MATH240: Linear Algebra Exam #1 solutions 6/12/2015 Page 1

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1 MATH4: Linear Algebra Exam # solutions 6//5 Page Write legibly and show all work. No partial credit can be given for an unjustified, incorrect answer. Put your name in the top right corner and sign the honor pledge at the end of the exam. If you need more room than what s given, please continue onto the back.. Let A be the 3 4 matrix A = (a) (Fifteen points) Use row operations to determine the reduced row echelon form which is row-equivalent to A. Solution. Here s one possible sequence of operations. Remember the rref is unique, but the steps to get there are not necessarily (Start by switching rows around) (Replace R with R R) (Replace R3 with R3 R) (Divide R3 by three) (Replace R with R + R3 replace R with R R3) (b) (Five points) Do the columns of A span R 3? Briefly explain why or why not. Solution. They do. This is because there s a pivot position in every row of the matrix. According to Theorem 4 ( cow theorem ), the columns of an m n matrix span R m exactly when there s a pivot in each row.

2 MATH4: Linear Algebra Exam # solutions 6//5 Page. (a) (Five points) What is the definition of the span of a set of vectors { v,..., v n }? Solution. The span is the set of vectors that can be written as a linear combination of v,..., v n. That is to say, it s the set of all expressions of the form (b) (Fifteen points) Is the vector Explain why or why not. c v + c v c n v n. 5 6 in the span of 4 and Solution. One possible answer here is that the bottom entry of [ 5 ] 6 is nonzero, while the other two vectors both have zero on the bottom. That means every vector in the span of those two vectors has to look like [ ]. Therefore the answer is no. A second possible answer is to notice that our vector is in the span of the other two when the equation c 4 + c 3 = has a solution, since this equation having a solution is exactly the same as saying that [ 5 ] 6 is expressible as a linear combination of the other two. But this vector equation can be represented by a system of linear equations which leads to the augmented matrix This system is inconsistent: the bottom row says =. So the answer is no, it s not in the span ?

3 MATH4: Linear Algebra Exam # solutions 6//5 Page 3 3. Suppose A is the following 4 4 matrix. I ve helpfully computed rref(a) for you. (Thank me later.) A = rref(a) = (a) (Six points) Solve the matrix equation A x =. If there are infinitely many solutions, give a set of vectors that spans the solution set. Solution. The linear equations that come from rref(a) are x + x 3 =, x x 3 =, x 4 =, and x 3 is free. (Note that x 4 is actually determined: every solution to A x = must have x 4 =. The generic solution is x x 3 x = x x 3 = x 3 x 3 = x 3. x 4 So the solution set is spanned by the one vector [ ]. (b) (Six points) Are the columns of A linearly independent? If not, tell me a nontrivial linear dependence relation they satisfy. Solution. The columns are not linearly independent because there s a free variable in the system. To find a linear dependence, remember that the previous part showed A [ ] =. Writing that as a vector equation, we get that (col ) + (col ) + (col 3) + (col 4) =. There s our linear dependence relation. (c) (Eight points) Consider the function x A x. This is a function from R 4 to R 4. It is neither one-to-one nor onto. Explain why each of these properties fail to hold for this function. Solution. It s not injective (one-to-one) because there s a free variable, that is, a column with no pivot in it (the third column). It s not surjective (onto) because there s a row with no pivot in it (the fourth column).

4 MATH4: Linear Algebra Exam # solutions 6//5 Page 4 4. (Five points each) Each part of this problem is a question whose answer is yes, always, sometimes yes but also sometimes no, and no, never. Circle the correct choice of the three. No work is required for these problems: I ll give full credit if you circle the correct choice. If you want to explain yourself with examples and/or counterexamples, they may be worth partial credit if your answer is incorrect. (a) Suppose M is a 5 9 matrix. Is x M x a function from R 5 to R 9? Explanation. It s actually a map from R 9 to R 5. (b) Suppose rref(a) has a zero row at the bottom. Does the equation A x = have infinitely many solutions? Explanation. It depends on whether or not there s a free variable, so we don t have enough information. If A is the matrix on the left, then A x = will have infinitely many, while if A is the matrix on the right, then A x = will only have the trivial solution.. (c) Suppose that M and N are two different matrices and that rref(m) = rref(n). Is there a sequence of row operations that transforms M to N? Explanation. Keep track of the row operations that go from N to rref(n). Each one of those steps can be reversed. So go from M to rref(m) = rref(n), then run those steps in reverse and you ll get to N. (d) Suppose T is a linear transformation from R 4 to R 3. Is T surjective? Explanation. Just like in part (b), the question is about whether or not T is represented by a matrix with a zero row at the bottom or not. If not, it s surjective. But it s not hard to come up with a matrix that does have a zero row, and is consequently not surjective. The left matrix is an example of a yes, the right matrix is an example of a no..

5 MATH4: Linear Algebra Exam # solutions 6//5 Page 5 5. (a) (Ten points) Write down the standard matrix for the transformation T : R R which consists of the following operations: scaling the x direction by a factor of 3, scaling the y direction by a factor of 4, and then performing a rotation of degrees counterclockwise. Solution. Sorry to say I m too lazy to figure out all of the pictures, so you ll have to settle for a verbal description. The vector e is represented by the point (, ) in the xy-plane. Under the scaling operation it becomes (3, ). Then we rotate the point and get (3 cos( π 3 ), 3 sin( π 3 )) = ( 3, 3 3 ). Similarly, following e = (, ) step by step: after scaling it becomes (, 4), and then after rotating that we get ( 4 sin( π 3 ), 4 cos( π 3 )) = ( 3, ). Therefore the standard matrix for this transformation is [ 3/ 3 3 3/ ]. It s equally valid to say that the second column is gives you the matrix above. [ 4 cos( 7π 6 ) 4 sin( 7π 6 ) ]. Either approach (b) (Ten points) Suppose that the set { u, v, w} of three vectors in R 3 is linearly dependent. Let T : R 3 R 3 be a linear transformation. Explain why the set {T ( u), T ( v), T ( w)} is also linearly dependent. Solution. Since the set { u, v, w} is linearly dependent, there exist constants c, c, c 3 (not all zero) such that c u + c v + c 3 w =. Apply the transformation T to both sides of this equation. We get T ( c u + c v + c 3 w ) = T ( ). One property of linear transformations is that they take to, so the right side of this equation is still. On the other side we can break the expression up over addition, then pull out the scalars; doing that gives us c T ( u) + c T ( v) + c 3 T ( w) =, which is a linear dependence among T ( u), T ( v), and T ( w). Therefore the second set is linearly dependent as well. Rotating the point (,) gives (cos π 3, sin( π 3 )), so if we start with three times that point we ll get three times the result.