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Chapter 1: Q1E (page 1)

In Exercise 1, compute \(u + v\) and \(u - 2v\).

  1. \(u = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\), \(v = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\).

Short Answer

Expert verified

The vectors are \(u + v = \left[ {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right]\), and \(u - 2v = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]\).

Step by step solution

01

Write the condition to add the two vectors

From the given vectors, it is observed that both vectors contain two entries. So, the vectors can be denoted as \({\mathbb{R}^2}\).

Add the corresponding terms of \(u\)and \(v\)to compute \(u + v\).

02

Compute the sum of the vectors

Obtain vector \(u + v\)by using vectors \(u = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\),and \(v = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\).

\(\begin{aligned}{c}u + v &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 + \left( { - 3} \right)}\\{2 + \left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 - 3}\\{2 - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right]\end{aligned}\)

Thus, the vector is \(u + v = \left[ {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right]\).

03

Write the condition for scalar multiple of a vector by a constant

Multiply each entry of a \({\mathbb{R}^2}\) vector by a scalar number to obtain the scalar multiple of a vector by a constant.

04

Compute the vector

The vector \(u - 2v\)can be written as\(u + \left( { - 2} \right)v\).

Obtain the scalar multiple of vector \(v\)by scalar \(\left( { - 2} \right)\), and then add the resultant vector with vector \(u\).

\(\begin{aligned}{c}u + \left( { - 2} \right)v &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left( { - 2} \right)\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{\left( { - 2} \right)\left( { - 3} \right)}\\{\left( { - 2} \right)\left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 + 6}\\{2 + 2}\end{array}} \right]\end{aligned}\)

Solve further to get:

\(u + \left( { - 2} \right)v = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]\)

Thus, \(u - 2v = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]\).

Therefore, \(u + v = \left[ {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right]\), and \(u - 2v = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]\).

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