed as linear combinations of each other).
• Once confirmed as a basis, the coordinate vector relative to \( \mathcal{B} \) is the set of coefficients used in the linear combination of \( \mathbf{w} \).

In this problem, the computed coordinate vector \( \begin{bmatrix} 7 \ -4 \end{bmatrix} \) answers the core question, evidencing the positioning and weight of \( \mathbf{w} \) in relation to \( \mathcal{B} \). Understanding the notion of basis ensures a solid grasp of vector reconstruction and space comprehension, vital for adept linear algebra understanding.