F07 - Matrices and Linear Spaces - Tutorial Set 1 (Solutions)

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 1 Solutions) Page 1 of 12 National Junior College 2016 – 2017 H2 Further Mathematics Topic F7: Matrices and Linear Spaces (Tutorial Set 1 Solutions) Basic Mastery Questions 1 2 3 3 25 1 1.5 1.5 12.5 1 1.5 1.5 12.5 1 1.5 1.5 12.5 3 2 3 24 3 2 3 24 0 2.5 1.5 13.5 0 1 0.6 5.4 4 1 2 21 4 1 2 21 0 5 4 29 0 5 4 29 1 1.5 1.5 12.5 1 1.5 1.5 12.5 0 1 0.6 5.4 0 1 0.6 5.4 0 0 1 2 0 0 1 2                                                          1 0 0.6 4.4 1 0 0 3.2 0 1 0.6 5.4 0 1 0 4.2 0 0 1 2 0 0 1 2                            2 1 1 03 2 4 9 130 3 30 1 6 12 3 92 0 2 T a a a                        AB .   4 12 9 3 13 9 TT T a a             BA AB . 4 2016 2017 2016 2019 2017 2018 22018 2019      . 1 2019 2017 2 22016 2017 2019 20171 2018 2019 2018 2016 1009 10082                   . 5         1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0. 0 0 0 0 0 a b c b c a c a b b a c a b a c a b a c a a c b c a b b c a b a c                      6 (i)(a)       1 1 0 1det 1 3 1 0 7 3 0 2 17 0 2 0        A . (i)(b)      det 1 1 0 0 7 0 2 3 1 2 1 0 7 1 1 0 0 3 1                    A . (i)(c) 1 3 0 1 3 0 1 3 0 0 1 1 0 1 1 0 1 1 1 1 1 1 2 7 0 0 1 0 0 0 1          . www.KiasuExamPaper.com 626

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 1 Solutions) Page 2 of 12 (ii)(a) 1 3 0 1 0 0 1 3 0 1 0 0 1 3 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 2 7 0 0 0 1 0 1 0 2 0 1 0 0 1 2 1 1                                    1 0 3 1 3 0 1 0 0 7 0 3 0 1 1 0 1 0 0 1 0 2 0 1 0 0 1 2 1 1 0 0 1 2 1 1                          . Thus 1 7 0 3 2 0 1 2 1 1           A . (ii)(b) 11 1 1 77 0C   , 12 0 1 22 0C   , 13 0 1 22 7C   , 21 3 0 07 0C   , 22 1 0 02 0C   , 23 1 3 12 7C   , 31 3 0 31 1C   , 32 1 0 10 1C    , 33 1 3 10 1C   . Thus, 7 2 2 7 0 3 0 0 1 2 0 1 3 1 1 2 1 1 T adj                         A ,   1 7 0 3 1 2 0 1det 2 1 1 adj            A A A . 7(a) From Q1, we know that 2 3 3 25 3 2 3 24 4 2 21 x y z x y z x y z          can be reduced to 1.5 1.5 12.5 0.6 5.4 2 x y z y z z       . By backward substitutions, we have 2, 5.4 0.6 2 4.2, 12.5 1.5 4.2 1.5 2 3.2z y x           . 7(b) From Q1, we know the linear system can be further reduced to 3.2 4.2 2 x y z    , which is the solution. 7(c) 2 3 3 25 3 2 3 24 4 2 21 x y z x y z x y z          can be rewritten as 2 3 3 25 3 2 3 24 4 1 2 21 x y z                        . Thus 1 2 3 3 25 3.2 3 2 3 24 4.2 4 1 2 21 2 x y z                                        . 7(d) 2 3 3 3 2 3 5 4 1 2  , 25 3 3 24 2 3 16 21 1 2  , 2 25 3 3 24 3 21 4 21 2  , 2 3 25 3 2 24 10 4 1 21  . By Cramer’s Rule, 16 21 10, , 25 5 5x y z    . The three planes intersect at exactly one point, with coordinates  3.2, 4.2, 2 . www.KiasuExamPaper.com 627

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 1 Solutions) Page 3 of 12 8(a) 2 3 4 5 3 5 7 11 3 2 7 x y z w x y z w x z w             2 3 4 5 2 3 6 2 6 6 12 x y z w y z w y z w             2 3 4 5 2 3 6 2 0 x y z w y z w z           2 3 4 5 2 3 6 0 x y z w y z w z          2 7 2 3 6 0 x z w y z w z         2 7 3 6 0 x w y w z      Let w t , we have 7 2 , 6 3 , 0, , .x t y t z w t t        . 9 1 1 2 2 1 2 3 2 3 1 0 1 0 1 0 21 0 2 3 0 3 0 1R R R R R                               . The corresponding elementary matrices are 0 1 1 0      , 1 0 2 1      and 1 3 1 0 0      . Then 1 3 1 0 1 0 0 1 2 3 0 2 1 1 0 1 0                I . Thus, 11 1 1 3 1 02 3 0 1 1 0 0 1 1 0 1 0 01 0 1 0 2 1 1 0 2 1 0 3                                    I . Practice Questions 10                       2 2 2 5 1 1 1 2 2 1 10 1 3 5 1 3 1 1 5 2 2 10 1 3 10 2 4 10 15 3 10 20 4 7 15 4 5 3 a a a a a a a a a a a a a                            When the matrix is singular, 3a  or 5 4a . When 3b  and 5 4b , the coefficient matrix is invertible and there is exactly one solution, i.e. the three planes intersect at exactly one common point. When 3b , the augmented matrix is 2 5 1 0 1 3 1 2 3 10 6 10           , whose reduced row-echelon form is 1 0 8 10 0 1 3 4 0 0 0 0        . There are infinitely many solutions, the three planes intersect in a common line. www.KiasuExamPaper.com 628

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 1 Solutions) Page 4 of 12 When 5 4b , the augmented matrix is 5 4 5 2 2 5 1 0 1 1 2 3 10 10         , whose reduced row -echelon form is 1 2 2 5 1 0 0 0 1 0 0 0 0 1         . There is no solution, the three planes have no point in common. Since the planes are parallel pairwise, they form a triangular prismatic surface. 11                  1 2 3 2 1 3 1 2 1 3 1 2 2 2 2 2 2 2 2 2 2 2 1 2 3 2 1 3 1 2 1 3 1 2 1 3 1 2 1 3 1 3 1 3 1 3 1 2 1 3 1 3 2 2 1 3 1 3 2 2 1 3 2 3 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 1 a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a                                12(a)          3 33 3 2 2 3 3 2 2 3 3 2 2 3 2 3 2 2 2 3 2 2 2 3 9 6 3 3 6 12 8 9 9 9 a a b a b a b a b a a a b a b a a b a b a b a a b a a b ab a a a b ab b a a b ab b ab b b a b                               If a b  , then 0a b  , the determinant is 0. If the determinant   29 0b a b   , then 0a b  since 0b  , i.e. a b  . (b)   1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 1 b c c a a b b c a b c b b a a b a b b c c a a b b c a b c b b a a b a b b c c a a b b c a b c b b a a b a b c c a c c b c a a c a b c c a c c b c a a c a b c c a c c b c a a c a b                      1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 1 1 1 2 2 2 3 3 3 0 0 0 0 0 0 1 2 a b c a b c a b c a b c a b c a b c a b c a b c a b c          www.KiasuExamPaper.com 629

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Matrices and Linear Spaces (Tutorial Set 1 Solutions) Page 5 of 12 13      det det det 0 nn   A A O for some pos itive integer n, so  det 0 A and A is not invertible. 14    det det T a d g a d g a d g b e h b e h b e h c f i a c d f g i a b c d e f g h i              A A 0 0 0 0 a d g b e h  . Thus A is not invertible. 15     22det det   A A ,      det 1 det 1 n    I I when n is odd. Since   2 det 0 1     A ,     2det det  A I . So 2 A I , i.e. 2  A I O has no solution when n is odd. 16 When A is singular,    1 4 3 0 2 8 24 6 4 24 24 24 6 4 6 4 1 3 a b a ab b a ab a b b                , so 6a or 4b . 17 (i) Show by induction starting from 1m . (ii) Using the result that m m  D