MATH240: Linear Algebra Review for exam #1 6/10/2015 Page 1

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1 MATH24: Linear Algebra Review for exam # 6//25 Page No review sheet can cover everything that is potentially fair game for an exam, but I tried to hit on all of the topics with these questions, as well as show you some of the different kinds of things that I could ask.. Let A be the 4 matrix A = Row reduce an appropriate augmented matrix to solve the equation A x =. Go all the way to the reduced row-echelon form. Solution. Here is a step by step row reduction: (Start) (R2 becomes 5R R2) (I also switched R and R4) (R becomes 2R R) (R becomes R R2) (R4 becomes R R4) (R2 becomes R2 + 2R) From this we see that the vector that solves the equation A x = [ 2 2 [ 2 2. is x = [ 2 2. Corrections to the initially-posted version are in blue. The mistake here was that I had x as a 4 vector, but if you re going to multiply A by a vector it should be. Oops.

2 MATH24: Linear Algebra Review for exam # 6//25 Page 2 2. (a) Explain what it means for a set of vectors to be linearly independent. (b) Are the vectors [ 4 [ [ 2 2,, and linearly independent? Explain why or why not. Solution. (a) There are many ways to say what this means. I ll offer a couple. The first one is the original definition. Suppose the vectors are named v, v 2,..., v n. Then these vectors are linearly independent if and only if the equation c v + c 2 v c n v n = can only be solved by the constants c = c 2 =... = c n =. Another way to say it is that there are no free variables in the system v v 2 v n x =, or put a third way, the reduced row-echelon form of the matrix in the centered equation should have a pivot in every column. (b) We can answer this from the definition: suppose that 4 2 c 2 + c 2 + c = The left side of this equation is the vector [ 4c +c 2 2c 2c. If this is the zero vector, then c the bottom entry must be zero, so c =. Then the middle entry is zero too, so c 2 =, and then the top entry being zero means 4c + c 2 2c =, but c 2 = c = already, so c = as well. Therefore the only possible solution is c = c 2 = c =, and they re linearly independent. It s equally possible to answer this question by row reducing the matrix you get by making the three vectors the columns; the reduced row-echelon form is, so. [ there s a pivot in each column, so there s no free variables, so the vectors are linearly independent.

3 MATH24: Linear Algebra Review for exam # 6//25 Page. Suppose A is the following 4 6 matrix. The reduced row-echelon form of A is given alongside it. 2 A = rref(a) = (a) Solve the matrix equation A x =. If there are infinitely many solutions, give a set of vectors that spans the solution set. [ [ 6 [ 6 4 (b) Determine a linear dependence among the vectors 9, 6, and [Hint. It s easy to do too much work for this question. (c) Consider the function x A x. This is a function from R 6 to R 4. Is it injective? Is it surjective? Discuss. Solution. (a) The system of equations that we get from rref(a) is x = x 2 x 4 = x x 4 = x 6 =. Since the pivot columns are the first, second, third, and sixth, x 4 and x 5 (which doesn t even appear in these equations) are both free. We get that the generic solution vector looks like x x 2 x 4 x = x x 4 = x 4 x 4 = x 4 + x 5, x 5 x 5 x 6 so the solution set is spanned by those two vectors. (b) These vectors are the second, third, and fourth columns of A. Looking at the corresponding columns of rref(a), we see that there s a linear dependence relation C2 + C + C4 =. That relation holds if we use the columns of A as well. Hence, [ [ 6 [ [ = If you didn t realize this fact, you can also solve this problem just like #2(b) above. (c) It is not injective because there are two free variables in the system. (There are columns that are not pivot columns.) However, the function is surjective because there is a pivot in each row.

4 MATH24: Linear Algebra Review for exam # 6//25 Page 4 4. When butane (C 4 H ) is burned, it combines with oxygen (O 2 ) to form carbon dioxide (CO 2 ) and water (H 2 O). Determine how many of each molecule are needed to balance the chemical equation. Solution. The slots in my vectors will mean, from top to bottom, carbon, hydrogen, and oxygen. Then the system that results from this chemical reaction is 4 x + x 2 + x + x 4 2 =. 2 2 Remember, the reactants and the product should have opposite sign since they re on opposite sides of the original chemical equation. We therefore should row-reduce the matrix I ll spare you the intermediate steps and tell you that the rref is This means that the general solution vector is. Picking x 4 = will clear all [ x4 /5 x 4 / 4x 4 /5 x 4 the denominators, and so we get our balanced chemical reaction 2C 4 H + O 2 8CO 2 + H 2 O. 5. Suppose that v is a vector that satisfies the equation A x = b. (a) If A w =, show that v + w also solves A x = b. (b) Suppose that A u = b, but u v. solutions. Give three such solutions. Explain why A x = has infinitely many Solution. (a) Plugging in v + w for x gives A( v + w) = A v + A w = b + = b. (b) We have the two equations A u = b and A v = b. We re supposed to get A times something equals. One way to make is to do b b, so suppose we subtract the two equations we have? That would give us A u A v = b b =. But the left side of this equation is A( u v)! Since u and v are assumed to be different, their difference is not the zero vector. Now I can write down three different vectors that solve A x = : u v, 2( u v), and ( u v) for example. 2 Indeed any multiple of u v works, since A ( c ( u v) ) = c ( A ( u v) ) = c =. 2 What s the principle at work here? The solution set to a homogeneous equation is a plane that goes through the origin. So once you know a nonzero vector that solves the system, everything in the span of that vector will be in the solution set. And there are infinitely many vectors in the span since there are infintely many possible different constants.

5 MATH24: Linear Algebra Review for exam # 6//25 Page 5 6. 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. (a) Suppose x is a 4 vector which has a zero as its top entry (so x = suppose M is any 4 4 matrix. Does M x also have a zero as its top entry? [ ), and Basically if this works out, it s a fluke. Here s an example that shows that it could be yes: = (Okay now that I ve just written that out, I don t know why I didn t make the matrix have every entry equal to zero. Maybe that would have made it more obvious that you ll get a zero at the top because you d just get all zeros.) Anyway, it could equally well be no, say if I just change the 2 to a in the top row: = What is actually true always is that if the top row of the matrix is all zeros, then no matter what you multiply it by, the top entry of the resulting vector will be zero (b) Let T be a linear transformation from R m to R n. Can the value of T on any vector of R m be calculated if you know its value on the standard vectors? Since if you know T ( e ),..., T ( e n ), then T is the same as the matrix transformation x T ( e ) T ( e 2 ) T ( e n ) x, and therefore you can calculate T ( v) for any v by multiplying this matrix centered above times v.

6 MATH24: Linear Algebra Review for exam # 6//25 Page 6 (c) Let S be a set of five vectors in R 4. Is S linearly independent? If you want to know whether or not the vectors are linearly independent, you should load them into a matrix, then row reduce that matrix and check whether there s a pivot in every column. In this case you re going to get a 4 5 matrix, that is, one with more columns than rows. That means there has to be a free variable in the system, so they cannot be linearly independent. (d) Suppose { u, v, w} is a linearly dependent set. Is u in the span of v and w? If the set is linearly dependent, then there is a linear dependence relation of the form c u + c 2 v + c w =. Now if c, then we can rearrange this equation to have it say ( u = c ) ( 2 v + c ) w, c c and that would show that u span( v, w). The issue here is that doing that involves dividing by c, and if c = then u doesn t appear in the equation so you can t solve for it. That s how I ll create my example where it can t happen. It could be yes: u = [, v = [, and w = [. The set is linearly dependent because u v w =. And u span( v, w) because u = v + w. It could be no: u = [ [ [, v =, and w =. The set is linearly dependent because v + w =. But u can t be in the span of v and w because everything in the span of those two vectors looks like c + d = c d =, but u doesn t have a zero as its top entry.

7 MATH24: Linear Algebra Review for exam # 6//25 Page 7 7. Write down the standard matrix for the transformation T : R 2 R 2 which consists of the following operations: reflecting across the line y = x, scaling the x direction by a factor of 2, then rotating 45 degrees counterclockwise. Solution. We have to track where the standard basis vectors go; that will tell us what the columns of the matrix will be. Let s start with e = (, ). When you flip it across the line y = x, you get the point (, ). Then scaling the x direction by a factor of two doesn t do anything you still have (, ). The rotation will move the point to ( 2, 2 ). 2 2 For e 2 = (, ), flipping across the line y = x turns the point into (, ). Then scaling makes it ( 2, ). When we rotate this vector, we get ( 2, 2). standard matrix for this transformation is [ So the 8. The set { e, e 2, e } of standard basis vectors in R is a linearly independent set. Suppose that T : R R is a linear transformation. Show by example that the set {T ( e ), T ( e 2 ), T ( e )} could be linearly dependent. (So to answer this problem you have to choose a linear transformation T, and then demonstrate a linear dependence relation among the vectors T ( e ), T ( e 2 ), and T ( e ).) Solution. One option is to{ choose} the{ linear } transformation that takes everything to. 4 Then {T ( e ), T ( e 2 ), T ( e )} =,, =, and a set with the zero vector in it is linearly dependent since =. (This is a linear dependence: we found a way to solve c = but c.) Another option is to have T be a matrix transformation that has two columns equal, that is, T ( e ) = T ( e 2 ) for instance. So if T is x for instance, then T ( e ) = T ( e 2 ) = [, and therefore the set {T ( e ), T ( e 2 ), T ( e )} satisfies the linear dependence relation T ( e ) T ( e 2 ) + T ( e ) =. It s on the circle of length two and it s also on the line y = x, so it s either ( 2, 2) or ( 2, 2), and it s pretty clear that it s in the third quadrant if you re drawing pictures. 4 This corresponds to the matrix transformation x x.