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### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# In Exercise 2, compute $$u + v$$ and $$u - 2v$$.2. $$u = \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right]$$, $$v = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]$$.

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

See the step by step solution

## Step 1: 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 $$u + v$$.

## Step 2: Compute the sum of the vectors

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

\begin{aligned}{c}u + v &= \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{3 + \left( 2 \right)}\\{2 + \left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{3 + 2}\\{2 - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}5\\1\end{array}} \right]\end{aligned}

Thus, the vector is $$u + v = \left[ {\begin{array}{*{20}{c}}5\\1\end{array}} \right]$$.

## Step 3: 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.

## Step 4: Compute the vector

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

Solve further to get:

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

Thus, $$u - 2v = \left[ {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right]$$.

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