Question

Let \(\mathbf{u}=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right], \mathbf{v}=\left[\begin{array}{l}0 \\ 1 \\ 2\end{array}\right],\) and \(\mathbf{w}=\left[\begin{array}{r}1 \\ 0 \\ -1\end{array}\right] .\) In each case, find numbers \(a, b,\) and \(c\) such that \(\mathbf{x}=a \mathbf{u}+b \mathbf{v}+c \mathbf{w}\) a. \(\mathbf{x}=\left[\begin{array}{r}2 \\ -1 \\ 6\end{array}\right]\) b. \(\mathbf{x}=\left[\begin{array}{l}1 \\ 3 \\ 0\end{array}\right]\)

Short answer

a: \( a = 8, b = -9, c = -6 \); b: \( a = -5, b = 8, c = 6 \).

Step-by-step solution

Represent the Problem as a System of Equations for Part a

Given \( \mathbf{x} = a \mathbf{u} + b \mathbf{v} + c \mathbf{w} \), substitute \( \mathbf{u}, \mathbf{v}, \mathbf{w} \) and set the corresponding components equal to the components of \( \mathbf{x} = \begin{bmatrix} 2 \ -1 \ 6 \end{bmatrix} \): 1. \( a \cdot 1 + b \cdot 0 + c \cdot 1 = 2 \) 2. \( a \cdot 1 + b \cdot 1 + c \cdot 0 = -1 \) 3. \( a \cdot 2 + b \cdot 2 + c \cdot (-1) = 6 \). These create a system of three equations.

Solve the System of Equations for Part a

Address each equation:1. From \( 1a + 0b + 1c = 2 \), we have \( a + c = 2 \).2. From \( 1a + 1b + 0c = -1 \), we have \( a + b = -1 \).3. From \( 2a + 2b - c = 6 \), we have \( 2a + 2b - c = 6 \).By solving:- From equation 1: Substitute \( c = 2 - a \) into other equations.- From equation 2: Substitute \( b = -1 - a \) into other equations.Substitute into equation 3:\( 2a + 2(-1-a) - (2-a) = 6 \), solve for \( a \):- \( 2a - 2 - 2a - (2-a) = 6 \)- \( -2 + a = 6 \)- \( a = 8 \)Then find \( b \) and \( c \) using expressions:- \( b = -1 - a = -1 - 8 = -9 \)- \( c = 2 - a = 2 - 8 = -6 \) Thus, \( a = 8, b = -9, c = -6 \).

Set Up System for Part b

Similarly, for \( \mathbf{x} = \begin{bmatrix} 1 \ 3 \ 0 \end{bmatrix} \): 1. \( a \cdot 1 + b \cdot 0 + c \cdot 1 = 1 \) 2. \( a \cdot 1 + b \cdot 1 + c \cdot 0 = 3 \) 3. \( a \cdot 2 + b \cdot 2 + c \cdot (-1) = 0 \). These set the equations to find \( a, b, c \).

Solve the System for Part b

Apply the same process:- From equation 1: \( a + c = 1 \) implies \( c = 1 - a \).- From equation 2: \( a + b = 3 \) implies \( b = 3 - a \).Substitute into equation 3:\( 2a + 2(3-a) - (1-a) = 0 \)Simplify:- \( 2a + 6 - 2a - 1 + a = 0 \)- \( a + 5 = 0 \), so \( a = -5 \).Use this value to find \( b \) and \( c \):- \( b = 3 - a = 3 + 5 = 8 \)- \( c = 1 - a = 1 + 5 = 6 \).Thus, \( a = -5, b = 8, c = 6 \).

Key concepts

System of Equations

Understanding a system of equations is crucial when dealing with vector decomposition. In the exercise, we're given a situation where a vector \( \mathbf{x} \) can be expressed as a combination of other vectors \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w} \). To solve this, we set up a system of equations corresponding to each component of the vectors. Imagine breaking down the vector \( \mathbf{x} \) into its x-coordinate, y-coordinate, and z-coordinate. Each coordinate gives us a separate equation. For example, the equations from Part a are:- \( a + c = 2 \)- \( a + b = -1 \)- \( 2a + 2b - c = 6 \)These equations share variables \( a, b, \) and \( c \). By solving them, you can find the specific values that represent how much of each vector \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w} \) are needed to make up the vector \( \mathbf{x} \). Each equation is interconnected, representing a different dimension of the vector.

Linear Combinations

A linear combination refers to an expression made up of scaled vectors. This exercise challenges us to express a vector as a linear combination of other vectors. In simple terms, it's like making a delicious smoothie by mixing certain amounts (or coefficients) of different ingredients.In mathematical terms, if we have the vectors \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \), and we want to create the vector \( \mathbf{x} \), we can write it as:- \( \mathbf{x} = a \mathbf{u} + b \mathbf{v} + c \mathbf{w} \)Here, \( a, b, \) and \( c \) are the coefficients that indicate how much of each vector is included in the mix. By solving the system of equations mentioned earlier, we determine the exact values for these coefficients. Vectors \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \) act like building blocks, and the coefficients \( a, b, \) and \( c \) tell us how to use these blocks to construct \( \mathbf{x} \). It's a process of blending different vectors to achieve a desired outcome.

Vector Components

Vector decomposition involves breaking down a vector into its components along different direction vectors, such as \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w} \). When we decompose a vector into these components, we essentially express it in a space defined by these direction vectors.Think of vector components as the ingredients of a recipe. Each component corresponds to a part of the vector along a particular direction. It tells you how much of each direction vector is needed to form the overall vector. The calculation involves considering:

• The component from vector \( \mathbf{u} \) which is \( a \cdot \mathbf{u} \)
• The component from vector \( \mathbf{v} \) which is \( b \cdot \mathbf{v} \)
• The component from vector \( \mathbf{w} \) which is \( c \cdot \mathbf{w} \)

Adding these together gives you the resulting vector \( \mathbf{x} \).In the exercise, decomposing \( \mathbf{x} \) means determining how much of \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \) are required to recreate \( \mathbf{x} \). By solving the system of equations, we're effectively finding these vector components or ingredients for \( \mathbf{x} \). This process highlights the interconnectedness and directionality inherent in vector spaces.