MATH 15a: Linear Algebra Exam 1, Solutions

Transcription

1 MATH 5a: Linear Algebra Exam, Solutions. Let T : R 3 R 4 be the linear transformation with T( e ) = 2 e + e e 3 4 e 4, T( e 2 ) = e e 2 +2 e 3 +6 e 4, and T( e 3 ) = 4 e e 2 +7 e 3 +8 e 4. (a) (6 points) Write the standard matrix for T. Denote this matrix A. Answer: Since it s a transformation R 3 R 4, the matrix needs to be 4 3. It is: 2 4 A = (b) (4 points) What is the rank of A? Answer: We compute the RREF of A: Since this has two pivots, the rank is 2. (c) (4 points) Find all solutions to the equation T( x) = Answer: Use the computation above:

2 Thus, the general solution is: x = 2 x 3 x 2 = 3 2x 3 x 3 = free. (d) (4 points) For what values of c does the equation T( x) = 5 0 c have at least one solution? Answer: We can follow through the row-reduction exactly the same way: c c c c 20 Whenever c 26, we will get a row that looks like [ ], which means that the system is inconsistent. Thus c = 26 is the only value for which there is a solution. 2. Let T : R n R m be a linear transformation. (a) (8 points) Complete the definition: The image of T is... Answer: The image of T is the set of all vectors v R m such that v = T( u) for some u R n. (b) (8 points) Prove that the image of T is a subspace of R m. Answer: We need to check three things. Since T( 0) = 0, we have 0 im(t). If v, v 2 im(t), then v = T( u ) and v 2 = T( u 2 ) for some u, u 2 R n. Since T is linear, T( u + u 2 ) = T( u )+T( u 2 ) = v + v 2. Hence, v + v 2 im(t). If v im(t), then v = T( u) for some u R n. For any scalar k, we have T(k u) = kt( u) = k v. Hence, k v im(t). 3. (a) (6 points) What is the matrix for counterclockwise rotation of R 2 through an angle θ? 2

3 Answer: cosθ sinθ sinθ cosθ (b) (6 points) For any angle θ, let L θ denote the line in R 2 obtained by rotating the x-axis counterclockwise about the origin through the angle θ. As we saw in class, the matrix for reflection across L θ is cos2θ sin2θ. sin2θ cos2θ (You don t have to prove this.) Now, for two angles α, β, let T : R 2 R 2 be the linear transformation consisting of reflection across L α followed by reflection across L β. Find the matrix for T. Answer: The matrix for T is cos2β sin2β cos2α sin2α = sin2β cos2β sin2α cos2α cos2βcos2α+sin2βsin2α cos2βsin2α sin2βcos2α sin2βcos2α cos2βsin2α cos2βcos2α sin2βsin2α (c) (3 points) T is a rotation through some angle. What angle? (Hint: Use the angle addition formulas for sine and cosine. If you can t remember them, use your answer for (a) to figure them out, as we ve discussed in class.) Answer: The angle addition and subtraction formulas are: cos(x+y) = cosxcosy sinxsiny sin(x+y) = cosxsiny +sinxcosy cos(x y) = cosxcosy +sinxsiny sin(x y) = cosxsiny +sinxcosy If you can t remember these, you can derive the addition formula from the fact that rotation by x followed by rotation by y is the same as rotation by x+y, so cos(x+y) sin(x+y) cosy siny cosx sinx = sin(x+y) cos(x+y) siny cosy sinx cosx cosxcosy sinxsiny cosxsiny sinxcosy = cosxsiny +sinxcosy cosxcosy sinxsiny The subtraction formulas follow from replacing y by y and using the facts that cos( y) = cosy and sin( y) = siny. Using these, we recognize the solution to (b) as so T is rotation by 2β 2α. [ cos(2β 2α) sin(2β 2α) sin(2β 2α) cos(2β 2α) ], 3

4 (d) (3 points) Draw a sketch illustrating where T sends the standard basis vectors in the case where α = π/4 and β = π/2. Answer: The following picture shows what happens to e and e 2 after the two reflections. From the third picture, we can see that the composition of the two reflections is rotation by π/2. Note that 2β 2α = π/2, as expected. L π/4 Ref π/4 ( e ) Ref π/2 (Ref π/4 ( e )) e 2 Ref π/4( e 2 ) e L π/2 Ref π/2 ( Ref π/4 ( e 2 )) 4. (8 points each) Say whether each of the following functions is linear or not. If it is linear, find its matrix. If it is not linear, give an example that shows why not. (a) (b) x x +2x 3 T x 2 = 2x 4x 2 +x 3 x 3 00x 2 +7x 3 Answer: It is linear, and its matrix is x T x 2 2x +x = 3 x x 3 +3x2 2 3 Answer: This function is not linear. For instance, let s notice that T 0 2 =, 0 while 2 T 0 = 0 4 2T (8 points each) True or false? If the statement is true, justify it. If it is false, provide a counterexample. (a) (8 points) If A is an n n matrix, and there is a vector b R n for which A x = b has a unique solution, then the equation A x = c has a unique solution for every c R n. 4

5 Answer: True. Since there is some b for which A x = b has exactly one solution, the rank of A must equal the number of columns, which is n. Since this is also the number of rows, for any c, the RREF of [A c] has a pivot in every row to the left of the bar. This means that the system A x = c is always consistent and there can never be any free variables. x (b) (8 points) The set consisting of all vectors y R 3 with xyz = 0 is a subspace z of R 3. 0 Answer: False. For example, the vectors and 0 are both in this set, but 0 their sum is not. 6. (a) (8 points) Find the inverse of the matrix 6 A = Answer: We do this by row-reducing [A I]: /2 0 / /2 0 / /2 0 / /2 3 5/ Therefore, A =

6 (b) (8 points) Without doing any additional Gauss-Jordan elimination, find all solutions to the system of equations x+6y z = 3 2y +z = 3x+3y +5z = 2. Answer: ThesolutiontoA x = bcanbefoundeasilyonceyouknowa : namely, x = A b. In this case, we have x y = 3 8 = 9 z so x = 80, y = 9, z = 37. You should check your work: (80)+6( 9) ( 37) = 3 2( 9)+( 37) = 3(80)+3( 9)+5( 37) = 2 6