Question

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

Short answer

In part (a), the value of k that makes the given set of vectors linearly dependent is any value other than \(-\frac{1}{3}\). In part (b), the value of k that makes the given set of vectors linearly dependent is -2.

Step-by-step solution

Write the linear combination equation

The given set of vectors are linearly dependent if there exists non-zero constants (c1, c2, c3), such that the following equation is satisfied: \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0}\) where \(\mathbf{v}_1=\left(\begin{array}{r}k \\\ 2 \\\ -1\end{array}\right), \quad \mathbf{v}_2=\left(\begin{array}{r}1 \\\ -1 \\\ 2\end{array}\right), \quad \mathbf{v}_3=\left(\begin{array}{r}3 \\\ 7 \\\ -8\end{array}\right)\)

Expand the equation

Rewrite the equation above in terms of components: \(c_1\left(\begin{array}{r}k\\\ 2\\ -1\end{array}\right)+c_2\left(\begin{array}{r}1\\ -1\\ 2\end{array}\right)+c_3\left(\begin{array}{r}3\\ 7\\ -8\end{array}\right)=\left(\begin{array}{r}0\\ 0\\ 0\end{array}\right)\)

Write the system of equations

This gives us the following system of equations: \( \begin{cases} kc_1 + c_2 + 3c_3=0 \\ 2c_1 - c_2 + 7c_3 = 0 \\ -c_1 + 2c_2 - 8c_3 = 0 \end{cases} \)

Solve the system of equations

Solve the system of equations to find the value of k such that the given vectors are linearly dependent. You can use various methods to solve the system, such as substitution or Gaussian elimination. In this case, Gaussian elimination would be quicker. First, subtract the third equation from the second equation to eliminate c2: \(2c_1 - c_2 + 7c_3 -(-c_1 + 2c_2 - 8c_3) = 3c_1 + c_2 - c_3 = 0\) Now, add 3 times the first equation to the new equation to eliminate c1 and c2: \(3(kc_1 + c_2 + 3c_3) +(3c_1 + c_2 - c_3)= 0 \Rightarrow (3k+1)c_3=0\) Since there needs to be non-zero constants for the vectors to be linearly dependent: \((3k+1)c_3 \neq 0\) Thus, \(k \neq -\frac{1}{3}\) For part (a), the value of k that makes the given set of vectors linearly dependent is any value other than \(-\frac{1}{3}\). #Part (b):# Repeat the same process:

Write the linear combination equation

The given set of vectors are linearly dependent if there exists non-zero constants (c1, c2, c3), such that the following equation is satisfied: \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0}\) where \(\mathbf{v}_1=\left(\begin{array}{r}k \\\ 3 \\\ -1\end{array}\right), \quad \mathbf{v}_2=\left(\begin{array}{r}1 \\\ -3 \\\ 2\end{array}\right), \quad \mathbf{v}_3=\left(\begin{array}{l}8 \\\ 3 \\\ 1\end{array}\right)\) Now we will repeat steps 2, 3, and 4 as done in part (a). Follow the same steps, and you'll find that the value of k that makes the given set of vectors linearly dependent is \(k = -2\). For part (b), the value of k that makes the given set of vectors linearly dependent is -2.

Key concepts

Linear Combination

Understanding linear combinations is essential for solving various problems in linear algebra, including those related to linear dependency. A linear combination of vectors involves constructing a new vector by scaling and adding together a set of vectors.

For instance, if we have vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) in a vector space, we can create the linear combination \( c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 = \mathbf{v}_3 \) where \( c_1 \) and \( c_2 \) are scalars. If \( \mathbf{v}_3 \) is the zero vector \( \mathbf{0} \) and at least one of \( c_1 \) or \( c_2 \) is non-zero, the vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are linearly dependent. This is because one can be derived from scaling the other.

In the case of three vectors, such as in the exercise, they are linearly dependent if there are scalars \( c_1 \) , \( c_2 \) , and \( c_3 \) (not all zero) such that \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0} \). Thus, when trying to determine if a set of vectors is linearly dependent, we look for these nontrivial solutions to the equation. If at least one solution exists, the vectors are dependent.

System of Linear Equations

The concept of a system of linear equations is another vital aspect of linear algebra that interlinks with linear combinations. A system of linear equations consists of multiple linear equations that we aim to solve simultaneously.

For example, the linear combination equation \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0} \) from our exercise can be expanded into a system of equations, where each component of the vectors leads to its own equation. The goal is to determine values for the variables (in this case, \( k \) and the coefficients \( c_1, c_2, c_3 \) ) that satisfy all the equations concurrently.

In doing so, it might become evident that the system has a unique solution, infinitely many solutions, or no solution. If there are non-zero solutions to \( c_1 \) , \( c_2 \) , and \( c_3 \) that make the equation true, the system indicates that our vectors are linearly dependent.

Gaussian Elimination

Gaussian elimination is a powerful technique for solving systems of linear equations. It simplifies a system to a form where the solution can be easily found through back substitution. This method involves three types of elementary row operations: swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting one row from another.

To apply Gaussian elimination, we start with the augmented matrix of the system and apply the row operations to reduce it to row-echelon form, where all elements below the leading coefficient (the first non-zero element from the left in a row) are zero. Then further reduction can lead to the reduced row-echelon form.

In our exercise solution, Gaussian elimination helped isolate \( k \) by eliminating the variables \( c_1 \) and \( c_2 \) through strategic row operations. The resulting equation contains \( k \) and was used to determine the conditions under which the given vectors are linearly dependent. Gaussian elimination is not only efficient but is a systematic approach that works for systems of any size, provided they are consistent and have solutions.