Linear algebra with applications jeffrey holt pdf download

Linear algebra with applications jeffrey holt pdf download

Linear algebra with applications 2nd edition holt solutions manual,More from lindan2018

Linear algebra has applications in science, computer science, social science, business, and other fields. Engineering is filled with visible applications of linear algebra, with examples all around us of infrastructure that affects our everyday lives Web15/08/ · Download Linear Algebra Solution Manual Jeffrey Holt File Type Pdf Download Linear Algebra Solution Manual Jeffrey Holt File Type free pdf, Download Linear Linear WebHolt's Linear Algebra with Applications, Second Edition, blends computational and conceptual topics throughout to prepare students for the rigors of conceptual thinking in Web[REQUEST] Linear Algebra with Applications by Jeffrey Holt (2nd Edition) ISBN: 3 3 3 comments Best Add a Comment [deleted] • 3 yr. ago Found it: Web5/12/ · Mathematics Books Collection of Refere ... read more

a If a general solution has free parameters, then there must be infinitely many solutions. Assume both equations have variables. A total of people attend the premiere of a new movie. A plane holds passengers. How many of each type of ticket were sold? Referring to Example 4, suppose that the minimum outside temperature is 10°F. In this case, how much of each type of solution is required? How much should be invested in each type of bond? A gallon bathtub is to be filled with water that is exactly ° F. The hot water supply is ° F and the cold water supply is 60° F. When mixed, the temperature will be a weighted average based on the amount of each water source in the mix. How much of each should be used to fill the tub as specified? The hot water supply is ° F and the cold water supply is 70° F.

Pure water freezes at 32° F and 0° C, and boils at ° F and ° C. Use this information to find a and b. For tax and accounting purposes, corporations depreciate the value of equipment each year. This problem requires 8 nickels, 8 quarters, and a sheet of 8. The goal is to estimate the diameter of each type of coin as follows: Using trial and error, find a combination of nickels and quarters that, when placed side by side, extend the height long side of the paper. Then do the same along the width short side of the paper.

Use the information obtained to write two linear equations involving the unknown diameters of each type of coin, then solve the resulting system to find the diameter for each type of coin. Bixby Creek Bridge. Suppose that a bag of concrete is projected downward from the bridge deck at an initial rate of 5 meters per second. After 3 seconds, the bag is Use the model in Example 6 to find a formula for H t , the height at time t. Exercises 71— Use computational assistance to find the set of solutions to the linear system. In this section, we develop a method for converting any linear system into a system in echelon form, so that we can apply back substitution.

To get us started, consider the following projectile motion problem. Suppose that a cannon sits on a hill and fires a ball across a flat field below. Figure 1 shows the elevation of the ball at three separate places. This system is not in echelon form, so back substitution is not easy to use here. We will return to this system shortly. The primary goal of this section is to develop a systematic procedure for transforming any linear system into a system that is in echelon form. The key feature of our transformation procedure is that it produces a new linear system that is in echelon form hence solvable using back substitution and has exactly the same set of solutions as the original system. Two linear systems are said to be equivalent if they have the same set of solutions. Elementary Operations Elementary Operations We can transform a linear system using a sequence of elementary operations.

Each operation produces a new system that is equivalent to the old one, so the solution set is unchanged. There are three types of elementary operations. Interchange the position of two equations. This amounts to nothing more than rewriting the system of equations. For example, we exchange the places of the first and second equations in the following system. Multiply an equation by a nonzero constant. Add a multiple of one equation to another. For this operation, we multiply one of the equations by a constant and then add it to another equation, replacing the latter with the result. It is similar to the method used in the first three examples of Section 1. Note that this is exactly what happened here, with the lower left coefficient becoming zero, transforming the system closer to echelon form. This illustrates a single step of our basic strategy for transforming any linear system into a system that is in echelon form. Our goal is to transform the system to echelon form, so we want to eliminate the x1 terms in the second and third equations.

This will leave x1, as the leading variable in only the top equation. Going forward we identify coefficients using the notation for a generic system of equations introduced in Section 1. We need to transform a21 and a31 to 0. We do this in two parts. Next, we focus on the x2 coefficients. Since our goal is to reach echelon form, we do not care about the coefficient on x2 in the top equation, so we concentrate on the second and third equations. Here we need to transform a32 to 0. To check our solution, we plug these values into the original system. Thanks to the previous step, we need only add the second equation to the third to transform a32 to 0. Figure 2 shows a graph of the model together with the known points.

Figure 2 Cannonball data and the graph of the model. The Augmented Matrix Matrix Augmented Matrix When manipulating systems of equations, the coefficients change but the variables do not. We can simplify notation by transferring the coefficients to a matrix, which for the moment we can think of as a rectangular table of numbers. When a matrix contains all the coefficients of a linear system, including the constant terms on the right side of each equation, it is called an augmented matrix. Elementary Row Operations The three elementary operations that we performed on equations can be translated into equivalent elementary row operations for matrices. Interchange two rows. Multiply a row by a nonzero constant.

Replace a row with the sum of that row and the scalar multiple of another row. Equivalent Matrices Borrowing from the terminology for systems of equations, we say that two matrices are equivalent if one can be obtained from the other through a sequence of elementary row operations. Hence equivalent augmented matrices correspond to equivalent linear systems. Zero Row, Zero Column When discussing matrices, the rows are numbered from top to bottom, and the columns are numbered from left to right. A zero row is a row consisting entirely of zeros, and a nonzero row contains at least one nonzero entry.

The terms zero column and nonzero column are similarly defined. In the examples that follow, we transfer the system of equations to an augmented matrix, but our goal is the same as before, to find an equivalent system in echelon form. We focus on the first column of the matrix, which contains the coefficients of x1. Although this step is not required, exchanging Row 1 and Row 2 will move a 1 into the upper left position and avoid the early introduction of fractions. To transform the system to echelon form, we need to introduce zeros in the first column below Row 1. This requires two operations. With the first column complete, we move down to the second row and to the right to the second column.

We now extract the transformed system of equations from the matrix. We can substitute into the original system to verify our solution. The resulting matrix is said to be in echelon form or row echelon form and will have the properties given in Definition 1. In the definition, the leading term of a row is the leftmost nonzero term in that row, and a row of all zeros has no leading term. b Any zero rows are at the bottom of the matrix. Echelon form is the counterpart to echelon systems from Section 1. Note that the first condition in the definition implies that a matrix in echelon form will have zeros filling out the column below each of the leading terms. The entries in the pivot positions for the matrices in 2 are shown in boldface.

The pivot columns are the columns that contain pivot positions, and a pivot is a nonzero number in a pivot position. In what follows, it will be handy to have a general augmented matrix when referring to entries in specific positions. We adopt a notation similar to that for a general system of equations given in 7 of Section 1. It is named in honor of German mathematician Carl Friedrich Gauss, who independently discovered the method and introduced it to the West in the nineteenth century. Next, we need zeroes down the first column below the pivot position. This makes a good pivot, so we move to the elimination step. Down the remainder of the second column, we need only introduce a zero at a32 by using the operation shown in the margin. However, there are nonzero terms down the first column, so we interchange Row 1 and Row 3 to place a 1 in the pivot position.

Next, we need zeros down the first column below the pivot position. This will undo the zeros in the first column. Since all the entries below a22 are also zero, interchanging with lower rows will not put a nonzero term in the a22 position. Thus a22 cannot be a pivot position, so we move to the right to the third column to determine if a23 is a suitable pivot position. From the Row 2 pivot position, we move down one row and to the right one column to a This entry is 0, as is the entry below, so interchanging rows will not yield an acceptable pivot. As we did before, we move one column to the right. We introduce a zero in the a45 position by using the operation shown in the margin.

Since Row 4 is the only remaining row and consists entirely of zeros, it has no pivot position. The matrix is now in echelon form, so no additional row operations are required. Gaussian elimination can be applied to any matrix to find an equivalent matrix that is in echelon form. If matrix A is equivalent to matrix B that is in echelon form, we say that B is an echelon form of A. Different sequences of row operations can produce different echelon forms of the same starting matrix, but all echelon forms of a given matrix will have the same pivot positions. We introduce zeros down the first column with the row operations shown in the margin. Thus this system has no solutions, and so is inconsistent. The preceding example illustrates a general principle. When applying row operations to an augmented matrix, if at any point in the process the matrix has a row of the form […0 c] 3 where c is nonzero, then stop.

The system is inconsistent. Gauss, and Wilhelm Jordan — , a German engineer who popularized this method for finding solutions to linear systems in his book on geodesy the science of measuring earth shapes. After extracting the linear system from this matrix, we back substituted and simplified to find the general solution. We can make it easier to find the general solution by performing additional row operations on the matrix. Specifically, we do the following: 1. Multiply each nonzero row by the reciprocal of the pivot so that we end up with a 1 as the leading term in each nonzero row. Use row operations to introduce zeros in the entries above each pivot position. Picking up with our matrix, we see that the first and third rows already have a 1 in the pivot position. When implementing Gaussian elimination, we worked from left to right. To put zeros above pivot positions, we work from right to left, starting with the rightmost pivot, which in this case appears in the fifth column.

Two row operations are required to introduce zeros above this pivot. One row operation is required to introduce a zero in the a13 position. Changing the order can result in a circular sequence of operations that lead to endless misery. Note that when the system is expressed in this form, the leading variables appear only in the equation that they lead. Thus during back substitution we need only plug in free parameters and then subtract to solve for the leading variables, simplifying the process considerably. The matrix on the right in 4 is said to be in reduced echelon form. b All pivot positions contain a 1. c The only nonzero term in a pivot column is in the pivot position.

The combination of the forward and backward phases is referred to as Gauss—Jordan elimination. Although a given matrix can be equivalent to many different echelon form matrices, the same is not true of reduced echelon form matrices. The proof of Theorem 1. Next, we implement the backward phase to transform the matrix to reduced echelon form. If there are additional solutions, they are called nontrivial solutions. We determine if there are nontrivial solutions in the usual way, using elimination methods. To find the other solutions, we load the system into an augmented matrix and transform to reduced echelon form. Proof of Theorem 1. Recall the statement of the theorem. Proof We can take any linear system, form the augmented matrix, use Gaussian elimination to reduce to echelon form, and extract the transformed system. In this case, the system has no solutions. If a does not occur, then one of b or c must: b The transformed system has no free variables and hence exactly one solution.

c The transformed system has one or more free variables and hence infinitely many solutions. Homogeneous linear systems are even simpler. Since all such systems have the trivial solution, a cannot happen. Therefore a homogeneous linear system has either a unique solution or infinitely many solutions. Which is better? For a system of n equations with n unknowns, Gaussian elimination requires approximately 23n3 flops i. Back substitution is slightly more complicated for Gaussian elimination than for Gauss—Jordan, but overall Gaussian elimination is more efficient and is the method that is usually implemented in computer software. When performing row operations by hand, partial pivoting tends to introduce fractions and leads to messy calculations, so we avoided the topic.

However, it is discussed Section 1. Counting flops gives a measure of algorithm efficiency. Transform the following matrices to echelon form. Transform the following matrices to reduced row echelon form. Convert the system to an augmented matrix, transform to echelon form, then back substitute to find the solutions to the system. Convert the system to an augmented matrix, transform to reduced row echelon form, then back substitute to find the solutions to the system. Determine if each statement is true or false, and justify your answer. a A matrix that has more columns than rows cannot be transformed to reduced row echelon form. b A matrix that has more rows than columns and is in echelon form must have a row of zeros. c Every elementary row operation is reversible. d A linear system with free variables cannot have one solution. Suppose that a matrix with four rows and nine columns is in echelon form.

a If the matrix has no row of all zeros, then how many pivot positions are there? b What is the minimum number of zeros in the bottom row? c What is the minimum number of zeros in the matrix? d If this is the augmented matrix for a linear system, then what is the minimum number of free variables? Exercises 1—4: Convert the augmented matrix to the equivalent linear system. Identify the row operation. Unfortunately, an error was made when performing the row operation. Identify the operation and fix the error. Which row operation will transform the matrix back to its original form? A matrix with three rows and five columns that is in echelon form, but not in reduced echelon form. A matrix with six rows and four columns that is in echelon form, but not in reduced echelon form.

An augmented matrix for an inconsistent linear system that has four equations and three variables. An augmented matrix for an inconsistent linear system that has three equations and four variables. A homogeneous linear system with three equations, four variables, and infinitely many solutions. Two matrices that are distinct yet equivalent. TRUE OR FALSE Exercises 43— Determine if the statement is true or false, and justify your answer. a If two matrices are equivalent, then one can be transformed into the other with a sequence of elementary row operations. b Different sequences of row operations can lead to different echelon forms for the same matrix. a Different sequences of row operations can lead to different reduced echelon forms for the same matrix. b If a linear system has four equations and seven variables, then it must have infinitely many solutions.

a If a linear system has seven equations and four variables, then it must be inconsistent. b Every linear system with free variables has infinitely many solutions. a Any linear system with more variables than equations cannot have a unique solution. b If a linear system has the same number of equations and variables, then it must have a unique solution. Exercises 47— The row operation shown is not an elementary row operation. If the given operation is the combination of elementary row operations, then provide them. Suppose that the echelon form of an augmented matrix has a pivot position in every column except the rightmost one. How many solutions does the associated linear system have? Justify your answer. Suppose that the echelon form of an augmented matrix has a pivot position in every column. Show that if a linear system has two different solutions, then it must have infinitely many solutions.

Show that if a matrix has more rows than columns and is in echelon form, then it must have at least one row of zeros at the bottom. Show that a homogeneous linear system with more variables than equations must have an infinite number of solutions. Show that each of the elementary operations on linear systems see page 16 produces an equivalent linear system. Recall two linear systems are equivalent if they have the same solution set. a Interchange the position of two equations. b Multiply an equation by a nonzero constant. c Add a multiple of one equation to another. Exercises 57— Find the interpolating polynomial f x , which is used to fit a function to a set of data. Figure 3 shows the plot of the points 1, 4 , 2, 7 , and 3, Figure 3 Exercise 57 data. Figure 4 shows the plot of the points 1, 8 , 2, 3 , 3, 9 , 5, 1 , and 7, 7. Figure 4 Exercise 58 data. Exercises 59— Refer to the cannonball scenario described at the start of the section.

For each problem, the three ordered pairs are x, E x , where x is the distance on the ground from the position of the cannon and E x is the elevation of the ball. Find a model for the elevation of the ball, and use the model to determine where it hits the ground. These are but a few of the many different possible applications that exist. This section is optional. Some applications presented here are referred to later, but can be reviewed as needed. Traffic Flow Example 1 Located on the northern coast of California is Arcata, a charming college town with a popular central plaza. Figure 1 shows a map of the streets surrounding and adjacent to the plaza.

As indicated by the arrows, all streets in the vicinity of the plaza are one-way. Traffic flows north and south on G and H streets, respectively, and east and west on 8th and 9th streets, respectively. The number of cars flowing on and off the plaza during a typical minute period on a Saturday morning is also shown. Find x1, x2, x3, and x4, the volume of traffic along each side of the plaza. Figure 1 Traffic volumes around the Arcata plaza. Solution The four intersections are labeled A, B, C, and D. At each intersection, the number of cars entering the intersection must equal the number leaving. There can be an arbitrary number of cars simply circling the plaza, perhaps looking for a parking space. The analysis performed above can be carried over to much more complex traffic questions, or to other similar settings, such as computer networks. Equilibrium Temperatures Example 2 Figure 2 gives a diagram of a piece of heavy wire mesh.

Each of the eight wire ends has temperature held fixed as shown. When the temperature of the mesh reaches equilibrium, the temperature at each connecting point will be the average of the temperatures of the adjacent points and fixed ends. Determine the equilibrium temperature at the connecting points x1, x2, x3, and x4. Figure 2 Grid Temperatures for Example 2. Solution The temperature of each connecting point depends in part on the temperature of other connecting points. Economic Inputs and Outputs Example 3 Imagine a simple economy that consists of consumers and just three industries, which we refer to as A, B and C. These industries have annual consumer sales of 60, 75, and 40 in billions of dollars , respectively. In addition, for every dollar of goods A sells, A requires 10 cents of goods from B and 15 cents of goods from C to support production. For instance, maybe B sells electricity and C sells shipping services.

Similarly, each dollar of goods B sells requires 20 cents of goods from A and 5 cents of goods from C, and each dollar of goods C sells requires 25 cents of goods from A and 15 cents of goods from B. What output from each industry will satisfy both consumer and between-industry demand? Example 3 is based on the work of Nobel prize winning economist Wassily Leontief — He divided the economy into sectors in developing his input-output model. Solution Let a, b, and c denote the total output from each of A, B, and C, respectively. The entire output for A is 60 for consumers, 0. Some values are rounded. Planetary Orbital Periods Example 4 Planets that are closer to the sun take less time than those farther out to make one orbit around the sun. Table 1 gives the average distance from the sun and the number of Earth days required to make one orbit for each planet.

Develop an equation that describes the relationship between the distance from the sun and the length of the orbital period. Table 1 Planetary Orbital Distances and Periods Distance from Sun × km Planet Mercury Orbital Period days There seems to be a pattern to the data. Here we proceed by substituting data to create a system of equations to solve. However, before doing that we note 3 is not linear in a and b, but is if we apply the logarithm function to both sides. Orbit Period. Table 2 Planetary Orbital Distances and Periods Planet Distance Actual Period Predicted Period The predictions are fairly good, suggesting that our formula is on the right track. However, the predictions become less accurate for those planets farther from the sun. Because we used the data for Mercury and Venus to develop our formula, perhaps this is not surprising. Table 3 Predicted Orbital Periods Planet Mercury Predicted Period Unfortunately, if we use more than two planets, we end up with a system that has no solutions.

Try it for yourself. Thus there are limitations to what we can do with the tools we currently have available. In Chapter 8 we develop a more sophisticated method that allows us to use all of our data simultaneously to come up with a formula that provides a good estimate for a range of distances from the sun. Balancing Chemical Equations Example 5 A popular chemical among college students is caffeine, which has chemical composition C8H10N4O2. When heated and combined with oxygen O2 , the ensuing reaction produces carbon dioxide CO2 , water H2O , and nitrogen dioxide NO2. No subscript indicates one atom. Balancing the equation involves finding values for x1, x2, x3, x4, and x5 so that the number of atoms of each element is the same before and after the reaction.

Most chemistry texts describe a method of solution that might be described as trial and error. However, there is no need for a haphazard approach. Use linear algebra to balance the equation. Figure 4 The caffeine molecule. On the left side there are 8x1 carbon atoms, while on the right side there are x3 carbon atoms. Any choice of s1 yields constants that balance our chemical equation, but it is customary to select the solution that makes each of the coefficients x1, x2, x3, x4, and x5 integers that have no common factors.

Find the equilibrium temperatures at x1, x2, and x3 for the heavy wires with endpoints held at the given temperatures. Economy has two industries A and B. These industries have annual consumer sales of 50 and 80 in billions of dollars , respectively. For every dollar of goods A sells, A requires 35 cents of goods from B. For each dollar of goods B sells, B requires 20 cents of goods from A. Determine the total output required from each industry in order to meet both consumer and between-industry demand. The equation below describes how carbon dioxide and water combine to produce glucose and oxygen during the process of photosynthesis. Balance the equation. Find a model for planetary orbital period using the data from Earth and Neptune.

Find the values of the unknown constants in the given decomposition. a Traffic flow problems must have a unique solution. b There are infinitely many ways to balance a chemical equation. c There is a unique parabola passing through any three distinct points in the plane. The volume of traffic for a collection of intersections is shown in the figure below. Find all possible values for x1, x2, and x3. What is the minimum volume of traffic from C to A? Find all possible values for x1, x2, x3, and x4. What is the minimum volume of traffic from C to D? Find all possible values for x1, x2, x3, x4, x5, and x6.

Exercises 5—8: Find the equilibrium temperatures for the heavy wires with endpoints held at the given temperatures. Exercises 9— Determine the total output required from each industry in order to meet both consumer and between-industry demand. See Example 3 for background information. These industries have annual consumer sales of 60 and 40 in billions of dollars , respectively. For every dollar of goods A sells, A requires 20 cents of goods from B. For each dollar of goods B sells, B requires 30 cents of goods from A.

These industries have annual consumer sales of 80 and 50 in billions of dollars , respectively. For every dollar of goods A sells, A requires 25 cents of goods from B. For each dollar of goods B sells, B requires 15 cents of goods from A. Economy has three industries A, B, and C. These industries have annual consumer sales of 30, 50, and 60 in billions of dollars , respectively. For every dollar of goods A sells, A requires 10 cents of goods from B and 15 cents of goods from C. For each dollar of goods B sells, B requires 15 cents of goods from A and 20 cents of goods from C. For each dollar of goods C sells, C requires 20 cents of goods from A and 10 cents of goods from B.

These industries have annual consumer sales of 40, 30, and 70 in billions of dollars , respectively. For every dollar of goods A sells, A requires 20 cents of goods from B and 10 cents of goods from C. For each dollar of goods B sells, B requires 25 cents of goods from A and 10 cents of goods from C. For each dollar of goods C sells, C requires 10 cents of goods from A and 15 cents of goods from B. Exercises 13— Balance the chemical equation. Earth and Mars. Mercury and Uranus. Venus and Neptune. Jupiter and Saturn. Exercises 25— The data given provides the distance required for a particular type of car to stop when traveling at various speeds. Use the data in the table to find values for a and k, and test your model. HINT: Methods similar to those used to find a model for planetary orbital periods can be applied here.

Speed MPH 10 20 30 40 Distance Feet 4. Speed MPH 10 20 30 40 Distance Feet 20 80 Exercises 27— When using partial fractions to find antiderivatives in calculus, we decompose complicated rational expressions into the sum of simpler expressions that can be integrated individually. Find the values of the missing constants in the provided decomposition. Where does the line cross the x-axis? Where does the line cross the y-axis? Where does this plane cross the z-axis? Find an equation for the parabola that passes through the given points. Calculus required Exercises 39— Find the values of the coefficients a, b, and c so the given conditions for the function f and its derivatives are met. This type of problem arises in the study of differential equations. And in practice elimination methods work fine as long as the system is not too large. However, when implemented on a computer, elimination methods can lead to the wrong answer due to round-off error. Furthermore, for very large systems elimination methods may not be efficient enough to be practical.

In this section we consider some shortcomings of elimination methods, and develop alternative solution methods. This section is optional and can be omitted without loss of continuity. In theory there is no difference between theory and practice. In practice there is. van de Snepscheut and physicist Albert Einstein. Round-off Error Sensible people do not spend their time solving complicated systems of linear equations by hand—they use computers. Although computers are fast, they have drawbacks, one being the round-off errors that can arise when using floating-point representations for numbers. For example, suppose that we have a simple computer that has only four digits of accuracy. The degree of error here is not too large, but this is a small system. Elimination methods applied to larger systems will require many more arithmetic operations, which can result in accumulation of round-off errors, even on a high-precision computer.

If the right combination of conditions exists, even small systems can generate significant round-off errors. Example 1 Suppose that we are using a computer with four digits of accuracy. For instance, a number such as , can be represented as 9. Solution We need only one row operation to put the system in triangular form. Although the approximation for x2 is fairly good, the approximation for x1 is off by quite a bit. During back substitution into this equation, the error in x2 is magnified by the coefficient and only can be compensated for by the 3x1 term. But since the coefficient on this term is so much smaller, the error in x1, is forced to be large. Partial Pivoting One way to combat round-off error is to use partial pivoting, which adds a step to the usual elimination algorithms. With partial pivoting, when starting on a new column we first switch the row with the largest leading entry compared using absolute values to the pivot position before beginning the elimination process.

Full Pivoting When Gaussian and Gauss—Jordan elimination are implemented in computer software, partial pivoting is often used to help control round-off errors. It is also possible to implement full pivoting, where both rows and columns are interchanged to arrange for the largest possible leading coefficient. However, full pivoting is slower and so is employed less frequently than partial pivoting. Jacobi Iteration It is not at all unusual for an application to yield a system of linear equations with thousands of equations and variables. In such a case, even if round-off error is controlled, elimination methods may not be efficient enough to be practical. Diverge, Converge Here we turn our attention to a pair of related iterative methods that attempt to find the solution to a system of equations through a sequence of approximations. These methods do not suffer from the round-off problems described earlier, and in many cases they are faster than elimination methods. However, they only work on systems where the number of equations equals the number of variables, and sometimes they diverge—that is, they fail to reach the solution.

In the cases where a solution is found, we say that the method converges. Our first approximation method is called Jacobi iteration. Table 1 shows the outcome from the first nine iterations, each rounded to four decimal places. Table 1 Jacobi Iterations n is the iteration number n 0 x1 x2 0 x3 0 0 1 0. In the next example we revisit an application introduced in Section 1. Figure 1 Grid Temperatures for Example 2. Example 2 Figure 1 gives a diagram of a piece of heavy wire mesh. But since each equation has one variable written in terms of the other variables, the problem sets up perfectly for Jacobi iteration. Table 2 Jacobi Iterations for Example 2 n x1 x2 x3 x4 4 To save space, only every fourth iteration is given in Table 2.

Gauss—Seidel Iteration At each step of Jacobi iteration we take the values from the previous step and plug them into the set of equations, updating the values of all variables at the same time. We modify this approach with a variant of Jacobi iteration called Gauss—Seidel iteration. With this method, we always use the current value of each variable. We encountered C. Gauss earlier. Ludwig Philipp von Seidel — was a German mathematician. Interestingly, Gauss discovered the method long before Seidel but discarded it as worthless. Nonetheless, the Gauss name was attached to the algorithm along with that of Seidel, who independently discovered and published it after Gauss died.

Table 3 gives the first six iterations of Gauss—Seidel applied to our system. Table 3 Gauss—Seidel Iterations n x1 x2 x3 0 0 0 0 1 0. Since Gauss—Seidel immediately incorporates new values into the computations, it seems reasonable to expect that it would converge faster than Jacobi. Most of the time this is true, but surprisingly not always—there are systems where Jacobi iteration converges more rapidly. Example 3 Imagine a simple economy that consists of consumers and just three industries, which we refer to as A, B, and C. This application is also discussed in Section 1. These match the solution found in Section 1. Table 4 Gauss—Seidel Iterations for Example 3 n a b c 0 60 75 40 1 85 On the other hand, as noted earlier, Jacobi and Gauss—Seidel are iterative methods and do not converge to a solution in all cases.

The values grow quickly in absolute value, and do not converge. Diagonally Dominant One case where we are guaranteed convergence is if the coefficients of the system are diagonally dominant. This means that for each equation of the system, the coefficient aii in equation i along the diagonal has absolute value larger than the sum of the absolute values of the other coefficients in the equation. In some instances convergence occurs without diagonal dominance. Example 4 Reverse the order of the equations in 4 to make the system diagonally dominant, and then find the solution using Gauss—Seidel iteration. The values from one iteration can be thought of as an initial guess for the next, so no accumulation of errors occurs.

For the same reason, if a computation error is made, the result still can be used in the next iteration. By contrast, if a computation error is made when using elimination methods, the end result is almost always wrong. For a system of n equations with n unknowns, Jacobi and Gauss—Seidel both require about 2n2 flops per iteration. As mentioned earlier, Gauss—Seidel usually converges in fewer iterations than Jacobi, so Gauss—Seidel is typically the preferred method. The rate of convergence of our iterative methods is influenced by the degree of diagonal dominance of the system. If the diagonal terms are much larger than the others, then iterative methods generally will converge relatively quickly. If the diagonal terms are only slightly dominant, then although iterative methods eventually will converge, they can be too slow to be practical.

There are other iterative methods besides those presented here that are designed to have better convergence properties. Sparse System, Sparse Matrix Iterative methods are particularly useful for solving sparse systems, which are linear systems where most of the coefficients are zero. The augmented matrix of such a system has mostly zero entries and is said to be a sparse matrix. Elimination methods applied to sparse systems have a tendency to change the zeros to nonzero terms, removing the sparseness. See Matrix Computations by G. Golub and C. Van Loan for a more extensive discussion of iterative methods and an explanation of why those described here work.

Solve the system as given with Gaussian elimination with three significant digits of accuracy. Then solve the system again, incorporating partial pivoting. Compute the first three Jacobi iterations for the given system, using 0 as the initial value for each variable. Then find the exact solution and compare. Compute the first three Gauss—Seidel iterations for the given system, using 0 as the initial value for each variable. Then compare to the results found using Jacobi iteration above. Exercises 1—4: Use partial pivoting with Gaussian elimination to find the solutions to the system. The system given in Exercise 9. The system given in Exercise Exercises 17— Determine if the system is diagonally dominant.

If not, then if possible rewrite the system so that it is diagonally dominant. Then rewrite the system so that it is diagonally dominant, set the value of each variable to 0, and compute 4 Jacobi iterations. Then rewrite the system so that it is diagonally dominant, set the value of each variable to 0, and compute four Gauss—Seidel iterations. Exercises 29— The values from the first few Jacobi iterations are given for an unknown system. Find the values for the next iteration. Exercises 31— The values from the first few Gauss—Seidel iterations are given for an unknown system. Exercises 33— Determine the total output required from each industry in order to meet both consumer and between-industry demand. These industries have annual consumer sales of 50 and 20 in billions of dollars , respectively. For every dollar of goods A sells, A requires 30 cents of goods from B. For each dollar of goods B sells, B requires 50 cents of goods from A. These industries have annual consumer sales of 40, 70, and 90 in billions of dollars , respectively.

For every dollar of goods A sells, A requires 20 cents of goods from B and 25 cents of goods from C. For each dollar of goods B sells, B requires 25 cents of goods from A and 30 cents of goods from C. For each dollar of goods C sells, C requires 10 cents of goods from A and 30 cents of goods from B. Exercises 35— Balance the given chemical equation. CHAPTER 2 Euclidean Space Shown here is a geothermal power station located in Iceland. Geothermal power uses steam generated and stored in the earth to produce electricity. The three basic types of geothermal power plants are dry steam, flash steam, and binary cycle.

Geothermal power is considered a sustainable, renewable source of energy that is cleaner than burning fossil fuels. However, gases from inside the earth can contribute to global warming and acid rain if they are released into the air, but with the binary cycle system there are no emissions. W e can think of algebra as the study of the properties of arithmetic performed on numbers. In linear algebra, we study the properties of arithmetic performed on objects called vectors. As we shall see, we can use vectors to express a system of linear equations and to compactly describe the set of solutions of a linear system. But vectors have many other applications as well. Section 2. The mixture of these nutrients varies from one type of fertilizer to the next. Organizing these quantities vertically in a matrix, we have [] This representation is an example of a vector. Using a vector provides a convenient way to record the amounts of each nutrient and also lends itself to compact forms of algebraic operations that arise naturally.

For instance, if we want to know the amount of nitrogen, phosphoric acid, and potash contained in a ton pounds of Ultra Turf, we just multiply each vector entry by The set of all vectors with n entries is denoted by Rn. Component, Column Vector, Row Vector Our convention will be to denote vectors using boldface, such as u. Each of the entries u1, u2, … , un is called a component of the vector. A vector expressed in the vertical form is also called a column vector, and a vector expressed in horizontal form is also called a row vector. It is customary to express vectors in column form, but we will occasionally use row form to save space. The fertilizer discussion provides a good model for how vector arithmetic works. Here we formalize the definitions. Euclidean space is named for the Greek mathematician Euclid, the father of geometry. Euclidean space is an example of a vector space, discussed in Chapter 7.

Although vectors with negative components and negative scalars do not make sense in the fertilizer discussion, they do in other contexts and are included in Definition 2. Two vectors can be equal only if they have the same number of components. Similarly, there is no way to add two vectors that have a different number of components. These are summarized in the next theorem. THEOREM 2. For b , suppose that a is a scalar. Proofs of the remaining properties have a similar flavor and are left as exercises. Note that it is possible for scalars to be negative or equal to zero. Linear combinations provide an alternate way to express a system of linear equations. The equation on the left is an example of a vector equation. The row operations are exactly the same. Thus 3 has no solutions, so the combination in b is not possible. Solution In Example 3 of Section 1.

A more complicated general solution arises in Example 5 of Section 1. We plot the vector [x1x2] by drawing an arrow from the origin to the point x1, x2 in the plane. Using an arrow to denote a vector suggests a direction, which is a common interpretation in physics and other sciences, and will frequently be useful for us as well. We call the end of the vector with the arrow the tip, and the end at the origin the tail. Figure 1 Vectors in R2. Note that the ordered pair for the point x1, x2 looks the same as the row vector x1, x2. The difference between the two is that vectors have an algebraic and geometric structure that is not associated with points. Most of the time we focus on vectors, so use that interpretation unless the alternative is clearly appropriate. Figure 2 a The vectors u and v. There are two related geometric procedures for adding vectors. The Tip-to-Tail Rule: Let u and v be two vectors. Translate the graph of v, preserving direction, so that its tail is at the tip of u.

The Tip-to-Tail Rule makes sense from an algebraic standpoint. When we add v to u, we add each component of v to the corresponding component of u, which is exactly what we are doing geometrically. We also see in Figure 2 b that we get to the same place if we translate u instead of v. The second rule follows easily from the first. The Parallelogram Rule: Let vectors u and v form two adjacent sides of a parallelogram with vertices at the origin, the tip of u, and the tip of v. Figure 3 illustrates the Parallelogram Rule. We see that the third and fourth sides of the parallelogram are translated copies of u and v, which shows the connection to the Tip-to-Tail Rule. Scalar multiplication and subtraction also have nice geometric interpretations. Figure 4 Scalar multiples of the vector u.

Figure 5 Subtracting vectors. We will consider how to find the length of a vector later in the book. Subtraction: Draw a vector w from the tip of v to the tip of u. Then translate w, preserving direction and placing the tail at the origin. The subtraction procedure is illustrated in Figure 5 and is considered in more detail in Exercise Express the vector equation as a linear system. Express the linear system as a vector equation. Express the general solution as a linear combination of vectors. Determine if b is a linear combination of the other vectors. If so, express b as a linear combination. a Addition of vectors is commutative but not associative. b The scalars in a linear combination can be negative or zero. c The set of solutions to a vector equation always can be expressed as a linear combination of vectors with the same number of components.

d The Parallelogram Rule gives a geometric interpretation of scalar multiplication. Exercises 7— Express the vector equation as a system of linear equations. Express this solution as a linear combination of vectors. Determine the total amount of nitrogen, phosphoric acid, and potash in the given mixture. Determine the amount of each type required to produce a mixture containing the given amounts of nitrogen, phosphoric acid, and potash. Exercises 47— One 8. One Determine the number of cans of each drink that when combined will contain the specified heart-pounding combination of sugar and caffeine. Determine the number of servings of each cereal required to create the given mix of nutrients.

An electronics company has two production facilities, denoted A and B. During an average week, facility A produces computer monitors and flat panel televisions, and facility B produces computer monitors and 10, flat panel televisions. Because many students learn by example, Linear Algebra with Applications provides a large number of representative examples, over and above those used to introduce topics. The text also has over exercises, covering computational and conceptual topics over a range of difficulty levels. Search Images Maps Play YouTube News Gmail Drive More Calendar Translate Books Shopping Blogger Finance Photos Docs. Account Options Sign in. Try the new Google Books. Check out the new look and enjoy easier access to your favorite features. Try it now. No thanks. Try the new Google Books My library Help Advanced Book Search. Get print book.

com Fishpond Whitcoulls Mighty Ape Find in a library All sellers ». Shop for Books on Google Play Browse the world's largest eBookstore and start reading today on the web, tablet, phone, or ereader. Go to Google Play Now ». Linear Algebra : with Applications. Jeffrey Holt. What people are saying - Write a review. Other editions - View all Linear Algebra with Applications Jeffrey Holt No preview available -

Thus, b is not a linear combination of a1 , a2 , and a3. a False. Addition of vectors is associative and commutative. b True. The scalars may be any real number. c True. The solutions to a linear system with variables x1 ,. d False. The Parallelogram Rule gives a geometric interpretation of vector addition. From row 11 2 Hence b is not a linear combination of a1 , a2 , and a3. Using vectors, we calculate. Hence we have 76 pounds of nitrogen, 31 pounds of phosphoric acid, and 14 pounds of potash. Hence we have pounds of nitrogen, pounds of phosphoric acid, and 58 pounds of potash. need to drink 1 can of Red Bull and 3 cans of Jolt Cola. Thus we cans of Red Bull and 4 cans of Jolt Cola. Thus we need 3 servings of Lucky Charms and 5 servings of Raisin Bran. Thus we need 3 servings of Lucky Charms and 2 servings of Raisin Bran. Thus we need 2 servings of Lucky Charms and 3 servings of Raisin Bran.

Thus we need 3 servings of Lucky Charms and 4 servings of Raisin Bran. flat panel televisions at facility B in 8 weeks. monitors and flat panel televisions at facilities A and B in 6 weeks. Thus we need 9 weeks of production at facility A and 2 weeks of production at facility B. We assume a 5-day work week. The company produces metric tons of PE, metric tons 50 of PVC, and metric tons of PS at facility C in 4 weeks. The company produces 10 40 50 metric tons of PE, metric tons of PVC, and metric tons of PS at facilities A,B, and C. d Let x1 be the number of days of production at facility A, x2 the number of days of production at facility B, and x3 the number of days of production at facility C.

Thus we need 4 days of production at facility A, 2 days of production at facility B, and 4 days of production at facility C. Let x1 , x2 ,and x3 be the mass of u1 , u2 , and u3 respectively. Let x1 , x2 , x3 ,and x4 be the mass of u1 , u2 , u3 , and u4 respectively. Vector components and scalars can be any real numbers. a True, by Theorem 2. b False. A vector can have any initial point. a True, by Definition 2. It works regardless of the quadrant, and can be established algebraically for vectors posi- tioned anywhere. Because vector addition is commutative, one can order the vectors in either way for the Tip-to-Tail Rule. b True, as long as the vectors have the same number of components. There is not a row of zeros, so every choice of b is in the span of the columns of the given matrix and, therefore, the columns of the matrix span R2. Since there is not a row of zeros, 2every choice of b is in the span of the columns of the given matrix, and therefore the columns of the matrix span R.

There is not a row of zeros, so every choice of b is in the span of the columns of the given matrix and, therefore, the columns of the matrix span R3. b Row-reduce to echelon form:. Because there is a row of zeros, there exists a vector3 b that is not in the span of the columns of the matrix and, therefore, the columns of the matrix do not span R. If the vectors span R3 , then vectors have three components, and cannot span R2. Every vector b in R2 can be written as b. which shows that {2u1 , 3u2 } spans R2. x3 d True. Thus b is not in the span of a1. Since there is a row of zeros, there exists a vector b which is not in the span of the columns of A, and therefore the columns of A do not span R2.

Row-reduce to echelon form: [. Since there is not a row of zeros, every choice of b is in the span of the columns of A, and therefore the columns of A span R2. Row-reduce to echelon form:. Since there is not a row of zeros, every choice of b is in the span of A, and therefore the columns of A span R2. Since there is not a row of zeros, every choice of b is in the span of the columns of A, and therefore the columns of A span R3. Since there is a row of zeros, there exists a vector b which is not in the span of the columns of A, and therefore the columns of A do not span R3. Since there is not a row of zeros, every choice of b is in the span of A, and therefore the columns of A span R3. Row-reduce A to echelon form. Row-reduce A to echelon form: [. b False, the zero vector can be included with any set of vectors which already span Rn.

a False, since every column of A may be a zero column. b False, by Example 5. b True, by Theorem 2. a True, the span of a set of vectors can only increase with respect to set containment when adding a vector to the set. The span of {u1 , u2 , u3 } will be a subset of the span of {u1 , u2 , u3 , u4 }. a True. See problem 61, and the solutions to problems 43 and 45 for examples. c Can possibly span R3. d Can possibly span R3. If b is in span{u1 ,. Hence the echelon form of the system will have free variables, and since the system is consistent this implies that it has infinitely many solutions. Now reverse the row operations used to transform A to B, and apply these to B1. Using a computer algebra system, the row-reduced echelon form of the matrix not have any zero rows. Hence the vectors span R3. Using a computer algebra system, the row-reduced echelon form of the matrix have a zero row. Hence the vectors do not span R3. Hence the vectors do not span R4.

Hence the vectors span R4. with the given vectors as columns does with the given vectors as columns does with the given vectors as columns does with the given vectors as columns does. The only solution is the trivial solution, so the vectors are linearly independent. The only solution is the trivial solution, so the columns of the matrix are linearly independent. a False, because is linearly independent in R3 but does not span R3. Since the only solution is the trivial solution, the vectors are linearly independent. Since there exist nontrivial solutions, the columns of A are not linearly independent.

Since the only solution is the trivial solution, the columns of A are linearly independent. We solve the homogeneous system of equations using the corresponding augmented matrix: [. Since there are trivial solutions, the columns of A are linearly dependent. We solve the homogeneous system of equations using the corresponding augmented matrix:. We solve the homogeneous equation using the corresponding augmented matrix:. Linearly dependent. Linearly independent. The vectors are not scalar multiples of each other. Apply Theorem 2. Any collection of vectors containing the zero vector must be linearly dependent. Since the only solution is the trivial solution, the columns of the matrix are linearly independent. By Theorem 2. Since there exist nontrivial solutions, the columns of the matrix are linearly dependent. We row—reduce to echelon form: [. Because the echelon form has a pivot in every row, by Theorem 2. We row—reduce to echelon form:.

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[REQUEST] Linear Algebra with Applications by Jeffrey Holt (2nd Edition) ISBN: 3 3 3 comments Best Add a Comment [deleted] • 3 yr. ago Found it: WebPDF. Read online LINEAR ALGEBRA WITH APPLICATIONS JEFFREY HOLT SOLUTION PDF book pdf free download link book now. All books are in clear copy here, and all files WebLinear Algebra With Applications 2nd Edition by Holt Test Bank - Free download as PDF File .pdf), Text File .txt) or read online for free. Link full download . Request a sample or Linear algebra has applications in science, computer science, social science, business, and other fields. Engineering is filled with visible applications of linear algebra, with examples all around us of infrastructure that affects our everyday lives Linear algebra has applications in science, computer science, social science, business, and other fields. Engineering is filled with visible applications of linear algebra, with examples all around us of infrastructure that affects our everyday lives WebLinear Algebra with Applications by Jeffrey Holt 2nd edition. [Request] I am looking for this book, but I can not find it anywhere. Does someone have access to a pdf or word ... read more

Linear Algebra With Application [PDF] Authors: Jeffrey Holt PDF Mathematics , Algebra: Linear Algebra Add to Wishlist Share. b An echelon system with five equations all with variables and eight variables must have three free variables. Four distinct nonzero vectors that do not span R3. The next theorem draws on ideas from Example 5 to provide a way to determine if a set of vectors spans Rn. The span of {u1 , u2 , u3 } will be a subset of the span of {u1 , u2 , u3 , u4 }.

Request Status. The subsection on the Shortcut Method has been moved up in section 5. Then rewrite the system so that it is diagonally dominant, set the value of each variable to 0, and compute four Gauss—Seidel iterations. W e can linear algebra with applications jeffrey holt pdf download of algebra as the study of the properties of arithmetic performed on numbers. The values from one iteration can be thought of as an initial guess for the next, so no accumulation of errors occurs. Figure 4 The caffeine molecule. Where does the line cross the y-axis?