Solutions to Math 51 First Exam October 13, 2015

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1 Solutions to Math First Exam October 3, 2. (8 points) (a) Find an equation for the plane in R 3 that contains both the x-axis and the point (,, 2). The equation should be of the form ax + by + cz = d. Show your work. Since the plane contains x-axis, it contains origin, so d =. And for any point (t,, ) on x-axis, at =, so a =. Since (,, 2) is on the plane, 2c = and hence c =. So the plane is by =, b which is equivalent to y =. (b) Let P be the plane through the origin in R 3 spanned by the vectors 2 and Let Q be the plane parallel to P, passing through the point (,, ). Write down an equation for the plane Q in the form ax + by + cz = d. Show your work. Since (,, ) is on the plane, we can let the plane to have form a(x ) + b(y ) + c(z ) =. And since the normal vector n = (a, b, c) is orthogonal to (,, 2) and (2, 3, 2), we have So we can solve n = (a, b, c) up to a scalar. a + b + 2c = and 2a + 3b + 2c =. a = 4, b = 2, c = So the plane is Rewrite this as 4(x ) + 2(y ) + (z ) =. 4x + 2y + z =.

2 Math, Autumn 2 Solutions to First Exam October 3, 2 Page 2 of 4 2. ( points) Consider the 4 4 matrix A: 2 3 A = (a) Find, showing all of your work, a basis for N(A), the null space of A. Find rref(a): A = = rref(a) Since N(rref(A)) = N(A), we obtain the following system of equations: x + 2x 2 + 2x 4 = x 3 + x 4 = = = Writing the pivot variables x, x 3 in terms of the free variables x 2, x 4, we obtain x 2x 2 2x x 2 x 3 = x 2 x 4 = x 2 + x 4. x 4 x 4 So a basis for N(A) is 2 2,. (b) Find, explaining your reasoning, a basis for C(A), the column space of A. A basis of C(A) is given by the columns of A corresponding to the pivot columns of rref(a). In particular, the first and the third columns of A gives a basis for C(A): 2, 4 3 (c) For the same matrix A consider the equation Ax = b where b b = b 2 b 3 b 4

3 Math, Autumn 2 Solutions to First Exam October 3, 2 Page 3 of 4 is a vector in R 4. Describe conditions on the components of b that are necessary and sufficient for there to be a solution x to Ax = b. Your answer should be one or more equations involving b, b 2, b 3, b 4. Now find the rref for the augmented matrix: 2 3 b b b b b b 2 b 3 3 b 3 b 2 2 b 4 b 2 2 2b b 2 b 2 b 2b 3b 2 + b 3 b 2b 2 + b 4 A necessary and sufficient condition for there to be a solution to Ax = b is that there is no inconsistency in the rref: { 2b 3b 2 + b 3 = b 2b 2 + b 4 =

4 Math, Autumn 2 Solutions to First Exam October 3, 2 Page 4 of 4 3. ( points) (a) Find the projection of the vector 4 onto the x-axis. Show your work. The projection of 4 onto the x-axis is just the vector in the x direction which goes the same distance as 4 in the x direction. More formally, let v = 4, and let w be the projection of v onto the x-axis. Then, w is defined to be the multiple of such that v w is orthogonal to. Thus, we can easily see by inspection that w =. Although the answer is geometrically clear in this case, it also works to use the projection formula. Let x = (we may also replace x with any vector spanning the x-axis without changing our calculation). Then, the problem asks for Proj x (v) = x v x 2 x = =. (b) What is the shortest distance between the point (, 4, ) and the x-axis? Show your work. The projection of a vector v onto a line can also be regarded as the closest point onthat line to v. Applying this principle together with the previous part, the closest point to 4 should be. Then, the distance is 4 2 = 4 2 = 2 + ( 4) = 7. To convince yourself of this geometrically, you may sketch these vectors, and you will see how to calculate the distance essentially by using the Pythagorean theorem. (c) Consider the line L parametrized by {(, 4, ) + t(,, ) t R} in R 3. Find the length of the shortest line segment joining a point on L and a point on the x-axis. Show your work. (Hints: Such a line segment must be perpendicular to each line. Alternatively, find an expression for the distance between each point of L and the x-axis in terms of t, and find its smallest value over all values of t.)

5 Math, Autumn 2 Solutions to First Exam October 3, 2 Page of 4 There are several ways to do this problem. + t Solution. Let v(t) = 4 t, so that the points on L are exactly those which take the form v(t) for some real number t. If we are given t, then we may calculate the projection of v(t) onto + t the x-axis to be, using the same procedure as in part (a). Then, by the same procedure as in part (b), the distance from v(t) to the x-axis is ( 4 t) () Thus, to calculate the minimum distance, we simply need to find the value of t which minimizes (). This can be done by calculus (taking a derivative, solving for its zeroes, etc.). However, in this case it is also easy to see that the expression ( 4 t) 2 is always non-negative and is minimized when t = 4. Thus, the minimum distance is. Solution 2. Suppose that a and b are points on L and the x-axis which achieve the minimum possible distance between the two lines. We may write + t s a = 4 t, b =, for some real numbers s and t. In order for a and b to achieve the minimal distance, it must be the case that a b is orthogonal to both L and the x-axis. Thus, we obtain the equations (a b) =, (a b) =. When expanded out in terms of S and t, this gives ( + t s) ( 4 t) =, + t s =. Solving for s and t, we find that we must have t = 4 and s =. Thus, a =, b =, and so the distance from a to b is. Solution 3. We can directly compute a vector v that is orthogonal to both L and the x-axis. Indeed, we must solve for coordinates v 2 such that v 3 v v v 2 =, v 2 =. v 3 v 3 Solving this system, we end up with a solution v v = v 2 =. v 3 v

6 Math, Autumn 2 Solutions to First Exam October 3, 2 Page 6 of 4 (Of course, any multiple thereof will also work.) Now, imagine two planes P and P 2, both orthogonal to v, with P containing L and P 2 containing the x-axis. (The fact that v is orthogonal to both L and the x-axis makes it possible to find such planes.) Then, the distance between L and the x-axis will be the distance between P and P 2. (If you cannot convince yourself of this, try to draw a picture. If you still can t, maybe you should focus on the other solutions.) Now, how do we compute the distance between P and P 2? Well, the shortest path from P to P 2 is by going along the direction of v. Next, we can pick any two points a P and b P 2. It is convenient to pick a to be a point on L, so let s pick a = 4, b =. Then, P 2 is a translate of P by b a, and so the distance between them is Proj v (b a) = Proj v 4 =. Note: In this case, P and P 2 were particularly simple; they were parallel to the xy plane. Thus, some of the steps above could have been done by inspection, but we have given an approach that would have worked in general.

7 Math, Autumn 2 Solutions to First Exam October 3, 2 Page 7 of 4 4. ( points) Suppose that a is an unspecified real number. Consider the linear system: 2 3 x = 2 2 a (a) Find, with justification, all values of a for which the above system has exactly one solution x. (b) Find, with justification, all values of a for which the above system has no solution x. (c) Find, with justification, all values of a for which the above system has an infinite number of solutions x. The linear system shown (of three equations in three variables) is represented by the augmented matrix a We row-reduce the matrix as follows: a 2 a 6 2 a 6 a 2 At this point we can see that if a = 2, the last equation is inconsistent, so the system has no solutions. (a 2)z = If a 2, then we can divide the last row by a 2 to get a 2 At this point the three pivot s have already appeared. We will be able to row-reduce the matrix with no further complications. There are three pivots and no free variables, so there is a unique solution. So, for all a 2, the equation has exactly one solution. Since all possible real a are accounted for, there are no values of a for which the equation has infinitely many solutions.

8 Math, Autumn 2 Solutions to First Exam October 3, 2 Page 8 of 4. (8 points) The solutions to a system of linear equations can be interpreted as the intersection of a set of (affine) subspaces of R n. Match each of the systems of equations to the corresponding picture of an (affine) subspace arrangement. (The axes might be rotated.) No justification required. Answer by writing the number (n o ) of the geometric picture to the right of the matrix. [ ] [ ] [ 2 3 n o 2 n o 2 4 n 2 o 2 [ ] n o n o n o ] 3 n o n o The question was graded out of 8, with one point per correctly labelled system of equations. A first observation: A system of m equations in 2 variables represents m lines in R 2 (with the possibility that some lines coincide). A system of m equations in 3 variables represents m planes in R 3 (again with the possibility that some planes coincide). This means we can immediately rule out the graphics 7, 8, and 9 as possible answers for any system, and narrow down the options for each remaining system of equations.

9 Math, Autumn 2 Solutions to First Exam October 3, 2 Page 9 of 4 The solution to the system the set of points that simultaneously satisfy each equation is the intersection of these lines or planes. One way to approach these problems is to determine the individual lines or planes represented by each row of each matrix (for example, the row [ 2 ] represents the equation x + y = 2, which is the line y = x+2. The row [ 2 4 ] represents the equation y +2z = 4, which represents a plane with normal vector [,, 2] T in R 3. The following solutions will not consider specific lines and planes, however, and instead look at qualitative features of the system, such as whether any lines or planes are parallel, and the dimension of the space of solutions. [ ] The system of equations represents two lines in R 3 2. Since the system is homogeneous, the solution space must be a subspace of R 2. Since the rows [ ] and [ 3 ] are (by inspection) linearly independent, these lines are not parallel, and they must intersect in a single point, the origin. The answer is. [ ] The system of equations also corresponds to two lines in R 2 2, but since the normal vectors [ ] and [ ] are collinear, these lines are parallel. Since the equations do not encode the same line, these lines are parallel and distinct; the system has no solutions, and the lines do not intersect. The answer is 2. The system 2 encodes three lines in R 2. None of the rows are collinear, so no two 2 lines are parallel. The system is homogeneous so all three lines must pass through the origin. The answer is 4. [ ] The system encodes two planes in R 2 3. Since (by inspection) the normal vectors [,, ] T and [,, 2] T are not collinear, these planes are not parallel. The answer is 3. [ ] The system also encodes two planes in R 3. Their normal vectors [,, ] T and [,, ] T are collinear, so these planes are parallel. The equations do not define the same plane (the equation y z = is not a scalar multiple of the equation y + z = ), so the planes are parallel and distinct. Alternatively, we can row-reduce the matrix to see that the system of equations has no solutions, so the planes cannot intersect. The answer is. 2 4 The system 2 4 gives three planes in R 3. Since no row is a scalar multiple of any 2 other row, no two planes coincide. Since no two normal vectors are collinear, no two planes are parallel. By row-reducing the matrix, we see that the system is consistent and has a -parameter space of solutions, so the planes must intersect in a line. The answer is. We can also complete this problem by noticing that the column space is 2-dimensional (since at most two columns are linearly independent), and the system is consistent since [4, 4, ] T is in the column space it is twice the third column. This tells us that the system has solutions, and the rank-nullity theorem tells us that the solutions are a line. Again we conclude that the answer is. The system also describes three planes in R 3, where no two planes coincide and no two planes are parallel. But this time, by row-reducing we find that the system is inconsistent, so the planes have no common points of intersection. The answer is 2. The system encodes three distinct, non-parallel planes in R 3. We can row-reduce

10 Math, Autumn 2 Solutions to First Exam October 3, 2 Page of 4 to find that these planes intersect in a point. Or we can observe that the columns are linearly independent, so the rank-nullity theorem implies that the nullspace (the intersection of the planes) is {}. The answer is.

11 Math, Autumn 2 Solutions to First Exam October 3, 2 Page of 4 6. ( points) (a) Complete the following sentence: A set of vectors {v, v 2,, v k } is defined to be linearly dependent if A set of vectors {v,, v k } is linearly dependent if we can find scalars c,, c k R not all zero such that the linear combination c v + + c k v k = is the zero vector. (b) Complete the following: A set V of vectors in R n to be is defined to be a subspace if the following three conditions hold The three conditions that must be satisfied in order for V R n a subspace are that it contain the zero vector, be closed under addition, and be closed under scalar multiplication. In other words, we must have (i) V, (ii) if v, w V, then v + w V, and (iii) if v V and c R is any scalar, then cv V. (c) Let A be a fixed n n matrix. Define S to be the set of all vectors x R n which satisfy: Show that S is a subspace of R n. Ax = 2x We need to show that the set of vectors x which satisfy Ax = 2x is a subspace. First, we check that is in this putative subspace. This claim is obvious: A = = 2. Next, if v and w satisfy the condition that Av = 2v, Aw = 2w, we can check that A(v + w) = Av + Aw = 2v 2w = 2(v + w), so v + w is also in the putative subspace. Nice. We re two-thirds of the way there. That s so cool. Finally, if Av = 2v and c R, then we check A(cv) = cav = c( 2v) = 2cv = 2(cv) so cv is also in the set of vectors satisfying the given condition. Hence, the set is in fact a subspace, as it satisfies the three properties listed in the previous part.

12 Math, Autumn 2 Solutions to First Exam October 3, 2 Page 2 of 4 7. (8 points) 7 (a) Find the coordinates (c, c 2 ) of the vector u = 3 with respect to the basis {v, v 2 } where 2 v = and v 2 =. 2 In other words, express u as a linear combination u = c v + c 2 v 2. Show your work. Set up the system which yields the linear system c c 2 = 3 2 c + 2c 2 = 7 c = 3 c + 2c 2 = The system is simple enough to solve it manually (without resorting to augmented matrices and RREF): the second equation gives c = 3 and substituting into either the first or third gives c 2 = 2. Thus = 3 2 We can indeed verify that this works. The coordinates are (c, c 2 ) = ( 3, 2). (b) Find the coordinates (c, c 2 ) of u with respect to the basis elements v and v 2. Justify your solution with a brief written explanation and/or a picture. Hint: The solutions are integers. The vectors appear to form a parallelogram with v the diagonal and u, v 2 the sides. By the

13 Math, Autumn 2 Solutions to First Exam October 3, 2 Page 3 of 4 parallelogram rule, we conclude This means so the answer is v = u + v 2 u = v v 2 = v + ( )v 2 (c, c 2 ) = (, )

14 Math, Autumn 2 Solutions to First Exam October 3, 2 Page 4 of 4 8. (8 points) Each of the statements below is either always true ( T ), or always false ( F ), or sometimes true and sometimes false, depending on the situation ( MAYBE ). For each part, decide which and circle the appropriate choice; you do not need to justify your answers. (a) Given a matrix A, then N(A) = N(rref(A)). T F SOMETIMES Row operations applied to the augmented matrix do not change the set of solutions. So the solutions to Ax = and rref(a)x = are the same. (b) Given a matrix A, then dim(c(a)) = dim(c(rref(a))). T F SOMETIMES The set of columns of A corresponding to the pivots of rref(a) form a basis for C(A). The pivot columns of rref(a) form a basis for C(rref(A)). The number of basis elements is the same. (c) Given a vector b R m and an m m (square) matrix A for which the system Ax = b has exactly one solution, then the system Ax = has infinitely many solutions. T F SOMETIMES If the system Ax = b has a solution x p then x p + x h, where x h N(A), is also a solution. So, if there is a unique solution, N(A) = {}. This is true for any shaped matrix A. (d) Given a vector b R m and an m m (square) matrix A for which the system Ax = b has no solutions, then the system Ax = has infinitely many solutions. T F SOMETIMES If Ax = b has no solution, C(A) R m and the rank of A is < m. But rank + nullity = m so nullity is > and Ax = has an infinite number of solutions. This uses the fact that A is square. The answer would be Maybe in the general m n case. (e) Given an m n matrix A with m < n, then dim(n(a)) >. T F SOMETIMES In this case there are more columns than rows, hence more than the number of pivots, so there is at least one free variable. (f) Given an m n matrix A with m > n, then N(A) = {}. T F SOMETIMES The conclusion N(A) = {} holds if and only the columns are linearly independent. That may or may not be the case. (g) Given an m n matrix A with m > n, then dim(c(a)) = m. T F SOMETIMES The dimension of C(A) equals the number of linearly independent columns of A. Since there are n of them and n < m there can never be m linearly independent ones. (h) Given an m n matrix A and a set {v,..., v k } R n that is linearly independent, then the set {Av,..., Av k } is linearly independent. T F SOMETIMES This depends on the matrix A and the linearly independent set. For example, if the linearly independent set equals {e,..., e n } then the set {Ae,..., Ae n } is the set of columns of A which are linearly independent if and only if N(A) = {}.