Problem 11 - Advanced Engineering Mathematics

Question

In Problems \(11-14\), determine whether the given set of vectors is linearly dependent or linearly independent. $$ \mathbf{u}_{1}=\langle 1,2,3\rangle, \mathbf{u}_{2}=\langle 1,0,1\rangle, \mathbf{u}_{3}=\langle 1,-1,5\rangle $$

Step-by-Step Solution

Answer

The vectors are linearly independent.

Step 1: Understand the Problem

We have three vectors \( \mathbf{u}_{1}=\langle 1,2,3\rangle \), \( \mathbf{u}_{2}=\langle 1,0,1\rangle \), and \( \mathbf{u}_{3}=\langle 1,-1,5\rangle \). We need to determine if these vectors are linearly dependent or independent. Vectors are linearly dependent if one of the vectors can be written as a combination of the others.

Step 2: Set Up the Linear Dependence Equation

For the vectors \(\mathbf{u}_1\), \(\mathbf{u}_2\), and \(\mathbf{u}_3\) to be linearly dependent, there must exist scalars \( a, b, c \), not all zero, such that: \[ a\mathbf{u}_1 + b\mathbf{u}_2 + c\mathbf{u}_3 = \langle 0, 0, 0 \rangle \] This expands to: \[ a\langle 1, 2, 3\rangle + b\langle 1, 0, 1\rangle + c\langle 1, -1, 5\rangle = \langle 0, 0, 0 \rangle \] Which results in the system of linear equations: \[ a + b + c = 0 \] \[ 2a + 0b - c = 0 \] \[ 3a + b + 5c = 0 \]

Step 3: Solve the System of Equations

We solve the system: \[ a + b + c = 0 \] \[ 2a - c = 0 \] \[ 3a + b + 5c = 0 \] - From the second equation, solve for \(c\): \( c = 2a \). - Substitute \( c = 2a \) into the first equation: \( a + b + 2a = 0 \Rightarrow 3a + b = 0 \). - Solve for \( b \): \( b = -3a \). - Substitute \( b = -3a \) and \( c = 2a \) into the third equation: \[ 3a - 3a + 5(2a) = 0 \Rightarrow 10a = 0 \] This implies \( a = 0 \).

Step 4: Conclude Linear Independence

Since \( a = 0 \), from \( b = -3a \) and \( c = 2a \), both \( b = 0 \) and \( c = 0 \). All the scalars \( a, b, \text{and} c \) in our equation are zero, which means the only solution is the trivial solution. Therefore, the vectors \( \mathbf{u}_1, \mathbf{u}_2, \text{and} \mathbf{u}_3 \) are linearly independent.

Key Concepts

Vectors

Vectors are fundamental in understanding linear algebra and play an important role in determining linear independence. A vector can be thought of as an arrow pointing in a specific direction, with a set magnitude or length. In mathematics, a vector is often represented as a list of numbers, such as \(\mathbf{u}_1 = \langle 1,2,3 \rangle\). Each number in this list is known as a component of the vector, corresponding to an entry in a multi-dimensional space.

In our specific problem, we're dealing with three vectors: \(\mathbf{u}_1 = \langle 1,2,3 \rangle\), \(\mathbf{u}_2 = \langle 1,0,1 \rangle\), and \(\mathbf{u}_3 = \langle 1,-1,5 \rangle\). To determine whether they are linearly independent, we need to see if one can be expressed as a combination of the others. This involves finding scalars (constant multipliers) for each vector.

• Linear independence means no vector in the set is a combination of the others.
• Linear dependence implies at least one vector is a combination of the others.

Evaluating these properties helps determine the uniqueness and span of the vector space formed by these vectors.

System of Equations

A system of equations is a collection of equations that are solved together. They are crucial for problems that involve multiple variables and their relationships. To determine if our vectors are linearly independent, we set up a system based on the condition for linear dependence: \(a\mathbf{u}_1 + b\mathbf{u}_2 + c\mathbf{u}_3 = \langle 0, 0, 0 \rangle\). This equation suggests that if it’s possible to find non-zero values for the scalars \(a, b,\) and \(c\) that satisfy the equation, then the vectors are dependent.

The system derived from our vectors is:

• \(a + b + c = 0\)
• \(2a - c = 0\)
• \(3a + b + 5c = 0\)

To solve this system, we look for values of \(a, b,\) and \(c\) that satisfy all equations simultaneously. This often involves substitution or elimination methods. For example, solving the second equation gives \(c = 2a\). Substituting into the other equations helps find the values of the others. Through this approach, we determined the trivial solution where all scalars are zero. This means the vectors are independent, highlighting how systems of equations are used to explore vector spaces.

Trivial Solution

The trivial solution in the context of linear algebra refers to the solution where all variables are zero. In our system of equations, we found that \(a = 0\), \(b = 0\), and \(c = 0\). This implies that no non-zero scalar multiples of the given vectors can sum to the zero vector unless all scalars are zero themselves, indicating linear independence.

In linear algebra, the trivial solution is significant because:

• If the only solution to the vector equation is trivial, the set of vectors is linearly independent.
• If non-trivial solutions exist (solutions other than all-zero), the vectors are linearly dependent.

Understanding the trivial solution helps comprehend the foundational concepts of vector spaces and why it's used to determine linear relationships between vectors. Linear independence results from the trivial solution, demonstrating the uniqueness and non-redundancy of the vector set.