Question

Exercise 2.2.1 In each case find a system of equations that is equivalent to the given vector equation. a. \(x_{1}\left[\begin{array}{r}2 \\ -3 \\\ 0\end{array}\right]+x_{2}\left[\begin{array}{l}1 \\ 1 \\\ 4\end{array}\right]+x_{3}\left[\begin{array}{r}2 \\ 0 \\\ -1\end{array}\right]=\left[\begin{array}{r}5 \\ 6 \\ -3\end{array}\right]\) b. \(x_{1}\left[\begin{array}{l}1 \\ 0 \\ 1 \\\ 0\end{array}\right]+x_{2}\left[\begin{array}{r}-3 \\ 8 \\ 2 \\\ 1\end{array}\right]+x_{3}\left[\begin{array}{r}-3 \\ 0 \\ 2 \\\ 2\end{array}\right]+x_{4}\left[\begin{array}{r}3 \\ 2 \\ 0 \\\ -2\end{array}\right]=\left[\begin{array}{l}5 \\ 1 \\ 2 \\\ 0\end{array}\right]\)

Short answer

a. \(2x_1 + x_2 + 2x_3 = 5\), \(-3x_1 + x_2 = 6\), \(4x_2 - x_3 = -3\). b. \(x_1 - 3x_2 - 3x_3 + 3x_4 = 5\), \(8x_2 + 2x_4 = 1\), \(x_1 + 2x_2 + 2x_3 = 2\), \(x_2 + 2x_3 - 2x_4 = 0\).

Step-by-step solution

Understand the Vector Equation (Part a)

Examine the vector equation given: \(x_1\left[\begin{array}{r}2 \ -3 \ 0\end{array}\right] + x_2\left[\begin{array}{l}1 \ 1 \ 4\end{array}\right] + x_3\left[\begin{array}{r}2 \ 0 \ -1\end{array}\right] = \left[\begin{array}{r}5 \ 6 \ -3\end{array}\right]\). This equation represents a system involving vector addition and scalar multiplication.

Break Down Vectors into Components (Part a)

Separate the vector equation into components based on matching each component on both sides of the equality: \[\begin{align*} 2x_1 + x_2 + 2x_3 &= 5 \ -3x_1 + x_2 &= 6 \ 4x_2 - x_3 &= -3 \end{align*}\]. This creates a system of linear equations.

Understand the Vector Equation (Part b)

For the second part, consider the given vector equation: \(x_1\left[\begin{array}{l}1 \ 0 \ 1 \ 0\end{array}\right] + x_2\left[\begin{array}{r}-3 \ 8 \ 2 \ 1\end{array}\right] + x_3\left[\begin{array}{r}-3 \ 0 \ 2 \ 2\end{array}\right] + x_4\left[\begin{array}{r}3 \ 2 \ 0 \ -2\end{array}\right] = \left[\begin{array}{l}5 \ 1 \ 2 \ 0\end{array}\right]\). Here, we have vectors with four components.

Break Down Vectors into Components (Part b)

Write down individual component equations based on vectors: \[\begin{align*} x_1 - 3x_2 - 3x_3 + 3x_4 &= 5 \ 8x_2 + 2x_4 &= 1 \ x_1 + 2x_2 + 2x_3 &= 2 \ x_2 + 2x_3 - 2x_4 &= 0 \end{align*}\]. These represent the system of equations corresponding to the vector equation.

Key concepts

System of Equations

A system of equations is a collection of equations that share variables. When solving these, the goal is to find values for the variables that satisfy all the given equations at the same time. In linear algebra, systems often appear from scenarios involving vectors and matrices.
For example, consider a vector equation that needs to be expressed as a system of equations. Each vector's components correspond to one equation in the set. By equating corresponding components on both sides, you can form a system of linear equations.

• In part a, each component creates a linear equation: \(2x_1 + x_2 + 2x_3 = 5\), \(-3x_1 + x_2 = 6\), and \(4x_2 - x_3 = -3\).
• In part b, the components form another system: \(x_1 - 3x_2 - 3x_3 + 3x_4 = 5\), \(8x_2 + 2x_4 = 1\), etc.

Solving these systems involves finding suitable values for the variables that satisfy all equations simultaneously.

Vector Addition

Vector addition is like adding two sets of directions. If you have two vectors, you add them by combining their corresponding components. This is useful in many applications, such as physics and computer graphics.
In the given vector equations, each term involving \(x_1\), \(x_2\), etc., is scaled and then added to form new components on the right-hand side.
Imagine moving from one point to another by taking several steps in different directions. Each step is a vector, and by adding them, you determine your final position. In algebra, this process helps to translate vector equations into linear systems.
For part a, vectors are added component-wise to equal \([5,6,-3]\). For part b, the process extends to four-component vectors.

Scalar Multiplication

Scalar multiplication involves multiplying each component of a vector by the same number, called a scalar. This operation stretches or shrinks the vector but keeps its direction the same unless the scalar is negative, which reverses the direction.
When you see terms like \(x_1[2, -3, 0]\) in the equations, \(x_1\) is a scalar that scales the vector \([2, -3, 0]\).

• This concept is crucial when forming equations: the scalar multiplies each vector component, contributing to the overall system.
• For example, in part a, multiplying \(x_1\) with \([2, -3, 0]\) results in the vector \([2x_1, -3x_1, 0]\).

This is the same for all terms in both given vector equations. Scalar multiplication is straightforward but forms the backbone of translating vector equations into systems.

Linear Combinations

A linear combination involves taking several vectors, scaling them by different amounts, and then adding the results. It's a fundamental concept in linear algebra because it shows how vectors relate to each other.
In our exercise, you see vectors being combined linearly to match a target vector. The equation represents combining different vector contributions through specific scalars to reach a desired result.
In part a, each vector is scaled by \(x_1\), \(x_2\), and \(x_3\). Together, they produce a target vector \([5, 6, -3]\). In part b, the goal is \([5, 1, 2, 0]\).
This approach helps to express relationships between vectors and find solutions in system equations, making it a key concept in understanding vector spaces and spanning sets.