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                                       . (d) No.      4 4 4T 1, T 2 4 2Tx x x   . (e) Yes.      5 1 2 5 1 5 2T T T     u u 0 0 0 u u .    5 1 5 1T Tm m m  u 0 0 u . 3 (i)   1 2 3 11 8 3T 1 T 1T7 3 4 13 11 2                                           . (ii) 1 2 3 44 24 41 1T 4T 3T0 3 4 52 33 517 17                                        , 0 2 3 33 16 11 1T 3T 2T1 3 4 39 22 117 17                                        . Thus 4 1 5 1      A 4 (i) nullity 3 2 1   .    2 2 1 2 1 1 2 1 1 2 1 2 1 2 0 3 0 0 3 0 3 2 4 0 2 6 4 3 0 0 1 3 a a a a a a a a a a a                                     . 1a . (ii) The range space is the column space of 1 2 1 2 2 2 3 2 3          with a basis 1 2 2 , 2 3 2                       . 5 (a) 1 4 5     e with 1 7  , 2 1 1     e with 2 2  . (working omitted) (b) 1 4 3 2         e with 1 5  , 2 1 0 2        e with 2 10  , 3 1 3 1        e with 3 10  . www.KiasuExamPaper.com 693

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 3 Solutions) Page 3 of 15 (c) 0 2 1 0 1 3 1 2 4 1                     3 2 3 2 2 3 1 4 4 2 3 2 1 4 2 3 8 2 6 2 2 4 2 1                                          Eigenvalues: 1 2 1   and 3 0  When 1 , we solve 1 2 1 0 1 2 1 0 2 4 2 0 x y z                          , i.e. 1 2 1 0 0 0 0 0 0 0 0 0 x y z                         . Let ,y z    , then 2x    . 2 0 1 1 0 1 x y z                               , so two corresponding vectors are 1 1 0 1        e and 2 2 1 0        e with 1 2 1   When 0 , we solve 0 2 1 0 1 3 1 0 2 4 1 0 x y z                          , i.e. 1 3 1 0 0 2 1 0 0 0 0 0 x y z                         . Let 2z  , then y  and x  , a corresponding vector is 3 1 1 2        e , 3 0  6 (i) The eigenvalue are –7 and 2 with eigenvectors 4 5      and 1 1      respectively (working omiited). Thus, 4 1 5 1     P and 7 0 0 2     D . (ii)  det 4 5 9  P , 1 1 11 5 49        P . Then 5 5 1 4 1 16807 0 1 11 5 1 0 32 5 49 67228 32 1 11 84035 32 5 49 67068 673561 84195 839079 7452 7484 9355 9323                                    A PD P www.KiasuExamPaper.com 694

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 3 Solutions) Page 4 of 15 Practice Questions 8 (i) Reduce the matrix 1 2 3 1 2 3 1 2 3 3 0 5 0 6 4 0 3 2 2 5 0 0 9 6 0 0 0                               , so  dim 2V  . Reduce the matrix 1 2 2 1 2 2 1 2 2 3 4 0 0 2 6 0 1 3 3 3 3 0 3 9 0 0 0                                 , so  dim 2W  . (ii) A basis for V is 1 2 3 , 0 2 5                       . Let x y W z         , then the rank of 1 2 3 0 2 5 x y z         must remain at 2.   1 2 1 2 1 2 3 0 0 6 3 0 6 3 2 5 0 9 2 0 0 2 1.5 3 x x x y y x y x z z x z x y x                                   . Therefore  2 1.5 3 0z x y x    , i.e. 5 3 2 0x y z   . (iii) A basis for W is 1 2 3 , 4 3 3                        , which is the null space of A, so the rank of A must be 1. Let  a b cA , substituting the vectors in this basis into AX 0 , we have 3 3 0 3 3 0 2 4 3 0 2 3 0 a b c a b c a b c b c               . We need a non-trivial solution. We may 2c , then 3b and 3a , so A can be  3 3 2  . (iv) For a vector u to be in V W , i.e. in both V and W, 1 2 1 2 3 0 3 4 2 5 3 3                                               u 1 2 1 2 0 3 0 3 4 0 2 5 3 3 0                               . 1 2 1 2 1 2 1 2 1 2 1 2 3 0 3 4 0 6 0 2 0 6 0 2 2 5 3 3 0 9 1 1 0 0 1 2                                       . Let t  , then 2t  . Thus 1 2 0 2 3 4 2 3 3 3 t t t                                u , so a basis for V W is 0 2 3             . www.KiasuExamPaper.com 695

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths