Question

Let \(\mathbf{x}=\left[\begin{array}{r}2 \\ 1 \\ -1\end{array}\right], \mathbf{y}=\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right],\) and \(\mathbf{z}=\left[\begin{array}{r}1 \\ 1 \\ -2\end{array}\right] .\) In each case, either write \(\mathbf{v}\) as a linear combination of \(\mathbf{x}, \mathbf{y},\) and \(\mathbf{z},\) or show that it is not such a linear combination. a. \(\mathbf{v}=\left[\begin{array}{r}0 \\ 1 \\ -3\end{array}\right]\) b. \(\mathbf{v}=\left[\begin{array}{r}4 \\ 3 \\ -4\end{array}\right]\) c. \(\mathbf{v}=\left[\begin{array}{l}3 \\ 1 \\ 0\end{array}\right]\) d. \(\mathbf{v}=\left[\begin{array}{l}3 \\ 0 \\ 3\end{array}\right]\)

Short answer

a. Yes, b. Yes, c. No, d. Yes.

Step-by-step solution

Set up the linear combination equations for part a

To determine if \( \mathbf{v} = \begin{bmatrix} 0 \ 1 \ -3 \end{bmatrix} \) can be expressed as a linear combination of \( \mathbf{x}, \mathbf{y}, \) and \( \mathbf{z} \), we write the equation: \[ a\begin{bmatrix} 2 \ 1 \ -1 \end{bmatrix} + b\begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} + c\begin{bmatrix} 1 \ 1 \ -2 \end{bmatrix} = \begin{bmatrix} 0 \ 1 \ -3 \end{bmatrix} \] This translates to the system of equations: 1. \( 2a + b + c = 0 \) 2. \( a + c = 1 \) 3. \( -a + b - 2c = -3 \)

Solve the system of equations for part a

By solving the system, we use the equations: 1. Substitute \( a \) from the second equation: \( a = 1 - c \) 2. Substitute into the third equation: \( -(1-c) + b - 2c = -3 \Rightarrow -1 + c + b - 2c = -3 \Rightarrow b - c = -2 \) 3. Use this simplification in the first equation: \( 2(1-c) + b + c = 0 \Rightarrow 2 - 2c + b + c = 0 \Rightarrow b - c = -2 \). The system checks out, meaning \( b, c \) can satisfy all, thus \( \mathbf{v} \) is a linear combination.

Set up and solve for part b

For \( \mathbf{v} = \begin{bmatrix} 4 \ 3 \ -4 \end{bmatrix} \), set the equation: \[ a\begin{bmatrix} 2 \ 1 \ -1 \end{bmatrix} + b\begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} + c\begin{bmatrix} 1 \ 1 \ -2 \end{bmatrix} = \begin{bmatrix} 4 \ 3 \ -4 \end{bmatrix} \] System: 1. \( 2a + b + c = 4 \) 2. \( a + c = 3 \) 3. \( -a + b - 2c = -4 \) Solving these equations proves consistent, meaning \( \mathbf{v} \) is a linear combination.

Set up and solve for part c

For \( \mathbf{v} = \begin{bmatrix} 3 \ 1 \ 0 \end{bmatrix} \): \[ a\begin{bmatrix} 2 \ 1 \ -1 \end{bmatrix} + b\begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} + c\begin{bmatrix} 1 \ 1 \ -2 \end{bmatrix} = \begin{bmatrix} 3 \ 1 \ 0 \end{bmatrix} \] System becomes: 1. \( 2a + b + c = 3 \) 2. \( a + c = 1 \) 3. \( -a + b - 2c = 0 \) Upon solving systems, it turns out inconsistent, so \( \mathbf{v} \) is not a linear combination.

Set up and solve for part d

For \( \mathbf{v} = \begin{bmatrix} 3 \ 0 \ 3 \end{bmatrix} \): \[ a\begin{bmatrix} 2 \ 1 \ -1 \end{bmatrix} + b\begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} + c\begin{bmatrix} 1 \ 1 \ -2 \end{bmatrix} = \begin{bmatrix} 3 \ 0 \ 3 \end{bmatrix} \] System of equations: 1. \( 2a + b + c = 3 \) 2. \( a + c = 0 \) 3. \( -a + b - 2c = 3 \) Solving gives consistency, indicating \( \mathbf{v} \) is indeed a linear combination.

Key concepts

Linear Combinations

Linear combinations form a foundational concept in linear algebra. They allow us to express a vector as a sum of scaled vectors. It's like mixing colors; by adjusting how much of each color you use, you can get new shades. Given vectors \( \mathbf{x} \), \( \mathbf{y} \), and \( \mathbf{z} \), they can form a new vector \( \mathbf{v} \) through a linear combination when you find scalars, often called coefficients, like \( a \), \( b \), and \( c \).

Mathematically, this is written as:

• \( \mathbf{v} = a \mathbf{x} + b \mathbf{y} + c \mathbf{z} \)

Here, the scalars \( a \), \( b \), and \( c \) determine how much of each vector \( \mathbf{x} \), \( \mathbf{y} \), and \( \mathbf{z} \) contributes to \( \mathbf{v} \).
Solving for these scalars involves setting up equations based on the components of the vectors. By finding \( a \), \( b \), and \( c \), you confirm whether \( \mathbf{v} \) is a linear combination of the given vectors. If you can't find suitable values that work in the equations, \( \mathbf{v} \) isn't a linear combination of the others.

Vector Equations

Vector equations are like a puzzle where vectors are combined to produce a specific result, often another vector. Each vector isn't just a direction but includes magnitude, giving us a defined point in space. When forming a vector equation, we use an equality involving vector addition and scalar multiplication.

This typically looks like:

• \( a \mathbf{x} + b \mathbf{y} + c \mathbf{z} = \mathbf{v} \)

Where \( a \), \( b \), and \( c \) are unknowns. Solving this equation finds the specific amounts of \( \mathbf{x} \), \( \mathbf{y} \), and \( \mathbf{z} \) needed to yield \( \mathbf{v} \). Each component of the vectors contributes one equation in the system. To solve the puzzle, look at each part and bring them together consistently.
Mastering vector equations helps in solving real-world problems where vectors model physical quantities.

Systems of Linear Equations

Systems of linear equations are a very structured way to solve multiple equations at once. Normally, when finding if one vector is a linear combination of others, we deal with several equations simultaneously, each representing a component of the vectors involved.

For a system like:

• \( 2a + b + c = 4 \)
• \( a + c = 3 \)
• \( -a + b - 2c = -4 \)

These equations need to be solved together to find \( a \), \( b \), and \( c \). Techniques such as substitution or elimination can simplify these systems. The goal is to isolate each variable to see if consistent values that satisfy all equations can be found.
The result tells us if a specific vector can be created using given vectors. If the system has a solution, the vector is a valid linear combination. If not, something has to give, meaning that vector can't be composed from the others. Once mastered, these tools open up vast possibilities, from geometry to more advanced applications in engineering and sciences.