/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 Go through each line of MATLAB c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Go through each line of MATLAB code and show the result, or indicate if error will occur by executing the commands. \(\mathrm{B}=[1 ; 0 ;-5 ; 4]\) \(\mathrm{D}=\mathrm{B}^{\prime}\)

Short Answer

Expert verified
The output of the MATLAB code results in two vectors. B is a column vector \(B = \begin{bmatrix} 1 \ 0 \ -5 \ 4 \end{bmatrix}\), and D is a row vector obtained from the transposition of B, \(D = \begin{bmatrix} 1 & 0 & -5 & 4 \end{bmatrix}\).

Step by step solution

01

Executing line 1: B=[1 ; 0 ;-5 ; 4]

MATLAB will create a column vector B, a vertical matrix with 4 rows and 1 column. It will look like this: \[B = \begin{bmatrix} 1 \ 0 \ -5 \ 4 \end{bmatrix}\].
02

Executing line 2: D=B'

MATLAB will execute this line applying the transpose operation, which interchanges the rows and columns in B, creating a matrix D. Hence, the column vector B will be transposed as a row vector. Here, no error will occur. The result will be \[D = \begin{bmatrix} 1 & 0 & -5 & 4 \end{bmatrix}\], which is a matrix with 1 row and 4 columns.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

MATLAB Coding
MATLAB (Matrix Laboratory) is a powerful numerical computing environment and programming language. Its functionality is designed specifically for engineers and scientists to express their ideas in a form that is both easy to write and understand. Coding in MATLAB revolves around vectors and matrices, where all elements are treated as matrix elements.

When engaging with MATLAB coding, one deals with variables, functions, and operations that are applied to arrays and matrices. For example, assigning a set of values to a vector or applying mathematical functions across vector elements. The simplicity of MATLAB coding is evidenced in creating a vector with a few elements, such as B = [1; 0; -5; 4], where each value represents an element in a column vector.
Column Vector
In MATLAB, a column vector is a matrix with a single column. It's analogous to a list of items, except that each item is aligned vertically. Creating a column vector in MATLAB is straightforward; vertical concatenation of elements using semicolons ; creates the vector, as demonstrated by B = [1; 0; -5; 4].

A column vector is a key part of linear albegra and MATLAB leverages this by allowing users to perform various types of computations and manipulations. The column vector created in the exercise B is a simple data structure used to perform operations such as scaling, addition of other vectors and can even be used in more complex matrix operations.
Transpose Operation
The transpose operation in MATLAB is symbolized by the apostrophe (') and it is an essential function that converts row vectors to column vectors and vice versa. In mathematical terms, transposing a matrix involves flipping over its diagonal, turning the row index of each element into the column index and the other way around.

Executing the transpose on the vector B as D = B' changes it from a 4-by-1 column vector into a 1-by-4 row vector. This operation does not alter the magnitude of the elements within the vector but rather switches their position within the matrix structure. Understanding how to transpose vectors and matrices is fundamental for matrix manipulations, such as solving systems of linear equations or performing matrix algebra.
Matrix Manipulation
Matrix manipulation is a fundamental aspect of MATLAB, given that it's designed for operations with arrays and matrices. This includes an array of operations such as addition, subtraction, multiplication, transposition, and more complex operations such as matrix inversion, determinant calculation, and eigenvalue decomposition.

Manipulating a matrix can be as simple as transposing it or may involve more complex equations and computations. For instance, adding two matrices of the same size involves adding their corresponding elements. Meanwhile, multiplying matrices is more involved, necessitating a dot product of the rows and columns of the involved matrices, adhering to matrix multiplication rules. Uncovering such manipulations is directly applicable in various mathematical, engineering, and scientific endeavors.

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

Go through each line of MATLAB code and show the result, or indicate if error will occur by executing the commands. \(\mathrm{X}=[0: 1: 5]\) \(\mathrm{Y}=\mathrm{X}^{\wedge} 2+4 ;\) plot \((\mathrm{X}, \mathrm{Y})\) xlabel(' \(\mathrm{X}^{\prime}\) ) ylabel('\(^{\prime} \mathrm{Y}^{\prime}\) )

As we explained in earlier chapters, the air resistance to the motion of a vehicle is something important that engineers investigate. The drag force acting on a car is determined experimentally by placing the car in a wind tunnel. The air speed inside the tunnel is changed, and the drag force acting on the car is measured. For a given car, the experimental data generally is represented by a single coefficient that is called drag coefficient. It is defined by the following relationship: $$ C_{\mathrm{d}}=\frac{F_{\mathrm{d}}}{\frac{1}{2} \rho V^{2} A} $$ where $$ \begin{aligned} C_{\mathrm{d}} &=\text { drag coefficient (unitless) } \\ F_{\mathrm{d}} &=\text { measured drag force }(\mathrm{N} \text { or } \mathrm{lb}) \\ \rho &=\text { air density }\left(\mathrm{kg} / \mathrm{m}^{3} \text { or } \mathrm{slugs} / \mathrm{ft}^{3}\right) \\ V &=\text { air speed inside the wind tunnel }(\mathrm{m} / \mathrm{s} \text { or } \mathrm{ft} / \mathrm{s}) \\ A &=\text { frontal area of the car }\left(\mathrm{m}^{2} \text { or } \mathrm{ft}^{2}\right) \end{aligned} $$ The frontal area \(A\) represents the frontal projection of the car's area and could be approximated simply by multiplying \(0.85\) times the width and the height of a rectangle that outlines the front of the car. This is the area that you see when you view the car from a direction normal to the front grill. The \(0.85\) factor is used to adjust for rounded corners, open space below the bumper, and so on. To give you some idea, typical drag coefficient values for sports cars are between \(0.27\) to \(0.38\) and for sedans are between \(0.34\) to \(0.5\). The power requirement to overcome air resistance is computed by $$ P=F_{\mathrm{d}} V $$ where $$ \begin{aligned} &P=\text { power }(\mathrm{W} \text { or } \mathrm{ft} \cdot \mathrm{lb} / \mathrm{s}) \\ &1 \text { horsepower }(\mathrm{hp})=550 \mathrm{ft} \cdot \mathrm{lb} / \mathrm{s} \end{aligned} $$ and 1 horsepower \((\mathrm{hp})=746 \mathrm{~W}\) The purpose of this problem is to see how the power requirement changes with the car speed and the air temperature. Determine the power requirement to overcome the air resistance for a car that has a listed drag coefficient of \(0.4\) and width of \(74.4\) inches and height of \(57.4\) inches. Vary the air speed in the range of \(15 \mathrm{~m} / \mathrm{s}

Solve the following set of equations using MATLAB. $$ \left[\begin{array}{rrr} 1 & 1 & 1 \\ 2 & 5 & 1 \\ -3 & 1 & 5 \end{array}\right]\left\\{\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right\\}=\left\\{\begin{array}{r} 6 \\ 15 \\ 14 \end{array}\right\\} $$

Find the equation that best fits the following set of data points. Compare the actual and predicted \(y\) values. Plot the data first. $$ \begin{array}{r|rrrrrrrrr} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline y & 5 & 8 & 15 & 32 & 65 & 120 & 203 & 320 & 477 \end{array} $$

In Chapter 10 , we discussed fluid pressure and the role of water towers in small towns. Use MATLAB to create a table that shows the relationship between the height of water above ground in the water to wer and the water pressure in a pipeline located at the base of the water tower. The relationship is given by $$ P=\rho g h $$ where \(\begin{aligned} P=& \text { water pressure at the base of the water tower } \\\ & \text { in pounds per square foot }\left(\mathrm{lb} / \mathrm{ft}^{2}\right) \end{aligned}\) \(\rho=\) density of water in slugs per cubic foot \((\rho=\) \(1.94\) slugs \(/ \mathrm{ft}^{3}\) ) \(g=\) acceleration due to gravity \(\left(g=32.2 \mathrm{ft} / \mathrm{s}^{2}\right)\) \(b=\) height of water above ground in feet (ft) Create a table that shows the water pressure in \(\mathrm{lb} / \mathrm{in}^{2}\) in a pipe located at the base of the water tower as you vary the height of the water in increments of \(10 \mathrm{ft}\). Also, plot the water pressure \(\left(\mathrm{lb} / \mathrm{in}^{2}\right)\) versus the height of water in feet. What should the water level in the water tower be to create 80 psi of water pressure in a pipe at the base of the water tower?

See all solutions

Recommended explanations on Physics Textbooks

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.