Math 314H EXAM I. 1. (28 points) The row reduced echelon form of the augmented matrix for the system. is the matrix

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1 Math 34H EXAM I Do all of the problems below. Point values for each of the problems are adjacent to the problem number. Calculators may be used to check your answer but not to arrive at your answer. That is, you must show all of the work necessary to derive your answers. Use the blank sides of the exam sheets if you need more space than what is provided. Good luck!. (28 points) The row reduced echelon form of the augmented matrix for the system x + x 2 +2x 3 + x 4 = 7 x + 4x 2 x 3 + 4x 4 +3x 5 = 2x + 5x 2 +x 3 + 5x 4 2x 5 = 3 x + x 2 +2x 3 + 6x 4 = is the matrix 3 3 In the following questions A denotes the coefficient matrix for the system. (a) What is the general solution to the system? From the rref of the system, we see that the general solution is x = 3s x 2 = 3 + s x 3 = s x 4 = x 5 = (b) What is the general solution to the homogeneous system Ax =? Replacing the last column of the rref with zeros, we find that the general solution to the homogeneous system is x = 3s x 2 = s x 3 = s x 4 = x 5 = (c) Are the columns of A linearly independent? Explain your answer. No, the oolumns of A are not linearly independent since the system Ax = has non-trivial solutions by part (b).

2 (d) Are the rows of A linearly independent? Why or why not? To answer this question, we take the transpose of A and put it in reduced row echelon form: A T = Since every column has a pivot, A T x = has only the trivial solution. Therefore, the columns of A T (and hence the rows of A) are linearly independent. (e) Do the columns of A span R 4? Again, justify your answer. Yes, the columns of A span R 4 since there is a pivot in every row of the rref of A. (f) True or false: Ax = b has infinitely many solutions for every vector b R 4. This is true. We know from part (e) that Ax = b is consistent for every b R 4. Since Ax = has infinitely many solutions, we have infinitely many solutions to Ax = b for every b. (See Theorem 6 on page 52 of your text.) 2. (5 points) Let S = {u, v, w} where u = 2 4 7, v = 5 3, and w = (a) Is S linearly independent? If not, find scalars c, c 2 and c 3 (not all zero) such that c u + c 2 v + c 3 w =. To answer this question we write the three vectors as the columns of a matrix and put the matrix in reduced row echolon form: 5 4 A = = R Now, the columns of A are linearly independent if and only if Ax = has only the trivial solution. But from R (the rref of A) we see that there are indeed non-trivial solutions to Ax =. Thus, the columns of A are linearly dependent. The general solution to Ax = is x = 4s, x 2 = 3s, and x 3 = s, where s is arbitrary. We 4 can get a particular solution if we let s =. Thus, if y = 3 then Ay =. From the way we multiply matrices, this means that 4u + 3v + w = (since u, v, and w are the columns of A). Thus, w = 4u 3v.

3 (b) Is the vector b = 9 in the span of S? Justify your answer. This question is the same as asking if b is in the span of the columns of A, where A is the matrix defined in part (a). Put another way, we are asking if the system Ax = b is consistent. To answer this, we form the augmented matrix for the system and row reduce it: At this point, we can see that the system is inconsistent, since the second and third equations (corresponding the the second and third rows of the last matrix) cannot simultaneously hold. Thus, b is not in the span of S. (c) True of false: Span{u, v, w} = Span{v, w}. This is true. First, it is clear that every linear combination of v and w is also a linear combination of u, v and w. (Just let the coefficient of u be zero.) Thus, Span{v, w} is a subset of Span{u, v, w}. The real question is, why is every linear combination of u, v, w a linear combination of v and w? From part (a) of this problem, we have the equation 4u + 3v + w =. We can solve for u in terms of v and w: u = 3 4 v + 4 w. Now let c u + c 2 v + c 3 w be an arbitrary element of Span{u, v, w}. Substituting our equation for u into this linear combination, we get c u + c 2 v + c 3 w = c ( 3 4 v + 4 w) + c 2v + c 3 w = ( 3 4 c + c 2 )v + ( 4 c + c 3 )w. This shows that every linear combination of u, v and w is also a linear combination of v and w. Hence, Span{u, v, w} = Span{v, w}. 3. (5 points) Let T : R 3 R 3 be the linear transformation which rotates all vectors 9 degrees counterclockwise about the z-axis. (a) Find the standard matrix for T. To find the standard matrix for T, we just have to find T (e ), T (e 2 ), and T (e 3 ), where {e, e 2, e 3 } is the standard basis for R 3. If we rotate e (which is just the unit vector pointing in the positive x-direction) 9 degrees counterclockwise about the z-axis, we just get e 2, the unit vector pointing in the positive y-direction. Therefore T (e ) = e 2 =.

4 Similarly, one finds T (e 2 ) = and T (e 3 ) =. Therefore, the standard matrix for T is A =. (b) Find T (u) where u = 2. 5 To find T (u), we just multiply u by the matrix A: 2 T (u) = Au = 2 =. 5 5 (c) Find the standard matrix for T. There are two ways one can do this. The first is to notice that the inverse of T is simply a 9 degree rotation clockwise about the z-axis and then repeat the procedure we did in part (a). The second method is to compute the inverse of the matrix A using row reduction: [A I 3 ] = = [I A ]. Therefore, A =. 4. (5 points) For each of the following matrices find its inverse or explain why it is not invertible. [ ] 3 4 (a) 5 6 Since ad bc = (3)(6) (5)(4) = 8, we know that this matrix is invertible. Using the formula for the inverse of a 2 2 matrix, we find that the inverse is: [ ]

5 (b) (c) [ ] For this matrix, ad bc = ( 8)( 6) (4)(2) =, so the matrix is not invertible. 5 3 As in Problem 3(c), we use the row reduction method: Since the first half of the matrix is the identity matrix, we know that our matrix is invertible and that the inverse is the second half of the matrix. 5. (2 points) Let T : R 5 R 4 be the linear transformation given by T (x) = Ax where A is the coefficient matrix from Problem. (a) Is T one-to-one? Explain your answer. No, T is not one-to-one since Ax = has nontrivial solutions by Problem (b). (b) Is T onto? Explain your answer. Yes, T is onto since the columns of A span R 4 by Problem (e). 6. (5 points) Let A and B be n n matrices. For each of the following, either prove your answer or give an counterexample. (a) Suppose AB is invertible. Must A be invertible? Yes, A must be invertible. For, let C be the inverse of AB. Then (AB)C = I. By associativity, we have A(BC) = I. Therefore, A is invertible and A = BC. (See Theorem 8(j) on page 2 of your text.) (b) Suppose AB =. Must A = or B =? No, both A and B can be nonzero. For example, let A = It is easily checked that AB =. (c) Suppose A and B are invertible. Must A + B be invertible? [ ] and B = [ ]. No, A+B does not have to be invertible. For example, let A = I (the 2 2 identity matrix) and B = I. Both A and B are clearly invertible, but A + B =, which is clearly not invertible.