Exam Solutions for Math 304 - Linear Algebra: Independence, Spans, and Vector Spaces, Exams of Linear Algebra

Download Exam Solutions for Math 304 - Linear Algebra: Independence, Spans, and Vector Spaces and more Exams Linear Algebra in PDF only on Docsity! Math 304 - 501 Exam 1 Solutions February 25, 2004 1. (15) Define the following: (a) The set of vectors {~x1, · · · , ~xk} is linearly independent. The set is linearly independent if whenever there are constants ci, 1 ≤ i ≤ k, such that c1~x1 + c2~x2 + · · ·+ ck~xk = ~0, then each ci must equal 0. (b) A is an invertible matrix. A square n× n matrix A is said to be invertible is there is another n× n matrix B such that AB = In = BA, where In denotes the n × n matrix with one’s down the main diagonal and zeros every where else. (c) The set of vectors {~x1, · · · , ~xk} spans the vector space V . The set is said to span V if for every ~x ∈ V there are constants ci, 1 ≤ i ≤ k, such that ~x = c1~x1 + c2~x2 + · · ·+ ck~xk . 2. (25) Let A =   −2 1 0 6 21 2 5 0 1 2 1 2 1 4   be the augmented matrix of a system of linear equations. (a) What is the system of linear equations? Use xi’s to designate the variables. −2x1 + x2 + 6x4 = 2 x1 + 2x2 + 5x3 = 1 2x1 + x2 + 2x3 + x4 = 4 (b) Find a row echelon form of the matrix A. Be sure to state which row operations you use, and the order of application. The following sequence of row operations transforms A into a row echelon form. R1 ←→ R2, 2R1 + R2, − 2R1 + R3 1 5 R2, 3R2 + R3, −1 2 R3. The row echelon form obtained by these elementary row operations is   1 2 5 0 10 1 2 6/5 4/5 0 0 1 −23/10 −11/5   (c) True or false, the solution set of this system is a vector space? The solution set is not a vector space. Set B =   −2 1 0 61 2 5 0 2 1 2 1  , then the solution set equals Solution set =  ~x ∈ R4 : B~x =   21 4     . This set is not a vector space as it is not closed under scalar multiplication or vector addition, nor does it contain the zero vector. 2