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Let \(B=\left\\{\left[\begin{array}{r}1 \\ -1 \\ 1 \\\ 0\end{array}\right],\left[\begin{array}{r}-1 \\ 1 \\ 0 \\\ 2\end{array}\right],\left[\begin{array}{r}1 \\ 1 \\ -1 \\\ -1\end{array}\right]\right\\} .\) For each of the following vectors, either express it as a linear combination of the vectors of \(B\) or show that it is not a vector in \(\operatorname{Span} B\). (a) \(\left[\begin{array}{r}3 \\ 2 \\ -1 \\ -1\end{array}\right]\) (b) \(\left[\begin{array}{r}-7 \\ 3 \\ 0 \\ 8\end{array}\right]\) (c) \(\left[\begin{array}{l}1 \\ 1 \\ 1 \\ 1\end{array}\right]\)

Short Answer

Expert verified
(a) In Span B; (b) In Span B; (c) Not in Span B.

Step by step solution

01

Understand the Problem

We need to determine if each given vector is a linear combination of the vectors in the set \( B \). This means we should find scalars \( a, b, \) and \( c \) such that \( a \left[ \begin{array}{r} 1\ -1\ 1\ 0 \end{array} \right] + b \left[ \begin{array}{r} -1\ 1\ 0\ 2 \end{array} \right] + c \left[ \begin{array}{r} 1\ 1\ -1\ -1 \end{array} \right] = \left[ \begin{array}{c} \, \sigma_1 \, \ \, \sigma_2 \, \ \, \sigma_3 \, \ \, \sigma_4 \, \end{array} \right] \), where \( \sigma_1, \sigma_2, \sigma_3, \sigma_4 \) are the components of the given vector.
02

Set Up Equations for Vector (a)

Vector \( \left[ \begin{array}{r} 3\ 2\ -1\ -1 \end{array} \right] \) is our target. We need to solve the system of equations: 1. \( a - b + c = 3 \) 2. \( -a + b + c = 2 \) 3. \( a + 0b - c = -1 \) 4. \( 0a + 2b - c = -1 \)
03

Solve System for Vector (a)

By solving the equations: 1. From (3), \( a - c = -1 \), thus \( c = a + 1 \) 2. Substitute \( c = a + 1 \) into (4), results in \( 2b - (a + 1) = -1 \) -> \( 2b - a = 0 \) -> \( a = 2b \) 3. Substitute \( c = a + 1 \) and \( a = 2b \) in (1), - \( 2b - b + 2b + 1 = 3 \), \( 3b = 2 \) -> \( b = \frac{2}{3} \) 4. Substitute \( b = \frac{2}{3} \) in \( a = 2b \), we get \( a = \frac{4}{3} \) 5. \( c = a + 1 = \frac{4}{3} + 1 = \frac{7}{3} \) Thus, vector (a) can be expressed as a combination of \( B \).
04

Set Up Equations for Vector (b)

The target vector is \( \left[ \begin{array}{r} -7\ 3\ 0\ 8 \end{array} \right] \). Model equations: 1. \( a - b + c = -7 \) 2. \( -a + b + c = 3 \) 3. \( a - c = 0 \) 4. \( 2b - c = 8 \)
05

Solve System for Vector (b)

Solve the equations: 1. From (3), \( a = c \) 2. Substitute into (1): \( c - b + c = -7 \), \( 2c - b = -7 \) 3. Solve (2): substitute \( a = c \), gives \( -c + b + c = 3 \), thus \( b = 3 \) 4. Substitute \( b = 3 \) in \( 2c - 3 = -7 \) from step 2, obtain: \( 2c = -4 \), \( c = -2 \) 5. With \( c = -2 \), \( a = c = -2 \) 6. Check (4) with \( b = 3 \) and \( c = -2 \), verify: \( 2(3) - (-2) = 8 \). Thus, vector (b) can be expressed in terms of \( B \).
06

Set Up Equations for Vector (c)

The target vector is \( \left[ \begin{array}{l} 1\ 1\ 1\ 1 \end{array} \right] \). Model equations are: 1. \( a - b + c = 1 \) 2. \( -a + b + c = 1 \) 3. \( a - c = 1 \) 4. \( 2b - c = 1 \)
07

Solve and Verify for Vector (c)

Solve the equations: 1. From (3), \( a - c = 1 \), thus \( a = c + 1 \). 2. Substitute into (1): \( (c+1) - b + c = 1 \), simplifying gives \( 2c - b = 0 \), then \( b = 2c \) 3. Substitute \( b = 2c \) into (4): \( 2(2c) - c = 1 \), gives \( 3c = 1 \), thus \( c = \frac{1}{3} \) 4. Substitute \( c = \frac{1}{3} \) into \( a = c + 1 \), obtaining \( a = \frac{4}{3} \) 5. Calculate \( b = 2c = \frac{2}{3} \) 6. These do not satisfy equation (2), so vector (c) is not in \( \operatorname{Span}(B) \).
08

Conclude the Results

In all previous steps, summarize: - Vector (a) \( \left[ \begin{array}{c} 3\ 2\ -1\ -1 \end{array} \right] \) is in \( \operatorname{Span}(B) \). - Vector (b) \( \left[ \begin{array}{c} -7\ 3\ 0\ 8 \end{array} \right] \) is in \( \operatorname{Span}(B) \). - Vector (c) \( \left[ \begin{array}{c} 1\ 1\ 1\ 1 \end{array} \right] \) is not in \( \operatorname{Span}(B) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Combination
When we discuss linear combinations in the context of linear algebra, we are essentially exploring the concept of combining several vectors to create a new vector. This new vector is formed by multiplying each vector in a set by a scalar and then summing the results. For example, given vectors \( \mathbf{v}_1, \mathbf{v}_2, \) and \( \mathbf{v}_3 \) and scalars \( a, b, \) and \( c \) respectively, a linear combination would look like:
  • \( a\mathbf{v}_1 + b\mathbf{v}_2 + c\mathbf{v}_3 \)

In our original exercise, the vectors from set \( B \) are combined using specific scalars to determine if another vector can be constructed from them. By solving for the scalars, we can tell if the target vector is attainable through a linear combination of the vectors in \( B \). Understanding linear combinations is key because it forms the foundation for more advanced concepts in linear algebra, such as span and vector spaces. It allows us to see connections between vectors and how they relate through these scalars. If a vector can be formed by the linear combination of other vectors, it illustrates dependency within the set of vectors.
Vector Space
A vector space is a collection of vectors that can be added together and multiplied by scalars, while adhering to a set of rules or axioms. These rules ensure that operations like vector addition and scalar multiplication are consistent and reliable. Some characteristics of vector spaces include:
  • Closure under addition: Adding any two vectors from the space results in another vector that also belongs to the space.
  • Closure under scalar multiplication: Multiplying a vector by a scalar results in another vector within the space.
  • Contains a zero vector: Every vector space includes a zero vector, which acts as the additive identity.

The vectors given in set \( B \) and the tasks we perform to verify if given vectors can be expressed as linear combinations essentially test the conditions of forming a vector space around \( B \). Understanding vector spaces helps in recognizing frameworks in higher-dimensional algebra, providing insight into the properties that define the space through concepts like basis, span, and dimension. The idea of a vector space is fundamental in many areas of mathematics and physics, providing a setting for the analysis of linear equations and transformations.
Span
The span of a set of vectors refers to the collection of all possible vectors that can be produced through linear combinations of those vectors. If we take our set \( B \) and create different linear combinations by adjusting the scalars, all resultants will lie in what we call the span of \( B \). Essentially, to find if a vector is in the span of another set of vectors, we try to express it as a linear combination of the vectors in that set.
  • If a solution exists, the vector lies in the span of the set.
  • Otherwise, it means the vector cannot be formed using the vectors in the set.

In our exercise, checking whether each target vector is in the span of \( B \) involves solving systems of linear equations to find the right scalars, if they exist. Spanning is crucial because it essentially tells us how 'large' or 'capable' a set of vectors is regarding linear combinations. In our problem, if a vector could not be expressed as a linear combination, it indicates that the given set \( B \) does not entirely cover that portion of the vector space.
System of Equations
When determining if a given vector can be expressed as a linear combination of a set of vectors, we often translate this challenge into a system of linear equations. This system arises from equating the target vector components with the expressions formed by the linear combinations of the set's vectors. Each equation corresponds to a component of the vectors, and the set of equations is solved simultaneously to find the scalars.
  • An equation is written for each component of the vectors' dimension.
  • The solutions to these equations reveal if and how the combination can be achieved.

In our exercise, each part involved setting up such a system of equations to find if the target could be expressed with the set \( B \). Solving these equations involves methods such as substitution, elimination, or matrix operations. Understanding systems of equations in this context illustrates their practical application in exploring vector properties such as dependencies and span within a vector space. They are fundamental not only in linear algebra but in numerous disciplines requiring quantitative analysis.

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Most popular questions from this chapter

A student is taking courses in algebra, calculus, and physics at a college where grades are given in percentages. To determine her standing for a physics prize, a weighted average is calculated based on \(50 \%\) of the student's physics grades, \(30 \%\) of her calculus grade, and \(20 \%\) of her algebra grade; the weighted average is 84 . For an applied mathematics prize, a weighted average based on one-third of each of the three grades is calculated to be \(83 .\) For a pure mathematics prize, her average based on \(50 \%\) of her calculus grade and \(50 \%\) of her algebra grade is \(82.5\). What are her grades in the individual courses?

Determine all values of \(k\) such that the given set is linearly independent. (a) \(\left\\{\left[\begin{array}{l}1 \\ 0 \\ 1 \\\ 0\end{array}\right],\left[\begin{array}{l}0 \\ 1 \\ 1 \\\ 1\end{array}\right],\left[\begin{array}{r}2 \\ -3 \\ -1 \\\ k\end{array}\right]\right\\}\) (b) \(\left\\{\left[\begin{array}{l}1 \\ 1 \\ 1 \\\ 2\end{array}\right],\left[\begin{array}{r}1 \\ -1 \\ 2 \\\ 0\end{array}\right],\left[\begin{array}{r}-1 \\ 2 \\ k \\\ 1\end{array}\right]\right\\}\)

Solve the following systems of linear equations by row reducing the coefficient matrix to RREF. Compare your steps with your solutions from the Problem 2.1.A5. (a) $$ \begin{array}{r} 3 x_{1}-5 x_{2}=2 \\ x_{1}+2 x_{2}=4 \end{array} $$ (b) $$ \begin{array}{r} x_{1}+2 x_{2}+x_{3}=5 \\ 2 x_{1}-3 x_{2}+2 x_{3}=6 \end{array} $$ (c) $$ \begin{array}{r} x_{1}+2 x_{2}-3 x_{3}=8 \\ x_{1}+3 x_{2}-5 x_{3}=11 \\ 2 x_{1}+5 x_{2}-8 x_{3}=19 \end{array} $$ (d) $$ \begin{array}{r} -3 x_{1}+6 x_{2}+16 x_{3}=36 \\ x_{1}-2 x_{2}-5 x_{3}=-11 \\ 2 x_{1}-3 x_{2}-8 x_{3}=-17 \end{array} $$ (e) $$ \begin{array}{r} x_{1}+2 x_{2}-x_{3}=4 \\ 2 x_{1}+5 x_{2}+x_{3}=10 \\ 4 x_{1}+9 x_{2}-x_{3}=19 \end{array} $$ (f) $$ \begin{aligned} x_{1}+2 x_{2}-3 x_{3} &=-5 \\ 2 x_{1}+4 x_{2}-6 x_{3}+x_{4} &=-8 \\ 6 x_{1}+13 x_{2}-17 x_{3}+4 x_{4} &=-21 \end{aligned} $$ (g) $$ \begin{array}{rr} 2 x_{2}-2 x_{3} & +x_{5}=2 \\ x_{1}+2 x_{2}-3 x_{3}+x_{4}+4 x_{5}= & 1 \\ 2 x_{1}+4 x_{2}-5 x_{3}+3 x_{4}+8 x_{5}= & 3 \\ 2 x_{1}+5 x_{2}-7 x_{3}+3 x_{4}+10 x_{5}= & 5 \end{array} $$

Solve the system \([A \mid \vec{b}]\) by row reducing the coefficient matrix to RREF. Then, without any further operations, find the general solution to the homogeneous \([A \mid \overrightarrow{0}]\). (a) \(A=\left[\begin{array}{rrr}2 & -1 & 4 \\ 1 & 3 & 0 \\ 1 & 1 & 2\end{array}\right], \vec{b}=\left[\begin{array}{l}1 \\ 0 \\\ 2\end{array}\right]\) (b) \(A=\left[\begin{array}{rrr}1 & 7 & 5 \\ 1 & 0 & 5 \\ -1 & 2 & -5\end{array}\right], \vec{b}=\left[\begin{array}{r}5 \\ -2 \\\ 4\end{array}\right]\) (c) \(A=\left[\begin{array}{rrrr}0 & -1 & 5 & -2 \\ -1 & -1 & -4 & -1\end{array}\right], \vec{b}=\left[\begin{array}{r}-1 \\\ 4\end{array}\right]\) (d) \(A=\left[\begin{array}{rrrr}1 & 0 & -1 & -1 \\ 4 & 3 & 2 & -4 \\ -1 & -4 & -3 & 5\end{array}\right], \vec{b}=\left[\begin{array}{l}3 \\ 3 \\\ 5\end{array}\right]\)

A fruit seller has apples, bananas, and oranges. Altogether he has 1500 pieces of fruit. On average, each apple weighs 120 grams, each banana weighs 140 grams, and each orange weighs 160 grams. He can sell apples for 25 cents each, bananas for 20 cents each, and oranges for 30 cents each. If the fruit weighs 208 kilograms, and the total selling price is \(\$ 380\), how many of each kind of fruit does the fruit seller have?

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