Linear Algebra with Applications, Global are linear algebra and numerical analysis. Leon and Publisher Pearson. Leon, Edition statement 9th edition; His areas of specialty are linear algebra and numerical analysis. Leon is This item is out of print and has been replaced with Linear Algebra with Applications, 9th Edition. Leon is currently. His areas of specialty are linear algebra and numerical analysis. Steven J. Leon The following pages include all the items of errata that have been uncovered so far.
In each case we include This thorough and accessible text, from one of the leading figures in the use of technology in linear algebra, gives students a challenging and broad understanding of … with Applications 9th edition by Steven J. Enter the email address you signed up with and we'll email you a reset link. Need an account? Click here to sign up.
Download Free PDF. Ivan Pixabaj. A short summary of this paper. Download Download PDF. Translate PDF. The answers in this manual supple- ment those given in the answer key of the textbook.
In addition this manual contains the complete solutions to all of the nonroutine exercises in the book. The questions in each Chapter Test A are to be answered as either true or false. Although the true- false answers are given in the Answer Section of the textbook, students are required to explain or prove their answers. This manual includes explanations, proofs, and counterexamples for all Chapter Test A questions. The chapter tests labelled B contain workout problems.
The answers to these problems are not given in the Answers to Selected Exercises Section of the textbook, however, they are provided in this manual. Complete solutions are given for all of the nonroutine Chapter Test B exercises. Con- sequently they have not been included in this solutions manual.
On the other hand, the text also includes questions related to the computations. The purpose of the questions is to emphasize the significance of the computations. The solutions man- ual does provide the answers to most of these questions. There are some questions for which it is not possible to provide a single answer.
For example, aome exercises involve randomly generated matrices. In these cases the answers may depend on the particular random matrices that were generated. Steven J. Leon sleon umassd. Two nonparallel lines intersect in a point. That point will be the unique solution to the system. Since parallel lines do not intersect, there is no point on both lines and hence no solution to the system.
The system must be consistent since 0, 0 is a solution. A linear equation in 3 unknowns represents a plane in three space. If the planes are parallel or one plane is parallel to the line of intersection of the other two, then the solution set will be empty.
The three equations could represent the same plane or the three planes could all intersect in a line. In either case the solution set will contain infinitely many points. If the three planes intersect in a point then the solution set will contain only that point. A homogeneous linear equation in 3 unknowns corresponds to a plane that passes through the origin in 3-space. Two such equations would correspond to two planes through the origin. If one equation is a multiple of the other, then both represent the same plane through the origin and every point on that plane will be a solution to the system.
If one equation is not a multiple of the other, then we have two distinct planes that intersect in a line through the origin. Every point on the line of intersection will be a solution to the linear system. So in either case the system must have infinitely many solutions. If one equation is a multiple of the other, then both represent the same plane and there are infinitely many solutions. If the equations represent planes that are parallel, then they do not intersect and hence the system will not have any solutions.
If the equations represent distinct planes that are not parallel, then they must intersect in a line and hence there will be infinitely many solutions.
At each intersection the number of vehicles entering must equal the number of vehicles leaving in order for the traffic to flow. Thus if no glucose is produced then there is no reaction. Note the ratios are still There are no other solutions since the echelon form of A is strictly triangular. There are many possible choices for A and B. For most pairs of symmetric matrices A and B the product AB will not be symmetric.
Section 4 9 Generally the product of two elementary matrices will be a matrix formed from the identity matrix by the performance of two row operations. But if a homogeneous system has a nonzero solution, then it must have infinitely many solutions. See Theorem 1. It follows form Theorem 1. If B is singular, then it follows from Theorem 1.
Row operation III can then be used to eliminate all of the entries above the diagonal. Thus U is row equivalent to I and hence is nonsingular. Row operation II applied to I will just change the values of the diagonal entries.
When the row operation III steps referred to in part a are applied to a diagonal matrix, the entries above the diagonal are filled in. Since A is nonsingular it is row equivalent to I.
Hence there exist elementary matrices E1, E2,. There- fore the reduced row echelon form of A B will be I C. Section 4 13 If A is row equivalent to B then there exist elementary matrices E1, E2,. If B is row equivalent to A then it follows from the result in Exercise 24 a that B is row equivalent to U. If B is row equivalent to A, then there exist elementary matrices E1, E2,.
Since M is nonsingular it is row equivalent to I. Thus there exist elementary matrices E1 , E2,. It is possible to perform both block multiplications. Let 0 denote the zero vector in Rn. Exact equality will not occur in parts c and d because of roundoff error.
The computation of x is also more efficient since the solution is computed using Gaussian elimination with partial pivoting and this involves less arithmetic than computing the inverse matrix and multiplying it times b.
By construction B is upper triangular whose diagonal entries are all equal to 1. Thus B is row equivalent to I and hence B is nonsingular. If E1,. Examining the last row of the reduced row echelon form of the augmented matrix A b , we see that the system is inconsistent. There is a free variable x3, so the system will have infinitely many solutions. The change in part b should not have a significant effect on the survival potential for the turtles. The change in part c will effect the 2, 2 and 3, 2 of the Leslie matrix.
With these values the Leslie population model should predict that the survival period will double but the turtles will still eventually die out. The statement is false in general. If the row echelon form has free variables and the linear system is consistent, then there will be infinitely many solu- tions. However, it is possible to have an inconsistent system whose coefficient matrix will reduce to an echelon form with free variables. The statement is true since the zero vector will always be a solution.
The statement is true. A matrix A is nonsingular if and only if it is row equivalent to the I the identity matrix. A will be row equivalent to I if and only if its reduced row echelon form is I.
The statement is false. An elementary matrix is a matrix that is constructed by performing exactly one elementary row operation on the identity matrix. The product of two elementary matrices will be a matrix formed by per- forming two elementary row operations on the identity matrix.
The row vectors of A are x1yT , x2yT ,. Note, all of the row vectors are multiples of yT. Since x and y are nonzero vectors, at least one of these row vectors must be nonzero. However, if any nonzero row is picked as a pivot row, then since all of the other rows are multiples of the pivot row, they will all be eliminated in the first step of the reduction process.
The resulting row echelon form will have exactly one nonzero row. However, if there is more than one solution, then the echelon form of A must involve a free variable. A consistent system with a free variable must have infinitely many solutions. If one equation is a multiple of the other then the equations represent the same plane and any point on the that plane will be a solution to the system. If the two planes are distinct then they are either parallel or they intersect in a line.
If they are parallel they do not intersect, so the system will have no solutions. If they intersect in a line then there will be infinitely many solutions.
It follows from part b then that a homogeneous sys- tem of 2 equations in 3 unknowns must have infinitely many solutions. Geometrically the 2 equations represent planes that both pass through the origin, so if the planes are distinct they must intersect in a line. Therefore it follows from Theorem 1. If this is done the second row of the augmented matrix will zero out in the elimination process and you will end up with one equation in 2 unknowns.
The reduced system will have infinitely many solutions. It follows from Theorem 1. In general the product of two symmetric matrices is not necessarily symmet- ric. If E and F are elementary matrices then they are both nonsingular and their inverses are elementary matrices of the same type.
Theorem 2. Proof: The proof is by induction on n. If E is an elementary matrix of type III formed from the identity matrix by adding c times its ith row to its jth row, then E T will be the elementary matrix of type III formed from the identity matrix by adding c times its jth row to its ith row 9.
Row operation III has no effect on the value of the determinant. Row operation I has the effect of changing the sign of the determinant. We are assuming the entries of A are all real numbers. Thus V will be nonsingular if and only if the three points x1 , x2 , x3 are distinct. Thus AB is nonsingular if and only if A and B are both nonsingular. To see this look at the cofactor expansion of the B along its last row.
Since 1! By the induction hypothesis the calculation of det M1j requires m! In the elimination method the matrix is reduced to triangular form and the determinant of the triangular matrix is calculated by multiplying its diagonal elements. The j, i entry of QT is qij.
The magic squares generated by MATLAB have the property that they are nonsingular when n is odd and singular when n is even.
The matrix U is very ill-conditioned. So in general one could not expect to get even a single digit of accuracy in the computed values of det U T and det U U T. On the other hand, since U is upper triangular, the computed value of det U is the product of its diagonal entries.
This value should be accurate to the machine precision. However, there may be no indication of this if the computations are done in floating point arithmetic. In this case the product u11u22u33u44u55 has magnitude on the order of MATLAB knows that if you started with an integer matrix, you should end up with an integer value for the determinant.
In general if the determinant is computed in floating point arithmetic, then you cannot expect it to be a reliable indicator of whether or not a matrix is nonsingular.
Hence the exact value of det B should be 0. By Theorem 1. It follows that if U is any row echelon form of A then U can have at most one nonzero row. If A has integer entries then adj A will have integer entries. To show that C is a vector space we must show that all eight axioms are satisfied. Let f, g and h be arbitrary elements of C[a, b].
The proof is exactly the same as in Exercise 5. Axiom 6 fails to hold. The system is not a vector space. Axioms A3, A4, A5, A6 all fail to hold. Axioms 6 and 7 fail to hold. To see this consider the following example.
Thus all eight axioms hold and hence S is a vector space. C n[a, b] is a nonempty subset of C[a, b]. Any scalar multiple of a continuous function is continuous. If f and g are vectors in C n[a, b] then both have continuous nth derivatives and their sum will also have a continuous nth derivative. Also, S2 is not closed under addition. To determine whether or not the system is consistent we can compute the row echelon form of the augmented matrix X x.
It depends on whether xk is in Span x1, x2 ,. The vectors x and y are elements of both U and V. For a counterexample, consider the vector space R2. For a counterexam- ple we can use the same subspaces S, T , and U of R2 that were used in part a.
Thus if x1,. To test whether or not x1 , x2 ,. Since the Wronskian is not identically zero the vectors are linearly independent. Thus x3 and x 3 are linearly independent. Let v1,. This would imply that there exist scalars c1 , c2,. Since v1 ,. Therefore x1 and x2 are linearly independent. Enlarge cover.
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