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Transcript

3 2 −1 2. The rank of A is 1. Let A = 1 3 4 5 1 (a) 2 (b) 3 (c) 0 (d) 4 (e) 1 2. Let P2 = {a0 +a1 t+a2 t2 } where {a0 , a1 , a2 } range over all real numbers, and let T : P2 → P2 be a linear transformation dedifined by T (a0 + a1 t + a2 t2 ) = a1 + 9a2 t � � If T p(t) = λp(t) for some non-zero polynomial and some real number λ, then p(t) is called an eigenvector corresponding to λ. The linear transformation T has an eigenvector: (a) 100 (b) 0 (c) t2 (d) t + 9t2 (e) t � � 1 6 3. Let {λ1 , λ2 } be two eigenvalues of . Then the product of two eigenvalues, λ1 λ2 , is 5 2 equal to (b) −28 (a) 28 (c) 3 (d) 4 (e) 7 � � 2 (d) 3 � � 3 (e) 2 � � 4 −3 5. The matrix A = has a complex eigenvector: 3 4 � � � � � � � � 3 4 1 4 (a) (b) (c) (d) 4i 3i −1 3 � � 1 (e) i � � � � 1 1 4. Let b1 = , Let b2 = , −1 2 � � 5 Let x = and B = {b1 , b2 } Find [x]B . 4 � � � � � � 1 1 5 (a) (b) (c) 2 −1 4 2 4 3 6. Let A = −4 −6 −3. The eigenvalues of A are 3 3 1 (a) 1, 2, 3 (b) 1, ±2 (c) 0, 1, 2 (d) 0, 1, −2 (e) 1, −2, −2 � � � � 5 3 7. Let S be the parallelogram determined by the vectors b1 = and b2 = and let 3 2 � � −2 99 A= . Compute the area of the images of S under the mapping v �→ Av. 0 1 (a) 3 (b) 99 (c) 2 (d) −2 (e) 5 (d) 4 (e) ±i � � 5 −5 8. Let A = . The complex eigenvalues of A are 1 1 (a) 1 ± 3i (b) 3 ± i (c) ±3i 1 9. For what value(s) of h will y be in the subspace spanned by {v1 , v2 , v3 }, if v1 = 2, 3 2 3 6 v2 = 1, v3 1 and y = 4. 3 4 h (a) 2 (b) 4 (c) 10 (d) 6 (e) 3 � � � � � � � � −9 −5 1 3 10. Let b1 = , b2 = , c1 = , c2 = , B = {b1 , b2 } and 1 −1 −4 −5 � � C = {c1 , c2 }. If P = [b1 ]C , [b2 ]C is the change-of-coordinates matrix from B to C, then find P . � � � � � � � � � � 1 1 −5 1 −5 3 −9 −5 6 4 1 3 (a) (b) (c) (d) (e) 9 4 1 1 −1 −5 −3 −4 −5 4 1 7 9a − 8b 11. Find a matrix A such that W = Col(A) where W = a + 2b and {a, b} range over −5a all real numbers. 9 −8 2 (a) 1 0 −5 1 0 (b) 0 1 0 0 9 −8 2 (c) 1 −5 0 � � � � 9 1 −5 0 1 0 (d) (e) −8 2 0 1 0 0 0 0 12. Let S = 0 , 2 . Then the subset S 1 −1 0 0 2 (a) 0 is orthogonal to (b) a basis of R3 1 −1 (c) spans R3 (d) a linearly dependent subset (e) a linearly independent subset 1 −4 2 0 8 −9 Use Cramer’s rule to solve Ax = 1. 13. Let A = −2 −1 7 0 0 14. Let P2 = {a0 +a1 t+a2 t2 } where {a0 , a1 , a2 } range over all real numbers, and let T : P2 → P1 be a linear transformation given by T (a0 +a1 t+a2 t2 ) = a1 +2a2 t. Suppose that B = {1, t, t2 } is a basis of P2 and C = {1, t} is a basis of P1 . (1) Find a matrix A such that [T v]C = A[x]B . (2) Find N ul(A) and Col(A). � � 1 7 2 15. Let A = . Find lim Ak . (Hint: Find the Diagonalization D of A use the formula −4 1 k→∞ 5 Ak = P Dk P −1 . Solutions 1. 3 2 −1 1 3 2 1 3 2 1 3 2 2 3 2 −1 0 −7 −7 0 1 1 There are Reduce to echelon form: 1 3 4 5 1 4 5 1 0 −7 −7 0 0 0 two pivots so rank is 2 2. 0 1 0 With the basis B = {1, t, t2 }, the matrix for T is 0 0 9. Because the matrix is triangular, 0 0 0 1 3 the characteristic polynomial is −λ and hence the eigenvalue is 0 and the eigenvector is 0, 0 2 so 100 = 100 + 0t + 0t is an eigenvector with eigenvalue 0. 3. The product of the roots, with multiplicities, of any polynomial with leading coeﬃcient 1 is (−1)n times the product of the roots where n is the degree: (t − λ1 )(t − λ2 ) · · · (t − λn ) = tn + · · · + (−1)n (λ1 · · · λn ). In this case λ1 λ2 = det A so the answer is −28. 4. � � 1 1 We need to solve Ay = x where A = . So reduce to reduced row echelon form −1 2 � � � � � � � � � � 1 1 5 1 1 5 1 1 5 1 0 2 2 Hence the solution is [x]B = −1 2 4 0 3 9 0 1 3 0 1 3 3 5. � � �4 − λ � −3 � = (4−λ)2 +9 = 16−8λ+λ2 +9 = λ2 −8λ+25. The characteristic equation is det �� 3 4 − λ� √ 8 ± 64 − 100 The roots are λ = = 4 ± 3i. The eigenspace for 4 + 3i is the null space for 2 � � � � −3i −3 1 and an eigenvector is . 3 3i i � � �2 − λ � � � 4 3 � � �−6 − λ � −3 �− −3�� = (2 − λ) det �� The characteristic equation is det �� −4 −6 − λ 3 1 − λ� � � 3 1−λ � � � � 3 �−4 � � � � � � −3� �−4 −6 − λ� = (2 − λ) (−6 − λ)(1 − λ) + 9 − 4(−4(1 − λ) + 9 + 4 det �� + 3 det � 3 3 1 − λ� 3� � � 3 −12 − 3(−6 − λ) and good luck to you! OR The sums of the five answers are all the diﬀerent and the sum of the eigenvalues is the trace of the matrix 2 + (−6) + 1 = −3. Hence the only choice for the eigenvalues in the given answers is 1, −2, −2. 6. 7. The area is the area of S times � the � absolute value of the determinant of A, which is 2. The �5 3� � = 10 − 9 = 1. area of S is det [b1 b2 ] = det �� 3 2� 8. 2 2 The characteristic equation is λ −trAλ+det A = λ −6λ+10 with roots λ = 3 ± i. 9. 1 2 We are looking for the h such that the equation Ax = y with A = 2 1 3 3 reduce the following matrix to reduced row echelon form: 1 2 3 6 1 2 3 6 1 2 3 6 1 2 1 1 4 0 −3 −5 −8 0 −3 −5 −8 = 0 3 3 4 h 0 −3 −5 h − 18 0 0 0 h − 18 + 8 0 To have a solution, h − 10 = 0. 10. We need to write b1 � � � 1 3 −9 1 −4 −5 1 0 � � � 1 3 −5 1 −4 −5 −1 0 6+ √ 36 − 40 = 2 3 1 so we need to 4 2 3 6 −3 −5 −8. 0 0 h − 10 � � � � � � � � � � −9 1 3 1 3 −9 = = ac1 + bc2 = a +b or solve x= . 1 −4 −5 −4 −5 1 � � � � � � � 3 −9 1 3 −9 1 0 6 6 so the first column of P is . 7 −35 0 1 −5 0 1 −5 −5 � � � � � � � 3 −5 1 3 −5 1 0 4 4 so the second column of P is . 7 −21 0 1 −3 0 1 −3 −3 � � 6 4 Hence P = . −5 −3 11. 9a − 8b 9 −8 9 −8 a + 2b = a 1 + b 2 so a matrix is 1 2 −5a −5 0 −5 0 12. Since there are only two vectors they can not be a basis for R3 . They can not even span R3 . If the two are dependent, one is a multiple of the other and this is clearly not the case. Hence they are independent. The dot product is −1 so they are not orthogonal. 13. To use� Cramer’s rule � we need to compute four determinants: � 1 −4 � � � � 2�� � �−2 8� � 1 −4� � � � � � � + 0 = −12 + 27 = 15 8 −9� = 2 det � det �−2 � − (−9) det �−1 � −1 7 7 �−1 � 7 0 � � �0 −4 � � � 2 � � �−4 2� � � � � + 0 = 14 8 −9� = 0 − 1 det � det �1 � 7 0 �0 � 7 0 � � � 1 0 � � � 2 � � � 1 2� � � � �−0=2 det �−2 1 −9� = −0 + 1 det � � −1 0 �−1 0 0� and� � � 1 −4 0� � � � � � 1 −4� � + 0 = −3 8 1�� = 0 − 1 det �� det ��−2 −1 7� �−1 � 7 0 14 15 2 Hence the solution vector is x = 15 . 3 − 15 14. A is a 2 × 3 matrix and the ith column is found by working out T on the ith basis element of B. � � 0 Column 1: T (1) = 0 so . 0 � � 1 Column 2: T (t) = 1 so . 0 � � 0 2 Column 3: T (t ) = 0 so . 2 � � 0 1 0 Hence A = . 0 0 2 The matrix A is already in echelon form and there are two pivot positions. Col(A) Hence 1 has dimension 2 and so is all of P1 . The space N ul(A) has dimension 1, 0 is clearly a 0 non-zero vector in the null space so it is a basis for it. 15. The characteristic equation for A is λ2 − trace(A)λ + det(A) = λ2 − 85 λ + 15 = (λ − 35 )(λ − 1). 25 3 Hence the eigenvalues are and 1. 5 � 4 2� � � 5 5 3 2 The eigenspace for is the null space for and this is spanned by . 4 2 −4 5 −5 −5 � 2 2� � � 5 5 1 The eigenspace for 1 is the null space for and this is spanned by 4 4 −1 −5 −5 � � �3 � � � 2 1 0 0 0 k −1 5 If P = and D = , then A = P DP . Note lim D = so −4 −1 0 1 0 1 k→∞ � �� � � � � �� � � � � � 1 1 2 1 0 0 1 −1 −1 2 1 0 0 4 2 2 1 k lim A = = = = . −4 −1 0 1 2 4 2 −2 −1 k→∞ 2 −4 −1 4 2 2 −4 −2 1. (•) (b) (c) (d) (e) 2. (•) (b) (c) (d) (e) 3. (a) (•) (c) (d) (e) 4. (a) (b) (c) (•) (e) 5. (a) (b) (c) (d) (•) 6. (a) (b) (c) (d) (•) 7. (a) (b) (•) (d) (e) 8. (a) (•) (c) (d) (e) 9. (a) (b) (•) (d) (e) 10. (a) (b) (c) (•) (e) 11. (a) (b) (•) (d) (e) 12. (a) (b) (c) (d) (•)