F07 - Matrices and Linear Spaces - Tutorial Set 3 (Solutions)

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 3 Solutions) Page 1 of 15 National Junior College 2016 – 2017 H2 Further Mathematics Topic F7: Matrices and Linear Spaces (Tutorial Set 3 Solutions) Basic Mastery Questions 1 (a) 1 3 1 3 1 3 2 1 0 7 0 1 1 1 0 4 0 0                               . A basis for the row space is     1 3 , 0 1 ; a basis for the column space is 1 3 2 , 1 1 1                       . rank 2 . nullity 2 2 0   . The basis for the null space is 0 0         . (b) 1 2 1 1 2 1 1 2 1 3 5 1 0 1 2 0 1 2 13 23 7 0 3 6 0 0 0                                    . A basis for the row space is     1 2 1 , 0 1 2  ; a basis for the column space is 1 2 3 , 5 13 23                        . rank 2 nullity 3 2 1   . To find the null space, let z  , then 2y  and 3x  , so a basis for the null space is 3 2 1              . (c) 2 1 3 3 2 1 3 3 2 1 3 3 0 3 1 2 0 3 1 2 0 3 1 2 4 5 5 8 0 3 1 2 0 0 0 0                                   . A basis for the row space is     2 1 3 3 , 0 3 1 2   ; a basis for the column space is 2 1 0 , 3 4 5                       ., rank 2 , nullity 4 2 2   . To find the null space, let 3 , 6z t    , then 4y    and 5 7x    , so a basis for the null space is 5 7 1 4 ,3 0 0 6                              . www.KiasuExamPaper.com 692

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 3 Solutions) Page 2 of 15 2 (a) Yes.  1 2 1 2 1 2 1 21 2 1 1 1 2 1 2 1 2 1 2 1 21 2 T T T Tx x x x kx kx x xk x x y y y y y y y yy y                                                         .. 1 1 1 1 1 1 1 1 1 1 2 1 1 T T Tx mx kmx kx xm m my my my y y                                   . (b) No. 2 2 2 2 2 2 0 1 1 1 1 0 1 2 0 1T ,T ,T T T T1 1 0 1 0 1 1 0 1 1                                                                    . (c) Yes. 1 1 2 2 1 2 1 2 1 2 1 2 3 3 1 1 2 2 1 2 1 2 1 2 1 2 T Ta b a b a a b b a a b b c d c d c c d d c c d d                                     , and 1 1 2 2 1 1 2 2 1 2 1 2 3 3 1 1 2 2 1 1 2 2 1 2 1 2 T Ta b a b a b a b a a b b c d c d c d c d c c d d                                        . 1 1 1 1 1 1 1 1 1 1 3 3 3 1 1 1 1 1 1 1 1 1 1 T T Ta b ma mb ma mb a b a bm m mc d mc md mc md c d c d                     