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

Expert verified

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

01

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.

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

02

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

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Most popular questions from this chapter

In Exercise 9 mark each statement True or False. Justify each answer.

9. a. In order for a matrix B to be the inverse of A, both equations \(AB = I\) and \(BA = I\) must be true.

b. If A and B are \(n \times n\) and invertible, then \({A^{ - {\bf{1}}}}{B^{ - {\bf{1}}}}\) is the inverse of \(AB\).

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ab - cd \ne {\bf{0}}\), then A is invertible.

d. If A is an invertible \(n \times n\) matrix, then the equation \(Ax = b\) is consistent for each b in \({\mathbb{R}^{\bf{n}}}\).

e. Each elementary matrix is invertible.

Let Abe an invertible \(n \times n\) matrix, and let B be an \(n \times p\) matrix. Show that the equation \(AX = B\) has a unique solution \({A^{ - 1}}B\).

Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?

[M] Suppose memory or size restrictions prevent your matrix program from working with matrices having more than 32 rows and 32 columns, and suppose some project involves \(50 \times 50\) matrices A and B. Describe the commands or operations of your program that accomplish the following tasks.

a. Compute \(A + B\)

b. Compute \(AB\)

c. Solve \(Ax = b\) for some vector b in \({\mathbb{R}^{50}}\), assuming that \(A\) can be partitioned into a \(2 \times 2\) block matrix \(\left[ {{A_{ij}}} \right]\), with \({A_{11}}\) an invertible \(20 \times 20\) matrix, \({A_{22}}\) an invertible \(30 \times 30\) matrix, and \({A_{12}}\) a zero matrix. [Hint: Describe appropriate smaller systems to solve, without using any matrix inverse.]

Prove the Theorem 3(d) i.e., \({\left( {AB} \right)^T} = {B^T}{A^T}\).

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