Question

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 \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).

Short answer

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

Step-by-step solution

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 Ais 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.

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

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