Introduction to Linear Algebra with Applications by Jim DeFranza, Daniel Gagliardi

By Jim DeFranza, Daniel Gagliardi

During the last few a long time, linear algebra has turn into extra proper than ever. purposes have elevated not just in volume but additionally in range, with linear structures getting used to unravel difficulties in chemistry, engineering, economics, meals, city making plans, and extra. DeFranza and Gagliardi introduce scholars to the subject in a transparent, enticing, and easy-to-follow demeanour. themes are built totally ahead of relocating directly to the following via a sequence of usual connections. the result's a high-quality creation to linear algebra for undergraduates' first course.

Outstanding good points include:

Early assurance of vector areas, supplying the summary conception essential to comprehend applications

routines that variety from regimen to more difficult, extending the recommendations and strategies by way of asking scholars to build entire arguments

a variety of examples designed to boost instinct and get ready readers to imagine conceptually approximately subject matters as they're introduced

truth summaries to finish every one bankruptcy that use nontechnical language to recapitulate info and formulas

Not-for-sale teacher source fabric to be had to school and college college simply; touch writer directly.

Brief desk of Contents

1. structures of Linear Equations and Matrices 2. Linear mixtures and Linear Independence three. Vector areas four. Linear adjustments five. Eigenvalues and Eigenvectors 6. internal Product areas

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0 1 −1 0 1 ⎦ c. ⎣ 0 3 −1 0 0 1 3 0 0 0 1 3 0 0 0 0 0 a. Reading directly from the augmented matrix, we have x3 = 3, x2 = 2, and x1 = 1. So the system is consistent and has a unique solution. b. In this case the solution to the linear system is x4 = 3, x2 = 1 + x3 , and x1 = 5. So the variable x3 is free, and the general solution is S = {(5, 1 + t, t, 3) | t ∈ ‫}ޒ‬. c. 2 Matrices and Elementary Row Operations 17 So the variables x3 and x4 are free, and the two-parameter solution set is given by �� � �� � 1 s 1 5s � − + t, + , s, t � s, t ∈ ‫ޒ‬ S= 3 3 3 3 EXAMPLE 2 Solution Write the augmented matrix and solve the linear system.

Let A = � a M= c 15. Find 2A(B − 3C). 16. Find (A + 2B)(3C). In Exercises 17–24, use the matrices � � 2 0 −1 A= 1 0 −2 � � −3 1 1 B= −3 −3 −2 � � 3 −1 C= −1 −3 Whenever possible, perform the operations. If a computation cannot be made, explain why. 17. 2At − 1 0 0 1 � � 2 1 . Find all matrices of the form 1 �1 b such that AM = MA. d 29. Find matrices A and B such that AB = 0 but BA �= 0. 30. Show there are no 2 × 2 matrices A and B such that � � 1 0 AB − BA = 0 1 31. Determine all values of a and b such that � � � �� � −5 6 3 b 1 2 = 12 16 −4 1 a 0 Bt 18.

This is true for any n × n matrix with two equal rows and provides an alternative method for concluding that such a matrix is not invertible. Theorem 9 gives a formula for the inverse of the product of invertible matrices. THEOREM 9 Let A and B be n × n invertible matrices. Then AB is invertible and (AB)−1 = B −1 A−1 Proof Using the properties of matrix multiplication, we have and (AB)(B −1 A−1 ) = A(BB −1 )A−1 = AI A−1 = AA−1 = I (B −1 A−1 )(AB) = B −1 (A−1 A)B = B −1 I B = BB −1 = I Since, when it exists, the inverse matrix is unique, we have shown that the inverse of AB is the matrix B −1 A−1 .

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