2018 Fall 2210Q Section 013 Midterm Exam I Solution

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1 8 Fall Q Section 3 Midterm Exam I Solution True or False questions ( points = points) () An example of a linear combination of vectors v, v is the vector v. True. We can write v as v + v. () If two matrices A and B are row equivalent, then the solution set of Ax = and the solution set of Bx = are the same. True. Row operators do not change the set of solutions. (3) A vector b is a linear combination of the columns of a matrix A if and only if the equation Ax = b has at least one solution. True. Ax exactly gives the linear combination of columns of A with weights x,..., x n. (4) A linear transformation T : R 3 R is always onto. False. For example, the zero map is not onto. (5) The solution set of Ax = b is obtained by translating the solution set of Ax =. True. (6) If a system of linear equations has no free variables, then it has a unique solution. False. The system may not be consistent. (7) The columns of any 4 5 matrix are linearly dependent. True. Since the number of vectors is more than the rows (5 > 4), these vectors are linearly dependent. (8) The transpose of a product of matrices equals the product of their transposes in the same order. False. It is equal to the product of their transposes in the reverse order. (9) If T : R R rotates the points about the fixed point ( ) through an angle θ, then T is a linear transformation. False. T does not send to. () If the matrix multiplication AB makes sense, then the matrix multiplication B T A T makes sense too. True. (AB) T = B T A T.

2 Problem I (5 points) Consider the traffic pattern as indicated by the following diagrams A x x B 6 x 4 C () (3 points) Write a system of equations that describes the relations among x, x,, x 4. () ( points) Give the augmented matrix of the system of linear equations. (3) (8 points) Perform a Gaussian elimination process to solve the system of linear equations. For this, you do not need to write which row operations you used at each step. (Solving the system with equations will only get partial credits.) Indicate which variable is the free variable following the rules of Gaussian elimination. (4) ( points) While thinking mathematically, the free variable can be an arbitrary number. Explain in practice what is the range of the free variable. Solution () By looking at the in-and-out at teach intersection, we get x + = x = 6 + x 4 = x + x () The augmented matrix is 6 (3) We solve it as follows: The solution is x = x = 6 + is free x 4 = 8. (4) Since the traffic flows are nonnegative, we must have.

3 Problem II ( points) Consider the vectors v =, v =, v 3 =. 3 () (8 points) Are the vectors {v, v, v 3 } linearly dependent or linearly independent? () (8 points) Which b = b b b 3 is in the span of {v, v, v 3 }? (3) (4 points) Consider the linear transformation T (x) = Ax with A = [ v v v 3 ]. Is T one-to-one; is T onto? Explain why. Solution () We test this by solving the following system: There is a free variable; so {v, v, v 3 } is linearly dependent. () For this, we need to solve the system as follows: b b b b 3 b b 3 b b 3 b 3 3 b 3 b 3 + b b So b is in the span of {v, v, v 3 } if and only if b 3 + b b =. (3) By (), the columns of A are not linearly dependent, so T is not one-to-one. By (), only certain b s are in the range of T. For example, if b 3 + b b, b is not in the range, and therefore T is not onto. 3

4 Problem III ( points) () ( points) Let T be the linear transformation from R 3 R, given by first projecting to the x x -plane, and then rotating counterclockwise π/. Compute the standard matrix for T (denoted by A). () (4 points) Explain why the columns of A are linearly ( dependent. ) (3) (6 points) Find a vector x in R 3 that satisfies T (x) =. How many such x are there? Solution () First understand the effect of T on the standard basis of R 3. ( ) ( ) ( ) ( ) T :,, So the standard matrix for T is A = ( ) ( ) ( ). () The columns of A are linearly dependent because the last column is the zero vector, and a set with a zero vector is linearly dependent. (3) Solve ( ) ( ) ( ) So we may take x = for any a R. There are infinitely many such x which can a already be seen from (). 4

5 Problem IV (points) ( ) Let A = 3 and B =. Determine if the following matrix exists or not. If exists, compute it. Otherwise, explain why. () AB. () A T + B. (3) A T B. Solution () AB = () (3) This does not exist. ( ) 3 = A T + B = + + ( ) ( ) 3 + =. + ( ) + ( )( ) 3 ( ) + A T B = ( ) 3 = ( ) ( ) 3 ( 3 ) 4 5

6 Problem V (5 points) Consider the linear system Ax = b with ( ) A = 3 and b = ( ). () (7 points) Write down the associated homogeneous equation, and solve it. () (4 points) Observe that b is the same as the first column of A. Which particular solution does this fact give you? (3) (4 points) Give the general solution of Ax = b based on () and (), and write it in the form of p + tv. Solution () The associated homogeneous equation is ( ) x = 3 x ( ) Solve it as follows: ( ) 3 ( ) ( ) We get x = x = or x h = =. is free () Since b is the first column of A, the system Ax = b has a solution. (3) Based on () and (), the solution to Ax = b is x = + t. 6