Math 2940: Prelim 1 Practice Solutions

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1 Math 294: Prelim Practice Solutions x. Find all solutions x = x 2 x 3 to the following system of equations: x 4 2x + 4x 2 + 2x 3 + 2x 4 = 6 x + 2x 2 + x 3 + x 4 = 3 3x 6x 2 + x 3 + 5x 4 = 5 Write your answer in parametric vector form, that is, x = v + c v + c 2 v c k v k, where the vectors v,..., v k are specific numerical vectors in R 4 that you must find, and any choice of values for the real numbers c,..., c k yields a valid solution to the system. Solution: Row-reduce the augmented matrix: The new equations are: This yields x + 2x 2 x 4 = 2 x 3 + 2x 4 = x = 2 2x 2 + x 4 x 2 = x 2 x 3 = 2x 4 x 4 = x 4 so the solutions are 2 2 x = +x 2 +x or, if you like, x = +c +c 2 2.

2 2. Determine whether or not the vectors 5 7 v =, v 2 = 2, v 3 = are linearly independent. If not, provide a linear dependence relation that they satisfy. Solution: Row-reduce the matrix A = [ v v 2 ] v Since there are only two pivots, there will be a nontrivial solution to Ax =, which we can use to find a linear dependence relation among v, v 2, v 3. Continue the row-reduction: The solutions x = x x 2 to Ax = satisfy the equations x 3 x x 3 = x 2 + 2x 3 = All we need is a single nontrivial solution, so we can choose any value we like for x 3 (except x 3 =, which leads to the trivial solution). Say x 3 =, then x = and x 2 = 2. The matrix equation Ax = is equivalent to the linear dependence relation =. 3. Let v, v 2, v 3 be in R n, and let A = [ v v 2 v 3 ]. Determine whether each if-then statement is true or false. If true, explain why. If false, provide a specific numerical example of vectors v, v 2, v 3 for which the if holds but the then does not. (a) If {v, v 2, v 3 } is linearly independent, then v 3 is not in Span(v, v 2 ). Answer: True. Suppose v 3 Span(v, v 2 ), then one can write v 3 as a linear combination of v and v 2, say v 3 = c v + c 2 v 2. This means c v + c 2 v 2 v 3 =

3 is a linear dependence relation among v, v 2, v 3. (Note that the coefficient of v 3 is nonzero.) (b) If v 3 is not in Span(v, v 2 ), then {v, v 2, v 3 } is linearly independent. Answer: False. Say v = but v, v 2, v 3 are linearly dependent. [ [ [ 2, v ] 2 =, v ] 3 =. Then v ] 3 / Span(v, v 2 ) (c) If {v, v 2, v 3 } is linearly independent, then the only solution to Ax = is x =. Answer: True, because if A x x 2 x 3 = with x, x 2, x 3 not all zero, then x v + x 2 v 2 + x 3 v 3 = is a linear dependence relation among the v i. (d) If the equation Ax = has at least one solution, then {v, v 2, v 3 } is linearly dependent. Answer: False. The correct statement would be If the equation Ax = has at least one nontrivial solution, then {v, v 2, v 3 } is linearly dependent. For an easy counterexample, let v, v 2, v 3 be respectively e, e 2, e 3 R 3. The equation Ax = does have at least one solution, namely x =, but the v i are linearly independent. (e) If {v, v 2, v 3 } is linearly independent, then so is {3v, 5v 2, 2v 3 }. Answer: True. We ll show that the only choice of c, c 2, c 3 that makes c (3v ) + c 2 (5v 2 ) + c 3 ( 2v 3 ) = is c =, c 2 =, c 3 =. Indeed, if the above equation is true, then (3c )v + (5c 2 )v 2 + ( 2c 3 )v 3 =. This is a linear combination of v, v 2, v 3 to get the zero vector. Since we are given that v, v 2, v 3 are linearly independent, it must be true that 3c =, 5c 2 =, 2c 3 = which immediately implies that c = c 2 = c 3 =. 4. (a) Let A be an n n matrix. Suppose that for a particular b in R n, the equation Ax = b has no solutions. Is A invertible? What can you say about the number of solutions to Ax =? Explain your reasoning. Solution: If A were invertible, then for every b R n the equation Ax = b would have exactly one solution. Therefore A isn t invertible. In particular, A

4 has fewer than n pivots, so when we solve Ax = there will be at least one free variable, leading to infinitely many solutions. (b) Let A be an n n matrix. Suppose that for a particular b in R n, the equation Ax = b has exactly one solution. What can you say about the number of solutions to Ax =? Is A invertible? Explain your reasoning. Solution: Since Ax = b has at least one solution, the solutions to Ax = b are a parallel translate of the solutions to Ax =. In this case there is exactly one solution to Ax = b, so there s also exactly one solution to Ax = (which has to be x = ). This implies that the columns of A are linearly independent, so A is invertible. 5. Consider the linear transformation T : R 2 R 2, T (x) = x. (a) Fill in the blanks: T represents a clockwise rotation of followed by a dilation by a factor of. degrees Since T (e ) = and T (e 2 ) =, the standard square grid determined by e, e 2 is sent by T to the diagonal grid shown below. Answer: The blanks are 45 degrees and 2. (b) Let C be the circle of radius centered at the area of T (C).. Use a determinant to find Solution: Let A =. The area of T (C) equals det A multiplied by the area of C. Since det A = 2 and C has area π, the area of T (C) is 2π.

5 (c) Draw C and T (C) on two separate graphs. Use the graph of T (C) to verify your answer from part (b). Solution: The grid lines are optional but they help you draw the graph of T (C). (The circular curves are supposed to be fully connected.) 3 3,. There- 2 T (C) is a circle centered at passing through fore its radius is 2 and its area is 2π. (d) What is the area of T (T (C))?,, Solution: The area of T (T (C)) is det A multiplied by the area of T (C), which is 2 2π = 4π. 6. (TRICK QUESTION) Consider the invertible matrix 2 4 A = Find all solutions to the equation A x =. Solution: A x = if and only if x = A. Therefore the unique solution 4 is x = A = 5.

6 7. Given the following determinants: = 9, 3 2 = 6 Compute the determinant Solution: Since [ 2 ] [ 38 = 2 ] [ ], we have = = 9 2 = Let A be a 3 4 matrix. Determine whether each statement is true or false. Explain your answers. (a) If b is in the span of the columns of A, then the equation Ax = b has a solution. Answer: True. Label the columns of A by a, a 2, a 3, a 4. If b = c a + c 2 a 2 + c 3 a 3 + c 4 a 4, then A c c 2 c 3 c 4 = b. (b) If thereduced row echelon form of A has two pivots, then the equation Ax = b b 2 b 3 has a solution if and only if b 3 =. Answer: False. When solving Ax = b, we set up the augmented matrix [ A b ]. Suppose that after row-reduction, we get the augmented matrix [ U d ], where U = rref(a). The first and second rows of U have pivots, while the third row is all zeros. Therefore, the original equation Ax = b has a solution if and only if the third coordinate of d is zero. However, the third coordinate of d will usually be different from the third coordinate of b, since we performed various row operations to get from b to d.

7 (c) The columns of A are linearly dependent. Answer: True. Any list of 4 vectors in R 3 is linearly dependent, since 4 > 3. Alternatively, the row-reduction of A must have at least one free column, which produces a nontrivial solution to Ax =. (d) The columns of A span all of R 3. Answer: False. (That is, it could be true, but it is not necessarily true.) For example, we could have A = Use the LU factorization = to solve the equation x x 2 = x 3 6 Solution: First solve The equations are y y 2 = y 3 6 Then solve y = 2 = y = 2 y + y 2 = 4 = y 2 = 2 2y + y 3 = 6 = y 3 = x x 2 = 2. 2 x 3 2 Writing the equations and working from the bottom up, 4x + 3x 2 5x 3 = 2 = x = /4 2x 2 + 2x 3 = 2 = x 2 = 2 2x 3 = 2 = x 3 =.

8 . Let S : R k R n and T : R n R m be linear transformations given by S(x) = Bx and T (y) = Ay, where B is an n k matrix and A is an m n matrix. Define R : R k R m by R(x) = T (S(x)) = ABx. (a) Explain why the range of R is contained in the range of T. R k S R n T R m Solution: The range of R is the set of all possible values of T (S(x)) for x R k. The range of T is the set of all possible values of T (y) for y R n. Suppose b R m is in the range of R. Then we can write b = T (S(x)) for some x. If we set y = S(x), then b = T (y), which confirms that b is in the range of T. (By the way: If S is onto, so that the range of S is all of R n, then the range of R is equal to the range of T. Even if S isn t onto, it is still possible for the ranges of R and T to be equal. The question is whether the extra values of y R n that can t be expressed as S(x) lead to new values of T (y).) (b) Denote the columns of A by a,..., a n, and denote the columns of AB by v,..., v k. What can you say about the relationship between Span(a,..., a n ) and Span(v,..., v k )? Solution: Span(a,..., a n ) equals the range of T, and Span(v,..., v k ) equals the range of R. Thus, Span(v,..., v k ) is contained within Span(a,..., a n ). Another way of seeing this is to note that every b Span(v,..., v k ) can be written as b = ABx for some x R k. If we write b = A(Bx) and define y = Bx, then the equation b = Ay expresses b as a linear combination of the columns of A. (c) Suppose that k = n = m, so that A, B, and AB are all square matrices. Prove that if AB is invertible, then A is invertible. Solution: By part (b), the span of the columns of AB is contained within the span of the columns of A. If AB is invertible, then the columns of AB span all of R m, which means that the columns of A must also span all of R m. This proves that A is invertible (by the Invertible Matrix Theorem). An alternative solution is to write (AB)(AB) = I, therefore A[B(AB) ] = I. Let C = B(AB), so that AC = I. The Invertible Matrix Theorem implies that also CA = I, so C = A.