MATH10212 Linear Algebra B Homework Week 3. Be prepared to answer the following oral questions if asked in the supervision class

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1 MATH10212 Linear Algebra B Homework Week Students are strongly advised to acquire a copy of the Textbook: D. C. Lay Linear Algebra its Applications. Pearson, ISBN Normally, homework assignments will consist of some odd numbered exercises from the Textbook. The Textbook contains answers to most odd numbered exercises. Be prepared to answer the following oral questions if asked in the supervision class 1. [Mostly Lay True or False? Justify each answer. 1. In some cases, a matrix may be row reduced to more than one matrix in reduced echelon form, using different sequences of row operations. 2. The row reduction algorithm applies only to augmented matrices for a linear system.. A basic variable in a linear system is a variable that corresponds to a pivot column in the coefficient matrix. 4. This question is temporarily omitted. 5. If one row in an echelon matrix is [ then the associated linear system is inconsistent. 6. The echelon form of the matrix is unique. 7. The pivot positions in a matrix depend on whether row interchanges are used in the row reduction process. 8. Whenever a system has free variables, the solution set contains infinitely many solutions. 9. Whenever a system has no free variable, it is inconsistent. 10. Every matrix has a pivot position. 11. Many different matrices can be reduced to the same reduced echelon form. 1. Suppose a 5 coefficient matrix for a system has three pivot columns. Is the system consistent? Why or why not? 2. Suppose a system of linear equations has a 5 augmented matrix whose fifth column is a pivot column. Is the system consistent? Why or why not?. Suppose that the coefficient matrix of a consistent linear system has a pivot in each column. Explain why a system has a unique solution. 4. Restate using the concept of pivot columns: If a linear system is consistent, then the solution is unique if only if What would you have to show about the pivot columns in an augmented matrix in order to know that the linear system is consistent has a unique solution? 6. A system of linear equations with fewer equations than unknowns is called underdetermined system. Suppose that such a system happens to be consistent. Explain why there must be an infinite number of solutions. 7. Give an example of an inconsistent underdetermined system of two equations in three unknowns. 8. A system of linear equations with more equations than unknowns is called overdetermined system. Can such a system be consistent? Give an example. 2.

2 MATH10212 Linear Algebra B Homework Week 2. [ True of False? Justify your answer. 1. Another notation for vector [ 4 is [ An example of a linear combination of vectors v 1 v 2 is the vector 1 2 v 1.. The solution set of the linear system whose augmented matrix is [ a1 a 2 a b is the same as the solution set of the equation a 1 + x 2 a 2 + x a = b. 4. Any list of five numbers is a vector in R The weights c 1,..., c p in a linear combination cannot all be zero. c 1 v c p v p 6. The vector u results when a vector u v is added to the vector v. 7. Asking whether the linear system corresponding to an augmented matrix [ a1 a 2 a b has a solution amounts to asking whether b is in Span{a 1, a 2, a }. Solve the following exercises (but do not submit them for marking the assignment is at the end of the paper!)) 5. For each of the two linear systems, write a vector equation which is equivalent to it. x 2 + 4x = 0 + 6x 2 2x = 0 + x 2 8x 2 = x 2 + x = x 2 5x = 5 7x 2 5x = 6 [1..11 Determine if b is a linear combination of a 1, a 2, a : a 1 = 2, a 2 = 1, a = 6, b = [1..1 Determine if b is a linear combination of the vectors formed from the columns of the matrix A: A = 0 5, b = [1..25 Let A = b = 1. 4 Denote the columns of A by a 1, a 2, a, let W = Span{a 1, a 2, a }. (a) Is b in { a 1, a 2, a }? (b) Is b in W? How many vectors are in W? (c) Show that a 1 is in W.

3 MATH10212 Linear Algebra B Homework Week Answers Solutions Answers to oral questions False. The reduced echelon form of a matrix is unique. 2. False. The row reduction algorithm could be applied to a matrix of any origin.. True. 4. This question is temporarily omitted. 5. False. For example, the system + x 2 + x + x 4 = 1 5x 4 = 0 has the augmented matrix in echelon form with a row [ is consistent because it has a solution (among many others) = 1, x 2 = x = x 4 = False. The echelon form of a matrix is not unique, it is the reduced echelon form of the matrix that is unique. 7. False this follows from the definition given in the textbook 8. False. For example, the augmented matrix of the system + x 2 = 1 + x 2 = 2 has the reduced echelon form [ 1 1 1, therefore has a free variable x 2 but is inconsistent. 9. False. For example the system = 1 x 2 = 1 has no free variables but is consistent. If you are looking for an even simpler example, the system made of a single equation in one variable = 0 is consistent has no free variables False: a matrix made only of zeroes, such as [ 0 0 0, has no pivot positions. 11. True. If a matrix A can be obtained from a matrix B by elementary row operations, then A B have the same reduced echelon form. 1. Yes, it is consistent. 2. No, the system is not consistent. The row containing the pivot element of the fifth column has the form [ b for some b 0, which corresponds to the inconsistent equation 0 = b.. Because the system has no free variables every column of the coefficient matrix is a pivot column. 5. Every column with the exception of the rightmost one should contain a pivot position, but the rightmost column should not. 6. Because there are columns which are not pivot columns; such columns correspond to free variables. 7. For example, + x 2 + x = 1 + x 2 + x = 2 8. Of course, it can. For example, 2 + 2x 2 + 2x = 2 + x 2 + x = 4 + 4x 2 + 4x = 4 1. False vectors in the textbook are defined as column vectors.

4 MATH10212 Linear Algebra B Homework Week 4 2. True.. True. 4. False. 5. False. 6. True. 7. True. Solutions for non-starred exercises x x 2 = x x 5 = Yes, b is a linear combination of a 1, a 2, a. For example, or b = 2 a 1 + a a b = a 1 1 a a as can be seen from solving the system x 1 x 2 = x 6 (which actually has infinitely many solutions). 7. No, b is not a linear combination of the columns of A. Hint: perform on the augmented matrix [ A b of the system Ax = b the row operation R R + 2R (a) No. The set { a 1, a 2, a } contains just three elements a 1, a 2, a, b is not equal to any of them. (b) Yes, b W = Span{a 1, a 2, a } because the system of equations Ax = b has a solution, therefore the vector equation a 1 + x 2 a 2 + x a = b has a solution, therefore a can be written as a linear combination of a 1, a 2, a. To see that the system Ax = b has a solution, it suffices to apply row operations to the augmented matrix of the system transform it to an echelon form R R + 2R R R + 2R There is no need to find the solution because it is already clear that the solution exists, is all that we wish to know. And W is infinite because it contains, for example infinitely many vectors of the form ca 1 = c 0 2c where c can be an arbitrary real number. (c) a 1 W because a 1 = 1 a a a. Submit for marking: 9*. Write a vector equation that is equivalent to the linear system 4 + 2x 2 + x = 9 7x 2 x = x 2 x = 1 10*. [1..12 Determine if b is a linear combination of a 1, a 2 a : a 1 = 2, a 2 = 5, a = 0, b =

5 MATH10212 Linear Algebra B Homework Week 5 11*. [1..14 Determine if b is a linear combination of vectors formed from the columns of the matrix A A = 0 7, b = Solutions for marked exercises: 9* x x 1 1 = *. No, it is not. The augmented matrix is converted by row operations R 2 R 2 +2R 1, ; R R 2R 1, R R R 2 into the matrix therefore the corresponding system of equations is inconsistent. 11*. Yes. There is no need to solve the corresponding system of linear equations completely; the Yes answer should be obvious after just one elementary row operation: R R R 1. Indeed the the question is equivalent to asking whether the system of linear equations with the augmented matrix is consistent. Applying a row operation we get the matrix R R R 1, which is in echelon form with a pivot position in every column of A. Hence the system is consistent, hence vector b is a linear combination of vectors formed from the columns of the matrix A. But we can say more: the same argument works for arbitrary vector b R, hence the columns of A span R.