Question

In each case, show that the given set of constant vectors is linearly dependent. (a) \(\mathbf{v}_{1}=\left(\begin{array}{r}3 \\ -1 \\ 2\end{array}\right), \quad \mathbf{v}_{2}=\left(\begin{array}{r}13 \\ 5 \\ -4\end{array}\right), \quad \mathbf{v}_{3}=\left(\begin{array}{r}2 \\ 4 \\ -5\end{array}\right)\). (b) \(\mathbf{v}_{1}=\left(\begin{array}{l}2 \\ 4 \\ 5\end{array}\right), \quad \mathbf{v}_{2}=\left(\begin{array}{l}2 \\ 6 \\ 8\end{array}\right), \quad \mathbf{v}_{3}=\left(\begin{array}{r}1 \\ -1 \\ -2\end{array}\right)\). (c) \(\quad \mathbf{v}_{1}=\left(\begin{array}{r}1 \\ 2 \\\ -3\end{array}\right), \quad \mathbf{v}_{2}=\left(\begin{array}{r}2 \\ 1 \\\ -3\end{array}\right), \quad \mathbf{v}_{3}=\left(\begin{array}{r}7 \\ -1 \\\ -6\end{array}\right)\).

Short answer

For each given set of constant vectors, we have found non-zero scalars \(c_1\), \(c_2\), and \(c_3\), such that the linear combination of the vectors is equal to the zero vector: (a) \(c_1 = 1\), \(c_2 = 0\), and \(c_3 = -\frac{3}{2}\) (b) \(c_1 = -\frac{3}{2}c_3\), \(c_2 = \frac{1}{2}c_3\), and any non-zero \(c_3\) (c) \(c_1 = 1\), \(c_2 = -2\), and \(c_3 = 0\) Thus, in all cases, the given sets of constant vectors are linearly dependent.

Step-by-step solution

Write the linear combination of vectors

Rewrite the given vectors as a linear combination: \[c_1\begin{pmatrix}3\\-1\\2\end{pmatrix} + c_2\begin{pmatrix}13\\5\\-4\end{pmatrix} + c_3\begin{pmatrix}2\\4\\-5\end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}\]

Solve the system of linear equations

Express it as a system of linear equations: \[ 3c_1 + 13c_2 + 2c_3 = 0 \] \[ -1c_1 + 5c_2 + 4c_3 = 0 \] \[ 2c_1 - 4c_2 - 5c_3 = 0 \]

Find the non-zero constants

We can find a solution to the system by multiplying the first equation by 2 and adding it to the third equation: \[\begin{cases} 6c_1 + 26c_2 + 4c_3 = 0\\ 2c_1 - 4c_2 - 5c_3 = 0 \end{cases}\] This gives us: \[ -c_2 = 0 \] Which means, \(c_2 = 0\). Now, substitute \(c_2 = 0\) back into the first equation from our system of linear equations: \[ 3c_1 + 2c_3 = 0 \] We can pick any value for \(c_1\) and find a corresponding value for \(c_3\). For example, let \(c_1 = 1\), then \(c_3 = -\frac{3}{2}\). Thus, we have found non-zero scalars: \(c_1 = 1\), \(c_2 = 0\), and \(c_3 = -\frac{3}{2}\), such that the linear combination of the vectors is equal to the zero vector. Therefore, the given set of constant vectors in (a) is linearly dependent. (b)

Write the linear combination of vectors

Rewrite the given vectors as a linear combination: \[c_1\begin{pmatrix}2\\4\\5\end{pmatrix} + c_2\begin{pmatrix}2\\6\\8\end{pmatrix} + c_3\begin{pmatrix}1\\-1\\-2\end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}\]

Solve the system of linear equations

Express it as a system of linear equations: \[ 2c_1 + 2c_2 + 1c_3 = 0 \] \[ 4c_1 + 6c_2 - 1c_3 = 0 \] \[ 5c_1 + 8c_2 - 2c_3 = 0 \]

Find the non-zero constants

Notice, that the third equation is a linear combination of the first and second equations: \(5 * \text{(first equation)} - \text{(second equation)} = \text{(third equation)}\) The third equation doesn't give any new information, meaning that we will have a free variable (let's say \(c_3\)) and we should be able to find non-zero constants for \(c_1\) and \(c_2\). By solving the first and second linear equations, we can find: \(c_1 = -\frac{3}{2}c_3\) and \(c_2 = \frac{1}{2}c_3\) As long as \(c_3\) is non-zero, the system has a non-trivial solution, implying that the set of constant vectors in (b) is linearly dependent. (c)

Write the linear combination of vectors

Rewrite the given vectors as a linear combination: \[c_1\begin{pmatrix}1\\2\\-3\end{pmatrix} + c_2\begin{pmatrix}2\\1\\-3\end{pmatrix} + c_3\begin{pmatrix}7\\-1\\-6\end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}\]

Solve the system of linear equations

Express it as a system of linear equations: \[ 1c_1 + 2c_2 + 7c_3 = 0 \] \[ 2c_1 + 1c_2 - 1c_3 = 0 \] \[ -3c_1 - 3c_2 - 6c_3 = 0 \]

Find the non-zero constants

Notice that we can eliminate \(c_1\) by subtracting two times the second equation from the first equation: \(\text{(first equation)} - 2 * \text{(second equation)} = 5c_3 = 0\) We have found that \(c_3 = 0\). By plugging this value back into the second equation, we get: \[2c_1 + c_2 = 0\] At this point, we can pick any value for \(c_1\) or \(c_2\) and find a corresponding non-zero value for the other constant (e.g., \(c_1 = 1\) and \(c_2 = -2\)). Thus, we have found non-zero scalars \(c_1 = 1\), \(c_2 = -2\), and \(c_3 = 0\), such that the linear combination of the vectors is equal to the zero vector. Therefore, the given set of constant vectors in (c) is linearly dependent.

Key concepts

Linear Combination

When we talk about a linear combination of vectors, we refer to an expression made up of several vectors multiplied by scalars and added together. For instance, if we have two vectors, \textbf{A} and \textbf{B}, a linear combination would look like: \( c_1\textbf{A} + c_2\textbf{B} \), where \( c_1 \) and \( c_2 \) are scalars.

Now, these vectors can represent various concepts in mathematics and physics, such as forces, velocities, or simply points in space. A very important application of linear combinations is in solving systems of linear equations, which we'll look at closely.

Linear combinations become especially interesting when we discuss sets of vectors in terms of their linear independence or dependence. If a non-trivial linear combination (where at least one scalar is non-zero) of vectors equals the zero vector, the set is said to be linearly dependent. On the other hand, if only the trivial combination (all scalars are zero) leads to the zero vector, the vectors are linearly independent.

This is crucial because it tells us about the 'span' of a set of vectors, or essentially, the dimensions or subspace they can cover in a vector space. If a set of vectors is linearly dependent, as in the textbook exercise, it implies that at least one of the vectors in the set can be expressed as a linear combination of the others, and therefore, they don't all point in independent directions.

System of Linear Equations

A system of linear equations consists of multiple linear equations that we solve together to find the values of unknown variables. Visually, we can interpret these equations as lines, planes, or hyperplanes, depending on the number of variables involved, and the solutions are the points where these geometric figures intersect.

In the case of vectors, when we set their linear combination equal to the zero vector, we create a system of equations. From here, we look for values of the scalars (which are the unknowns in this case) that satisfy all the equations simultaneously. Solving a system of linear equations is a fundamental skill in algebra and is applicable across numerous fields, from engineering to economics.

The method of solving the system varies depending on its properties. For instance, in the textbook example, the system is solved using a combination of substitution and elimination techniques. The overarching goal is to isolate the variables (in this case, the scalars \( c_1 \), \( c_2 \), and \( c_3 \) associated with the vectors) and find their values. If the system has no solution, the equations are inconsistent; if it has exactly one solution, it is consistent and independent; and if it has infinitely many solutions, it is consistent but dependent, which is what indicates our vectors are linearly dependent.

Non-Zero Constants

In the context of linear dependence and systems of equations, non-zero constants (or scalars) play a pivotal role. These are the values that we can multiply vectors by to obtain another vector within the same vector space. Specifically, with respect to our example, they refer to values like \( c_1 \) and \( c_3 \) that should not be zero if we're proving that vectors are linearly dependent.

Finding non-zero constants that make the linear combination of several vectors equal to the zero vector confirms that those vectors are indeed linearly dependent. This means we can express one or more vectors in the set as a combination of the others, which is profound because it implies redundancy within the set.

Understanding the concept of non-zero constants helps students grasp why certain solutions to systems of linear equations qualify as trivial or non-trivial. In the case of linear dependence, the goal is to demonstrate that a non-trivial solution exists, hence the importance of non-zero constants in the solution to the system as highlighted in the example. These constants are the 'witnesses' to linear dependence and are a testament to the interconnectedness of the vectors in a given set.