/*! 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 35 Solve for \(w,\) where \(u=(1,-1... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 35

Solve for \(w,\) where \(u=(1,-1,0,1)\) and \(v=(0,2,3,-1)\). $$3 w=u-2 v$$

Short Answer

Expert verified
The vector \(w\) based on the given vectors \(u\) and \(v\) and the described relation is \(w = (1/3, -5/3, -2, 1)\).

Step by step solution

01

Subtract 2v from u

To isolate \(3w\), we first do \(u - 2v\) to find the right side of the equation. This implies subtracting twice the values of each element of vector \(v\) from the corresponding element in vector \(u\). Which results in the vector: \((1-0, -1-4, 0-6, 1 - (-2)) = (1, -5, -6, 3)\).
02

Divide by 3

To find \(w\), we need to divide the result of Step 1 by 3, because the equation says that \(3w\), not \(w\), equals the result of step 1. So, we divide each element of the vector obtained in step 1 by 3 to obtain: \((1/3, -5/3, -6/3, 3/3) = (1/3, -5/3, -2, 1)\).
03

The solution

So, the solution of the vector equation is the vector \(w = (1/3, -5/3, -2, 1)\).

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 91影视!

Key Concepts

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

Vector Subtraction
Vector subtraction is vital for solving many vector algebra problems. It's the process of subtracting corresponding elements from two vectors. In our exercise, we subtract vector \(2v\) from \(u\), where \(v\) is first scaled. The vector \(u\) is \((1, -1, 0, 1)\) and \(2v\) is twice each element of vector \(v = (0, 2, 3, -1)\), which gives \((0, 4, 6, -2)\). Subtract each corresponding element in \(u\) and \(2v\):
  • From the first elements: \(1 - 0 = 1\)
  • From the second elements: \(-1 - 4 = -5\)
  • From the third elements: \(0 - 6 = -6\)
  • From the fourth elements: \(1 - (-2) = 3\)
Thus, the result of \(u - 2v\) becomes the new vector \((1, -5, -6, 3)\). Keep track of positive and negative signs while subtracting to ensure accuracy.
Scalar Multiplication
Scalar multiplication involves multiplying each component of a vector by a scalar (a constant number). It scales the vector but does not change its direction, unless the scalar is negative, which reverses the direction. In this problem, before subtracting the vectors, vector \(v\) is multiplied by \(2\). This means multiplying each component in vector \(v = (0, 2, 3, -1)\) by \(2\):
  • \(0 \times 2 = 0\)
  • \(2 \times 2 = 4\)
  • \(3 \times 2 = 6\)
  • \(-1 \times 2 = -2\)
After performing these multiplications, the new vector \(2v\) becomes \((0, 4, 6, -2)\). This scaled version of \(v\) is then used for the subtraction process to isolate \(3w\) in the original equation. Remember, scalar multiplication can only stretch or shrink vectors.
Linear Equations
Linear equations are foundational in mathematics; they involve variables with no exponents other than 1. Solving them consists of isolating the variable of interest, typically through operations like addition, subtraction, multiplication, or division. In this exercise, the linear equation is \(3w = u - 2v\). After you determine \(3w\) by subtracting \(2v\) from \(u\), divide by \(3\) on both sides to solve for \(w\). This process includes dividing each term in the vector \((1, -5, -6, 3)\) by \(3\):
  • \(1/3\)
  • \(-5/3\)
  • \(-6/3 = -2\)
  • \(3/3 = 1\)
Thus, you derive \(w = (1/3, -5/3, -2, 1)\). The key is maintaining balance in the equation by performing equivalent operations on both sides.

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

Is it possible for a transition matrix to equal the identity matrix? Explain.

True or False? In Exercises 49 and \(50,\) determine whether each statement is true or false. If a statement is true, give a reason or cite an appropriate statement from the text. If a statement is false, provide an example that shows the statement is not true in all cases or cite an appropriate statement from the text. A. A vector space consists of four entities: a set of vectors, a set of scalars, and two operations. B. The set of all integers with the standard operations is a vector space. C. The set of all ordered triples \((x, y, z)\) of real numbers, where \(y \geq 0,\) with the standard operations on \(R^{3}\) is a vector space.

Writing Explain why the row vectors of a \(4 \times 3\) matrix form a lincarly dependent set. (Assume all matrix entries are distinct.)

Prove that row operations do not change the dependency relationships among the columns of an \(m \times n\) matrix.

Perform a rotation of axes to eliminate the xy-term, and sketch the graph of the conic. $$ 9 x^{2}+24 x y+16 y^{2}+90 x-130 y=0 $$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation
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.