# LINEAR ALGEBRA WITH APPLICATIONS STEVEN J. LEON PDF

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Linear Algebra with Applications Eighth Edition Steven J. Leon University of Massachusetts, Dartmouth Boston Columbus Indianapolis New York San Francisco. Linear Algebra with Applications, 8th Ed. Steven J. Leon. The following pages include all the items of errata that have been uncovered so far. In each case we. Linear Algebra With Applications by Steven J. Leon 8th Edition With Applications by Steven J. Leon 8th Edition is available for free download in PDF format. Author: DARNELL WARNES Language: English, Indonesian, German Country: Japan Genre: Politics & Laws Pages: 394 Published (Last): 13.11.2015 ISBN: 827-4-43715-209-6 ePub File Size: 19.80 MB PDF File Size: 11.14 MB Distribution: Free* [*Registration needed] Downloads: 23604 Uploaded by: BRAD Linear Algebra with Applications (Steven Leon). Huajun Huang . If A and B are m × n matrices, then A = B if and only if aij = bij for each i and j. This solutions manual is designed to accompany the ninth edition of Linear Algebra with Applications by Steven J. Leon. The answers in this manual supplement. Find all the study resources for Linear Algebra with Applications by Steven J. Leon.

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. 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. 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.

It follows from Theorem 1. In general the product of two symmetric matrices is not necessarily symmetric. If E and F are elementary matrices then they are both nonsingular and their inverses are elementary matrices of the same type. Chapter Test B 25 Theorem 2. 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. 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.

This requires 1 division, It will require: Thus the calculation det A by the elimination method requires: 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.

## Linear Algebra With Applications by Steven J. Leon 8th Edition

This value should be accurate to the machine precision. 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. Hence the exact value of det B should be 0. By Theorem 1. In fact a i,: 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. Let A1. Let f, g and h be arbitrary elements of C[a, b].

Section 1 A2. 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. Section 1 41 So Axiom 6 fails. 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.

Section 3 45 b This distributive law also does not work in general. For a counterexample 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,. Section 3 47 This would imply that there exist scalars c1 , c2,. Since v1 ,. Therefore x1 and x2 are linearly independent. Therefore x1 and x2 cannot possibly span R3. A 2-dimensional subspace of R3 corresponds to a plane through the origin in 3-space.

Section 4 49 so x1 , x2, x3 are linearly dependent. Consider next the vectors x1 , x2, x4. It follows from Theorem 3. The vectors u1 ,. To show this we must show that the vectors are linearly independent. Section 6 51 is a basis for the row space. Since the reduced row echelon form of the matrix involves one free variable the nullspace will have dimension 1. The dimension of the column space equals the rank of the matrix which is 3. Thus a1 , a3 , a6 form a basis for the column space of A.

The remaining column vectors satisfy the following dependency relationships. There will be exactly one solution. The condition that b is in the column space of A guarantees that the system is consistent. If the column vectors are linearly independent, then there is at most one solution. Thus the two conditions together imply exactly one solution. If b is in the column space of A, then the system will be consistent, so there will be at least one solution. By part a , the column vectors are linearly independent, so there cannot Thus, if b is in the column space of A, then the system will have exactly one solution.

Since the dimension of the column space equals the rank it follows that the two column spaces will have the same dimension. The column vectors of A and U satisfy the same dependency relations. Since A has 8 columns, its reduced row echelon form will involve 3 free variables. The vectors y1 , y2, y3 do not span R5 and hence cannot form a basis for R5. Since A has rank n we know that its column vectors must be linearly independent. Therefore y cannot be equal to 0.

Therefore the column space of A b will equal the column space of A. Since the rank of a matrix is equal to the dimension of the column space it follows that the rank of A b equals the rank of A. Conversely if A b and A have the same rank, then b must be in the column space of A. Thus BA and A have the same nullspace. Corollary 3. Thus it follows from Theorem 1.

By part a the column space of B is a subspace of N A.

Thus, by part a , c must be in the column space of B T and hence cT must be in the row space of B. If the column vectors of C T are linearly dependent, then the row vectors of C must be linearly dependent. If the column vectors of B are linearly independent, then the row vectors of B T are linearly independent.

By Exercise 26, B T has a right inverse and consequently B must have a left inverse. If B has a left inverse, then B T has a right inverse. It follows from Exercise 25 that the column vectors of B T span Rm. Thus the rank of B T is m. The rank of B must also be m and consequently the column vectors of B must be linearly independent. Let u 1,: If c1 u 1,: Let us refer to the column containing the leading nonzero entry of u i,: Thus the rank of A is 1.

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## Linear Algebra With Applications by Steven J. Leon 8th Edition

Note the ratios are still 6: Therefore it follows from Theorem 1. Choose priority shipping upon checkout for fastest shipping speeds. 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.

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