/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 In Problems 29-34, let $$ \m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Problems 29-34, let $$ \mathbf{u}=\left[\begin{array}{l} 3 \\ 4 \end{array}\right], \quad \mathbf{v}=\left[\begin{array}{l} -1 \\ -2 \end{array}\right], \quad \text { and } \quad \mathbf{w}=\left[\begin{array}{r} 1 \\ -2 \end{array}\right] $$ $$ \text { Compute } \mathbf{u}+\mathbf{v} \text { and illustrate the result graphically. } $$

Short Answer

Expert verified
\( \mathbf{u} + \mathbf{v} = \left[ \begin{array}{c} 2 \\ 2 \end{array} \right] \). Graphically, it is a vector from the origin to (2, 2).

Step by step solution

01

Add the Vectors

To find \( \mathbf{u} + \mathbf{v} \), we add the corresponding components of \( \mathbf{u} \) and \( \mathbf{v} \). That means, add the first component of \( \mathbf{u} \) to the first component of \( \mathbf{v} \), and then add the second component of \( \mathbf{u} \) to the second component of \( \mathbf{v} \). Mathematically:\[\mathbf{u} + \mathbf{v} = \left[\begin{array}{c} 3 \ 4 \end{array}\right] + \left[\begin{array}{c} -1 \ -2 \end{array}\right] = \left[\begin{array}{c} 3 + (-1) \ 4 + (-2) \end{array}\right] = \left[\begin{array}{c} 2 \ 2 \end{array}\right]\]
02

Graphical Representation

To illustrate the result graphically, plot the vectors \( \mathbf{u} \), \( \mathbf{v} \), and their resultant \( \mathbf{u} + \mathbf{v} \) on an XY plane. Start \( \mathbf{u} \) at the origin and draw a vector to \((3, 4)\). Then, starting at \((3, 4)\), draw the vector \( \mathbf{v} \) falling to \((2, 2)\). The resultant vector \( \mathbf{u} + \mathbf{v} \) connects the origin directly to \((2, 2)\). This shows the tail-to-tip method and confirms that \( \mathbf{u} + \mathbf{v} = \left[ \begin{array}{c} 2 \ 2 \end{array} \right] \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Graphical Representation of Vectors
When working with vectors, visualizing them graphically can greatly enhance understanding. Imagine the XY plane as a blank canvas where every vector can be drawn as an arrow. The origin, point (0,0), is the starting point for our arrows.

For vector \( \mathbf{u} \), represented by \( [3, 4] \), we start drawing an arrow from the origin to the point (3,4). This arrow illustrates the direction and magnitude of \( \mathbf{u} \). Similarly, vector \( \mathbf{v} \) begins at the origin and points towards (-1, -2). However, when performing vector addition, we don't place every vector at the origin.

Instead, we utilize the tail-to-tip method. After plotting \( \mathbf{u} \), move to the tip of \( \mathbf{u} \) at (3,4) and start \( \mathbf{v} \) from there. This stepwise connection of vectors provides clear visual evidence of the operation and helps conclude the resultant vector \( \mathbf{u} + \mathbf{v} \) with ease, which in this exercise is \( [2, 2] \). By using graphical techniques alongside arithmetic solutions, we reinforce understanding with every vector operation.
Tail-to-Tip Method
The tail-to-tip method is a straightforward and valuable way to add vectors visually. It emphasizes the continuous path from one vector to the next, by connecting them in order.

To apply this method, start with the first vector, drawing it from the origin to its specific coordinates. When the first vector ends, the second vector begins, with its tail at the tip of the first.

Consider \( \mathbf{u} + \mathbf{v} \). Draw \( \mathbf{u} \) from (0,0) to (3,4), then draw \( \mathbf{v} \) from (3,4) to (2,2). This continuous line from the origin to \( (2, 2) \) represents the resultant vector. This method not only aids in visualizing results but also provides a geometric understanding of vector operations.
Component-Wise Addition
Component-wise addition is a basic but essential skill in vector mathematics. It involves adding corresponding components of two vectors, thus simplifying vector addition.

For vectors \( \mathbf{u} = [3, 4] \) and \( \mathbf{v} = [-1, -2] \), the process involves simple arithmetic:
  • First component: \( 3 + (-1) = 2 \)
  • Second component: \( 4 + (-2) = 2 \)
These calculations produce the resultant vector \( [2, 2] \). By understanding the arithmetic in component-wise addition, you gain insight into vector properties and operations. This breakdown not only simplifies calculations but also bridges the gap between numerical and graphical vector representations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 & 1 \\ 1 & -3 & -3 \end{array}\right], \\ \quad C=\left[\begin{array}{rrr} -2 & 0 & 4 \\ 1 & -3 & 1 \\ 0 & 0 & 2 \end{array}\right] \end{array} $$ $$ \text { Show that if } A+B=C, \text { then } A=C-B $$

(a) Show that if \(X=A X+D\), then $$ X=(I-A)^{-1} D $$ provided that \(I-A\) is invertible. (b) Suppose that $$ A=\left[\begin{array}{rr} 3 & 2 \\ 0 & -1 \end{array}\right] \text { and } \quad D=\left[\begin{array}{r} -2 \\ 2 \end{array}\right] $$ Compute \((I-A)^{-1}\), and use your result in (a) to compute \(X\).

Suppose that breeding occurs once a year and that a census is taken at the end of each breeding season. Assume that a population is divided into four age classes and that \(70 \%\) of the females age \(0,50 \%\) of the females age 1, and \(10 \%\) of the females age 2 survive until the end of the next breeding season. Assume further that females age 2 have an average of \(4.6\) female offspring and females age 3 have an average of \(3.7\) female offspring. If, at time 0, the population consists of 1500 females age 0,500 females age 1,250 females age 2, and 50 females age 3, find the Leslie matrix and the age distribution at time \(2 .\)

Suppose that $$ A=\left[\begin{array}{ll} a & 8 \\ 2 & 4 \end{array}\right], \quad X=\left[\begin{array}{l} x \\ y \end{array}\right], \quad \text { and } \quad B=\left[\begin{array}{l} b_{1} \\ b_{2} \end{array}\right] $$ (a) Show that when \(a \neq 4, A X=B\) has exactly one solution. (b) Suppose \(a=4 .\) Find conditions on \(b_{1}\) and \(b_{2}\) such that \(A X=\) \(B\) has (i) infinitely many solutions and (ii) no solutions. (c) Explain your results in (a) and (b) graphically.

Let $$ A=\left[\begin{array}{rr} 1 & 3 \\ 0 & -2 \end{array}\right] \text { and } I_{2}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] $$ Show that \(A I_{2}=I_{2} A=A\).

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.