Math 355: Linear Algebra: Midterm 1 Colin Carroll June 25, 2011

Transcription

1 Rice University, Summer 20 Math 355: Linear Algebra: Midterm Colin Carroll June 25, 20 I have adhered to the Rice honor code in completing this test. Signature: Name: Date: Time: Please read the following information: This exam is closed book, closed notes. Calculators are not allowed. There is a 4 hour time limit. The exam is worth 00 points total. Show all your work for full credit. Clearly indicate your final answers (box or circle them, or write Answer: ). Justify answers using complete sentences. No credit will be given for correct but unsupported answers. If you have a question, I will be available at (860) Have fun!

2 Question Points Score Total: 00 Page 2

3 . Suppose x = and x 2 = each solve Ax = b for some matrix A and some vector b. (a) (5 points) Find infinitely more solutions to Ax = b. Solution: Since we have Ax = b = Ax 2, A(x x 2 ) = 0. That is to say, x x 2 is in the null space of A. Since x is a particular solution, we have all being solutions. (b) (5 points) If b = ( 2 + t : t R ), then what size is the matrix A? Solution: Since x is 4 and b is 2, A must be 2 4. (c) (5 points) Find an A that satisfies all of the above (i.e., so that Ax = Ax 2 = (, 2) T for the vectors x and x 2 given above). Solution: We need the sum of columns through 4 to be (, 2) T, and need (0,, 2, 3) T to be in the null space. One example of such a matrix is ( ) A = Page 3

4 2. (a) (5 points) Give an example of a system of linear equations with 0 solutions. Solution: One example would be { x = x = 2 (b) (5 points) Give an example of a system of equations with (exactly) one solution. Solution: One example would be { x = (c) (5 points) Give an example of a system of equations with a 2 dimensional solution space. Solution: We need a 2 dimensional null space. One example would be { x + y + z =, whose solution would be x = t s, y = t and z = s, for any t, s R. Page 4

5 3. Give the dimension of each of the following subspaces. (a) (3 points) The null space of a 4 2 matrix with rank 2. Solution: 0. The dimension of the null space is the number of columns minus the rank. In this case, that is 2 2 = 0. (b) (3 points) The row space of a 20 8 matrix with rank. Solution:. The dimension of the row space is equal to the rank of a matrix. (c) (3 points) The space of polynomials of degree less than or equal to 34. Solution: 35. A basis for this space is given by {, x, x 2,..., x 34 }, which has 35 elements. (d) (3 points) The space of all 2 9 matrices. Solution: 8. A basis for this space consists of matrices with a in one spot, and 0 s in the other 7. There are 8 choices for the nonzero spot. Page 5

6 4. (0 points) Factor the matrix A = into the product of a lower triangular matrix and an upper triangular matrix. Solution: We need only subtract one of the first row from the second to put A in row echelon form. Hence A = Page 6

7 5. Consider again the matrix (a) (3 points) What is the rank of A? A = Solution: From the previous question, we have that (R E) = , so the rank is 2. (b) (4 points) Write down the dimension of the row space, column space, null space and left null space (please label clearly!) Solution: Since the rank is 2,. dim(c(a)) = 2, 2. dim(n(a)) = 3 2 =, 3. dim(n(a T )) = 3 2 =, 4. dim(c(a T )) = 2. (c) (4 points) Find a basis for C(A). Solution: so a basis for the column space is {(,, 0) T, (3,, 0) T }. (d) (4 points) Find a basis for N(A). Solution: From the above, we set x 2 =, and then solve x =, and x 3 = 0. Hence {(,, 0)} is a basis for N(A). (e) (4 points) Find a basis for C(A T ). Solution: The first two rows of R will form a basis: {(,, 0) T, (0, 0, ) T }. (f) (4 points) Find a basis for N(A T ). Solution: The the row of E corresponding to the zero row of R forms a basis: {(0, 0, )}. Page 7

8 6. True/False. Justify if true, provide a counterexample if false. (a) (5 points) If the columns of a matrix A are linearly independent, then Ax = b has a solution for every b. Solution: False. For example, the matrix A = has linearly independent rows, but there is no solution for b = (0, 0, ) T, for example. (b) (5 points) Every 3 3 matrix with rank 3 has an inverse. Solution: True. If A is such a matrix, then elimination will provide three pivots, so Gauss- Jordan elimination will go through without a hitch. (c) (5 points) The null space and row space of a matrix are subspaces of the same vector space. Solution: True. If A is m n, then the null space is a subspace of R n, as is the row space. Page 8

9 7. Suppose A is a 4 4 matrix. (a) (5 points) How would you use matrix multiplication to find the sum along each column of A? Solution: We would multiply A on the left by (,,, ). (b) (5 points) Suppose for this particular A, the sum along each column is. Prove that the matrix A I has a non-trivial nullspace, where I is the 4 4 identity matrix. (Partial credit will be awarded for demonstrating that this is true in particular cases). Solution: Since the sum along each column of A is, the sum along each column of A I is 0. Hence, (,,, ) N((A I) T ). This means that the dimension of the left null space is at least one, which in turn means that the rank of A I is at most 2. Hence, the null space of A I has dimension at least. Page 9