91Ó°ÊÓ

A system of linear equations is a collection of one or more linear equations involving the same set of variables. In our exercise, the problem is to check if the vector \( \mathbf{v} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} \) can be a linear combination of two other vectors \( \mathbf{u}_1 \text{ and } \mathbf{u}_2 \). To find this out, we set up a system of equations:- \( a \cdot 1 + b \cdot 0 = 1 \)- \( a \cdot 1 + b \cdot 1 = 2 \)- \( a \cdot 0 + b \cdot 1 = 3 \)These equations relate the scalars \( a \) and \( b \) to the original vectors, ensuring the linear combination equals the target vector \( \mathbf{v} \). Solving this system will determine if such a combination exists. However, as seen, solving these equations gives conflicting results, hence no consistent solution exists defined by this set of equations. A consistent solution would mean a single set of \( a \) and \( b \) that satisfies all equations.

Vectors are entities described by both a magnitude and a direction. In mathematics, they are typically represented as an array of numbers that signify their components in space. - The vector \( \mathbf{v} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} \) indicates a point in three-dimensional space: - Along the x-axis: 1 - Along the y-axis: 2 - Along the z-axis: 3 This representation simplifies complex concepts into manageable components that can be analyzed using operations such as scalar multiplication and addition.In our exercise, the task is to check if \( \mathbf{v} \) is a result of the combination of vectors \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \). These vectors form a basis that the target vector \( \mathbf{v} \) is supposed to be expressed in, using scalars \( a \text{ and } b \). Understanding this representation is crucial for maneuvering through linear algebra problems.

Scalar multiplication involves multiplying a vector by a scalar (a real number), which scales the vector's magnitude. The direction remains the same unless the scalar is negative, in which case the vector reverses direction.In terms of our exercise:- When you multiply vector \( \mathbf{u}_1 = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} \) by \( a \), each component of the vector is scaled by this number.- Similarly, multiplying \( \mathbf{u}_2 = \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} \) by \( b \) scales this vector.This operation is fundamental in forming a linear combination of vectors, which is precisely what we aim to achieve when expressing a vector like \( \mathbf{v} \) in terms of others, as indicated in our solution attempt. The right combination of scalar multiplications is key to solving the exercise, yet as demonstrated, the particular scalars derived led to an inconsistency, showing that \( \mathbf{v} \) is not a linear combination of \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \).