Page 35 - Past Year
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33 | P a g e

                                     1   1    
                                      4  2  0 
                                                                1 2 0
                                                                         
                                
                                 1
               6.       (a)    A     3  1  1      (b)    X     3 2 1
                                      4  2  4                         
                                                                  2 4 1  
                                      1  1  1 
                                        2  4  


               7.      (a)    p = 6, q = –4

               (b)        (i)     A    2

                                    (ii)      p = –2 ,q = 1

                                                                      5          1 
                                            5     4   1                2  2    2   
                                                           
                                            
                                                                -1
                                    (iii)    adjA =    2  2   2   ;  A  =    1   1  1  
                                                                                
                                             3    2   1              3  1    1 
                                                                      2          2 

                                                             4    1    4 
                                                                       
                                    1   0   0              5    5    5  
                                    
                                               
                                                      -1
               8.      (a)       AB= 5  0  1  0   ;      A  =        1    1  7    
                                                            5    5   10
                                     0  0    1                         
                                                               1  1   1    
                                                             5   5     5 
                    (b)        (i)  x + 2y +3z = 20

                                                         4z = x

                                              2y + 2z = 10
                                              1   2      3  x      20
                                                          
                                             
                                                                   
                                                                      
                                     (ii)      1  0   4   y   = 0
                                                                   
                                                                      
                                                                 
                                               0  2    2   z     10 
                       (iii)   x = 8, y = 3, z = 2


                                0  2   3
                                         
               9.      (a)      3  0   2          (b)    –Refer to Lecturer–
                                          
                                  2 1   1  
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