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Found in: Page 93

### 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 Exercises 27 and 28, view vectors in $${\mathbb{R}^n}$$ as $$n \times 1$$ matrices. For $${\mathop{\rm u}\nolimits}$$ and $${\mathop{\rm v}\nolimits}$$ in $${\mathbb{R}^n}$$, the matrix product $${{\mathop{\rm u}\nolimits} ^T}v$$ is a $$1 \times 1$$ matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product $${{\mathop{\rm uv}\nolimits} ^T}$$ is an $$n \times n$$ matrix, called the outer product of u and v. The products $${{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits}$$ and $${{\mathop{\rm uv}\nolimits} ^T}$$ will appear later in the text.27. Let {\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right) and {\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right). Compute $${{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits}$$, $${{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits}$$,$${{\mathop{\rm uv}\nolimits} ^T}$$, and $${\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}$$.

$${{\mathop{\rm u}\nolimits} ^T}v = {{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} = - 2a + 3b - 4c$$, {\mathop{\rm u}\nolimits} {{\mathop{\rm v}\nolimits} ^T} = \left( {\begin{aligned}{*{20}{c}}{ - 2a}&{ - 2b}&{ - 2c}\\{3a}&{3b}&{3c}\\{ - 4a}&{ - 4b}&{ - 4c}\end{aligned}} \right), and v{{\mathop{\rm u}\nolimits} ^T} = \left( {\begin{aligned}{*{20}{c}}{ - 2a}&{3a}&{ - 4a}\\{ - 2b}&{3b}&{ - 4b}\\{ - 2c}&{3c}&{ - 4c}\end{aligned}} \right).

See the step by step solution

## Step 1: Determine the product $${{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits}$$ and $${{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits}$$

It is known that the transpose of A is denoted by $${A^T}$$.

The matrix product $${{\mathop{\rm u}\nolimits} ^T}v$$ is a $$1 \times 1$$ matrix, commonly represented by a real number and written without the matrix brackets.

\begin{aligned}{c}{{\mathop{\rm u}\nolimits} ^T}v = \left( {\begin{aligned}{*{20}{c}}{ - 2}&3&{ - 4}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\\ = - 2a + 3b - 4c\\{{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a&b&c\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\\ = - 2a + 3b - 4c\end{aligned}

## Step 2: Determine the product $${\mathop{\rm u}\nolimits} {{\mathop{\rm v}\nolimits} ^T}$$ and $${\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}$$

\begin{aligned}{c}{{\mathop{\rm uv}\nolimits} ^T} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}a&b&c\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 2a}&{ - 2b}&{ - 2c}\\{3a}&{3b}&{3c}\\{ - 4a}&{ - 4b}&{ - 4c}\end{aligned}} \right)\\v{{\mathop{\rm u}\nolimits} ^T} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 2}&3&{ - 4}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 2a}&{3a}&{ - 4a}\\{ - 2b}&{3b}&{ - 4b}\\{ - 2c}&{3c}&{ - 4c}\end{aligned}} \right)\end{aligned}

Thus, the products is $${{\mathop{\rm u}\nolimits} ^T}v = {{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} = - 2a + 3b - 4c$$, {\mathop{\rm u}\nolimits} {{\mathop{\rm v}\nolimits} ^T} = \left( {\begin{aligned}{*{20}{c}}{ - 2a}&{ - 2b}&{ - 2c}\\{3a}&{3b}&{3c}\\{ - 4a}&{ - 4b}&{ - 4c}\end{aligned}} \right), and v{{\mathop{\rm u}\nolimits} ^T} = \left( {\begin{aligned}{*{20}{c}}{ - 2a}&{3a}&{ - 4a}\\{ - 2b}&{3b}&{ - 4b}\\{ - 2c}&{3c}&{ - 4c}\end{aligned}} \right).