Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 7: Problem 3

If \(\mathbf{A}=\left(\begin{array}{rrr}{-2} & {1} & {2} \\ {1} & {0} & {-3} \\\ {2} & {-1} & {1}\end{array}\right)\) and \(\mathbf{B}=\left(\begin{array}{rrr}{1} & {2} & {3} \\ {3} & {-1} & {-1} \\\ {-2} & {1} & {0}\end{array}\right),\) find (a) \(\mathbf{A}^{T}\) (b) \(\mathbf{B}^{T}\) (c) \(\mathbf{A}^{T}+\mathbf{B}^{T}\) (d) \((\mathbf{A}+\mathbf{B})^{T}\)

Short Answer

Expert verified
Question: Verify whether the transpose of the sum of two matrices A and B is equal to the sum of their transposes, that is, show that (A+B)^T = A^T + B^T. Solution: Based on the calculations above, (A+B)^T = \( \left(\begin{array}{rrr} -1 & 4 & 0 \\ 3 & -1 & 0 \\ 5 & -4 & 1 \end{array}\right) \) A^T + B^T = \( \left(\begin{array}{rrr} -1 & 4 & 0 \\ 3 & -1 & -2 \\ 5 & -2 & 1 \end{array}\right) \) Since (A+B)^T = A^T + B^T, we can conclude that this property of transposes holds true for matrices A and B.

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards
Start your free trial Skip

Over 22 million students worldwide already upgrade their learning with Vaia!

01

(a) Finding the transpose of matrix A)

To find the transpose of matrix A, just swap its rows with its corresponding columns. The transpose of matrix A, denoted by A^T, can be calculated as follows: $\mathbf{A}^{T} = \left(\begin{array}{rrr}{-2} & {1} & {2} \\\ {1} & {0} & {-3} \\\ {2} & {-1} & {1}\end{array}\right)^T = \left(\begin{array}{rrr}{-2} & {1} & {2} \\\ {1} & {0} & {-1} \\\ {2} & {-3} & {1}\end{array}\right)$
02

(b) Finding the transpose of matrix B)

Similar to the first step, we need to swap the rows with their corresponding columns of matrix B. The transpose of B, denoted by B^T, can be calculated as follows: \(\mathbf{B}^{T} = \left(\begin{array}{rrr}{1} & {2} & {3} \\\ {3} & {-1} & {-1} \\\ {-2} & {1} & {0}\end{array}\right)^T = \left(\begin{array}{rrr}{1} & {3} & {-2} \\\ {2} & {-1} & {1} \\\ {3} & {-1} & {0}\end{array}\right)\)
03

(c) Finding the sum of the transposed matrices A and B)

Add the corresponding elements of matrices A^T and B^T to obtain the sum: \(\mathbf{A}^{T}+\mathbf{B}^{T} = \left(\begin{array}{rrr}{-2+1} & {1+3} & {2+(-2)} \\\ {1+2} & {0+(-1)} & {(-3)+1} \\\ {2+3} & {(-1)+(-1)} & {1+0}\end{array}\right) = \left(\begin{array}{rrr}{-1} & {4} & {0} \\\ {3} & {-1} & {-2} \\\ {5} & {-2} & {1}\end{array}\right)\)
04

(d) Finding the transpose of the sum of matrices A and B)

First, find the sum of matrices A and B: \(\mathbf{A}+\mathbf{B} = \left(\begin{array}{rrr}{(-2)+1} & {1+2} & {2+3} \\\ {1+3} & {0+(-1)} & {(-3)+(-1)} \\\ {2+(-2)} & {(-1)+1} & {1+0}\end{array}\right) = \left(\begin{array}{rrr}{-1} & {3} & {5} \\\ {4} & {-1} & {-4} \\\ {0} & {0} & {1}\end{array}\right)\) Then, find the transpose of the sum: \((\mathbf{A}+\mathbf{B})^{T} = \left(\begin{array}{rrr}{-1} & {3} & {5} \\\ {4} & {-1} & {-4} \\\ {0} & {0} & {1}\end{array}\right)^T = \left(\begin{array}{rrr}{-1} & {4} & {0} \\\ {3} & {-1} & {0} \\\ {5} & {-4} & {1}\end{array}\right)\) We can now compare (c) and (d): \(\mathbf{A}^{T}+\mathbf{B}^{T} = \left(\begin{array}{rrr}{-1} & {4} & {0} \\\ {3} & {-1} & {-2} \\\ {5} & {-2} & {1}\end{array}\right)\) \((\mathbf{A}+\mathbf{B})^{T} = \left(\begin{array}{rrr}{-1} & {4} & {0} \\\ {3} & {-1} & {0} \\\ {5} & {-4} & {1}\end{array}\right)\) The two sums are equal, which showcases an important property of transposes: the transpose of the sum of two matrices is equal to the sum of their transposes.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transpose of a Matrix
Understanding matrix transpose is fundamental in linear algebra. When you transpose a matrix, you're essentially switching its rows with its columns. In simpler terms, the element at the ith row and jth column in the original matrix becomes the element at the jth row and ith column in the transposed matrix. The process can be visualized by flipping the matrix over its diagonal.

For instance, if you have a matrix \( \mathbf{A} \), its transpose, denoted as \( \mathbf{A}^{T} \), is performed by turning the first row of \( \mathbf{A} \) into the first column of \( \mathbf{A}^{T} \), the second row into the second column, and so on. To ensure clarity when practicing, you might want to work through the transformation step by step, rewriting each row as a column until the entire matrix has been transformed.
Matrix Addition
Moving onto another core concept, matrix addition, which is a straightforward operation if you remember one key rule: you can only add matrices of the same size. This is because the addition is done element by element; adding the entries in the same position together to produce a new matrix.

For example, if you want to add two matrices \( \mathbf{A} \) and \( \mathbf{B} \), you simply take each element \( a_{ij} \) from \( \mathbf{A} \) and add it to the corresponding element \( b_{ij} \) from \( \mathbf{B} \) to get the element in the resulting matrix, \( c_{ij} = a_{ij} + b_{ij} \). To prevent any confusion, double-check that the matrices are indeed the same size before attempting to add them. Being methodical in this way can help avoid errors and misunderstandings.
Properties of Matrix Transpose
The properties of matrix transpose are quite intriguing and are often tested in exercises. One fundamental property is that the transpose of a transpose brings you back to the original matrix (\( (\mathbf{A}^{T})^{T} = \mathbf{A} \) ). Moreover, we have an interesting relationship involving transpose and addition: the transpose of a sum of two matrices is equal to the sum of their transposes (\( (\mathbf{A}+\mathbf{B})^{T} = \mathbf{A}^{T} + \mathbf{B}^{T} \) ).

Understanding these properties isn't just about memorizing them; it's about recognizing patterns in matrix operations that can simplify complex problems. As you work through more examples, you'll start to see these properties manifest naturally, and they will become intuitive tools in your mathematical toolkit.
  • Always consider the dimensions of matrices when applying transpose operations.
  • Remember that individual elements are not affected, but their positions change.
  • Learning by doing is key—try transposing matrices and adding them to strengthen your grasp of these concepts.
Provided exercises typically encourage step-by-step solutions to ensure the student fully comprehends how the properties are applied in practice.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Find the eigenvalues of the given system. (b) Choose an initial point (other than the origin) and draw the corresponding trajectory in the \(x_{1} x_{2}\) plane. (c) For your trajectory in part (b) draw the graphs of \(x_{1}\) versus \(t\) and of \(x_{2}\) versus \(t .\) (d) For your trajectory in part (b) draw the corresponding graph in three- dimensional \(t x_{1} x_{2}\) space. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{-\frac{4}{5}} & {2} \\ {-1} & {\frac{6}{5}}\end{array}\right) \mathbf{x} $$

Show that if \(\mathbf{A}\) is a diagonal matrix with diagonal elements \(a_{1}, a_{2}, \ldots, a_{n},\) then \(\exp (\mathbf{A} t)\) is also a diagonal matrix with diagonal elements \(\exp \left(a_{1} t\right), \exp \left(a_{2} t\right), \ldots, \exp \left(a_{n} t\right)\)

Consider again the cliectric circuit in Problem 26 of Scction 7.6 . This circut is described by the system of differential equations $$ \frac{d}{d t}\left(\begin{array}{l}{I} \\\ {V}\end{array}\right)=\left(\begin{array}{cc}{0} & {\frac{1}{L}} \\\ {-\frac{1}{C}} & {-\frac{1}{R C}}\end{array}\right)\left(\begin{array}{l}{I} \\\ {V}\end{array}\right) $$ (a) Show that the eigendlucs are raal and equal if \(L=4 R^{2} C\). (b) Suppose that \(R=1\) ohm, \(C=1\) farad, and \(L=4\) henrys. Suppose also that \(I(0)=1\) ampere and \(V(0)=2\) volts. Find \(I(t)\) and \(V(t) .\)

Consider a \(2 \times 2\) system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\). If we assume that \(r_{1} \neq r_{2}\), the general solution is \(\mathbf{x}=c_{1} \xi^{(1)} e^{t_{1}^{\prime}}+c_{2} \xi^{(2)} e^{\prime 2},\) provided that \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent In this problem we establish the linear independence of \(\xi^{(1)}\) and \(\xi^{(2)}\) by assuming that they are linearly dependent, and then showing that this leads to a contradiction. $$ \begin{array}{l}{\text { (a) Note that } \xi \text { (i) satisfies the matrix equation }\left(\mathbf{A}-r_{1} \mathbf{I}\right) \xi^{(1)}=\mathbf{0} ; \text { similarly, note that }} \\ {\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(2)}=\mathbf{0}} \\ {\text { (b) Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(1)}=\left(r_{1}-r_{2}\right) \mathbf{\xi}^{(1)}} \\\ {\text { (c) Suppose that } \xi^{(1)} \text { and } \xi^{(2)} \text { are linearly dependent. Then } c_{1} \xi^{(1)}+c_{2} \xi^{(2)}=\mathbf{0} \text { and at least }}\end{array} $$ $$ \begin{array}{l}{\text { one of } c_{1} \text { and } c_{2} \text { is not zero; suppose that } c_{1} \neq 0 . \text { Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=\mathbf{0}} \\ {\text { and also show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=c_{1}\left(r_{1}-r_{2}\right) \boldsymbol{\xi}^{(1)} \text { . Hence } c_{1}=0, \text { which is }} \\\ {\text { a contradiction. Therefore } \xi^{(1)} \text { and } \boldsymbol{\xi}^{(2)} \text { are linearly independent. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Modify the argument of part (c) in case } c_{1} \text { is zero but } c_{2} \text { is not. }} \\ {\text { (e) Carry out a similar argument for the case in which the order } n \text { is equal to } 3 \text { ; note that }} \\ {\text { the procedure can be extended to cover an arbitrary value of } n .}\end{array} $$

In each of Problems 24 through 27 the eigenvalues and eigenvectors of a matrix \(\mathrm{A}\) are given. Consider the corresponding system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\). $$ \begin{array}{l}{\text { (a) Sketch a phase portrait of the system. }} \\\ {\text { (b) Sketch the trajectory passing through the initial point }(2,3) \text { . }} \\ {\text { (c) For the trajectory in part (b) sketch the graphs of } x_{1} \text { versus } t \text { and of } x_{2} \text { versus } t \text { on the }} \\ {\text { same set of axes. }}\end{array} $$ $$ r_{1}=1, \quad \xi^{(1)}=\left(\begin{array}{l}{1} \\ {2}\end{array}\right) ; \quad r_{2}=2, \quad \xi^{(2)}=\left(\begin{array}{r}{1} \\\ {-2}\end{array}\right) $$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free