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 