F07 - Matrices and Linear Spaces - Tutorial Set 2 (Solutions)

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 2 Solutions) Page 1 of 7 National Junior College 2016 – 2017 H2 Further Mathematics Topic F7: Matrices and Linear Spaces (Tutorial Set 2 Solutions) Basic Mastery Questions 1 (i) No. 0 1x S  , 0 1x S  , but    0 1 0 1 0 0x x x S      . (A1 fails) (ii) Yes. (To verify non-empty, A1 and A6 as T is a subset of 3 ) (iii) Yes. (To verify all the 10 axioms) (iv) Yes. (To verify all the axioms) (v) No.  1 1, 0, 0  0, A10 fails unless  1, 0, 0 is the zero vector.  1 2, 0, 0  0 , A10 fail unless  2, 0, 0 is the zero vector. However, there is only one zero vector in the vector space, so it is not a vector space. (vi) Yes. (To verify all the axioms) 2 (To verify non-emptiness, A1 and A6 as they are subsets of 3 ) (i)   1, 1, 2 . (ii)     2,1,0 , 3,0,1 . 3 (To verify non-emptiness, A1 and A6 as they are subsets of 2P ) (i)   2,1 2x x (ii)   2 2x x  4 (i), (ii), (iv) and (iv): Yes (iii) No.  1,1,0 the set but    2 1,1,0 2, 2,0 the set  as    2 2 2 5 0 2 1    . (v) No.  0,1,1 the set but    2 0,1,1 0, 2, 2 the set  . (vii) No. An empty set is not a vector space. (viii) No.  1,0,1 the set ,  1,0, 1 the set  but      1,0,1 1,0, 1 2,0,0 the set    . 5 To show linearly independent, let        1,0, 1 0,2,0 1,1,1 0,0,0a b c     . 1 0 1 0 1 0 1 0 0 2 1 0 0 2 1 0 1 0 1 0 1 0 1 0 a b c a a b c b a b c c                                      . Since the determinant of the coef ficient matrix is 4, the coefficient matrix is invertible and there is exactly one solution to the system which is the trivial solution. Thus, the set is linearly independent. To show spanning, let        1,0, 1 0, 2,0 1,1,1 , ,p q r x y z    . 1 0 1 1 0 1 0 2 1 0 2 1 1 0 1 1 0 1 p q r x p x p q r y q y p q r z r z                                      . Since the determinant of the coefficient matrix is 4, the coefficient matrix is invertible and there is exactly one solution to the system for any vector  , ,x y z in 3 . Thus, the set spans 3 . (i) Yes (ii) No. It is not linearly independent as      0,0,0 0 1,0, 1 0 0, 2,0   (iii) No. It is not linearly independent as        0, 2,0 0 1,0, 1 0 1,1,1 2 0,1,0    (iv) No.        0,0,1 0,2,0 1,1,1 , 2 ,k l l k l l    has no solution, it does not span 3 www.KiasuExamPaper.com 659

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 2 Solutions) Page 2 of 7 6 (a) Let  : ,U V W   v w v w . Since V0 and W0 , U  0 0 0 . Thus U is non-empty. 1 2, U u u , 1 2, V v v and 1 2, W w w s.t. 1 1 1 u v w , 2 2 2 u v w .        1 2 1 1 2