Question

Determine whether the given set of vectors is linearly independent. If linearly dependent, find a linear relation among them. The vectors are written as row vectors to save space, but may be considered as column vectors; that is, the transposes of the given vectors may be used instead of the vectors themselves. $$ \mathbf{x}^{(1)}=(2,1,0), \quad \mathbf{x}^{(2)}=(0,1,0), \quad \mathbf{x}^{(3)}=(-1,2,0) $$

Short answer

Short Answer: The given set of vectors is linearly dependent as there's a non-trivial solution for their linear combination equal to the zero vector: $$ -\frac{1}{2}\mathbf{x}^{(1)}+\frac{5}{2}\mathbf{x}^{(2)}+\mathbf{x}^{(3)}=\mathbf{0} $$

Step-by-step solution

Set up the linear equation

Write the linear combination of the given vectors equal to the zero vector: $$ a\mathbf{x}^{(1)} + b\mathbf{x}^{(2)} + c\mathbf{x}^{(3)} = \mathbf{0} $$ Where a, b, and c are the scalar coefficients and \(\mathbf{0}\) is the zero vector.

Write the linear system

Substitute the given vectors into the linear equation from step 1: $$ a(2,1,0) + b(0,1,0) + c(-1,2,0) = (0,0,0) $$

Convert to matrix form

Convert the linear system to an augmented matrix form: $$ \begin{bmatrix} 2 & 0 & -1 & 0 \\ 1 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $$

Perform Gaussian elimination

Now, we just need to get this matrix to its row-echelon form using Gaussian elimination: $$ \text{Swap Row 1 and Row 2}: \begin{bmatrix} 1 & 1 & 2 & 0 \\ 2 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $$ $$ \text{Replace Row 2 with Row 2 - 2*Row 1}: \begin{bmatrix} 1 & 1 & 2 & 0 \\ 0 & -2 & -5 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $$ $$ \text{Multiply Row 2 by -\frac{1}{2}}: \begin{bmatrix} 1 & 1 & 2 & 0 \\ 0 & 1 & \frac{5}{2} & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $$ $$ \text{Replace Row 1 with Row 1 - Row 2}: \begin{bmatrix} 1 & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & \frac{5}{2} & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $$ The matrix is now in row-echelon form.

Determine linear dependence and find the relation (if dependent)

Since the rows of the resulting matrix are non-zero, we have a unique solution for the system of equations. This solution is: $$ a=-\frac{1}{2}, \quad b=\frac{5}{2}, \quad c=1 $$ This non-trivial solution indicates that the vectors are linearly dependent. The linear relation among these vectors is given by their linear combination with these scalar coefficients: $$ -\frac{1}{2}\mathbf{x}^{(1)}+\frac{5}{2}\mathbf{x}^{(2)}+\mathbf{x}^{(3)}=\mathbf{0} $$

Key concepts

Vectors

Vectors are mathematical entities with both magnitude and direction, represented by coordinates in a space. For example, a vector in 3-dimensional space, like \( \mathbf{x}^{(1)}=(2,1,0) \), has three components. The position of each component corresponds to dimensions of the space.
Vectors are foundational in physics and engineering to represent quantities like force, velocity, or displacement. Beyond physical applications, vectors are crucial in linear algebra for solving systems of linear equations, investigating vector spaces, and more.
When multiple vectors are examined together as a set, we often look at their linear dependence or independence to understand their relationship. Linear dependence means that at least one vector is a linear combination of others in the set. Conversely, linear independence indicates no such linear combination exists. This property is key in determining the vectors' span, or the dimensions of the space they occupy.

Gaussian Elimination

Gaussian elimination is a method used in mathematics to solve systems of linear equations. It simplifies these systems to make it easier to find solutions. This involves three key steps:

• Swap rows to position a leading coefficient.
• Multiply rows to create leading ones.
• Add or subtract rows to form zeros below leading coefficients.

The process transforms matrices into a format that can be solved by back substitution, making complex systems manageable. This technique is critical for matrix manipulation and forms the basis for many algorithms used in computational applications.
In the context of linear dependence, Gaussian elimination helps find if a system of equations that represents the vector set has non-trivial solutions (solutions not all zero), indicating dependence.

Row-Echelon Form

Row-echelon form is a matrix format where each leading entry of a row is to the right of the leading entry of the previous row. Also, rows with all zero entries are placed at the bottom. This simplified structure makes it straightforward to solve linear systems.
The significance of row-echelon form lies in its capacity to reveal the linear relationships between vectors in a set. Through row operations, we transform a matrix which represents vector equations to this form. Once in row-echelon, it becomes easier to see if the matrix solutions are trivial (all zero) or non-trivial.
If the system reaches row-echelon form without free variables, this implies that there are no special solutions, indicating linear independence. If free variables are present or a non-trivial solution is found, the vectors are dependent. This test is core to understanding the structural composition of a vector space.