Ex06 Matrices Answer

Topic: Matrices Question 1 [2008 EOY, Q18 OR] marks (a) If  23a aa    is singular, find the possible values of a. Solution: 223det 0 2 3 0 (2 3) 0a aa a aaa         Hence  0a  or  3 2a   [3] (b) (i)   It is given that  23 1 5ax y  and  ay ax k , where a and k are real constants and  0a . Solution: Write  23 1 5ax y  and  ay ax k   in matrix form:  23 1 5ax aa y k                  For infinite many solutions,  3 2a  so  33 15 33 22 x y k           31 53: 15:22 kk       [2] (b) (ii)   Solve the simultaneous equations for x and y if  1a  and  5k  .  Solution: When  1a  and  5k  ,  23 1 5 11 5 x y                 hence      1 2 3 15 1 3 15 0 11 5 1 2 5 5 xx yy                           i.e.  0x  and  5y    [3]

Question 2 [2009 EOY, Q6] marks Solve the following simultaneous equations using the matrix method    531 0 773 0 862 9xy xy xy .  Solution: 24 4 0 22 0 ( 1 ) xy xy          1( 2 )xy            In Matrix Form:    12 2 0 11 1 x y                       1 12 2 0 11 1 x y            1 12 1 2 1 11 1 1 3                   12 2033 11 133 x y                 6 7 x y    => x = 6, y = 7      [5] Question 3 [2010 EOY, Q17 OR] marks (a) If  32a aa     is singular, find the possible values of a. Solution: 232det 0 3 2 0a aaaa                (3 2) 0aa         20 or 3aa      [3]

(b) It is given that  32 8ax y  and  ax ay k , where a and k are real constants and  0a .   (i) Determine the value of k if there are infinitely many solutions to the pair of simultaneous  equations.  Solution: Must be singular so  2 3a  (since  0a )   So  2828 33 kk    [2] (ii) Use the matrix method to solve the simultaneous equations for x and y if  1a  and  6k  . Solution: 32 8xy  and  6xy  so  32 8 11 6 x y                      Hence  1 32 8 11 6 x y                           12 81 13 65                     4 2         So  4x  and  2y    [3] Question 4 [2011 EOY, Q5] marks       Given that 71 1 1 1 1 1 0 0 30 3 4 3 0 1 0 30 1 3 3 4 0 0 1 m                       , show that  1m . Solution:         31 4 03 1 1 m m        [2]

Hence solve the following simultaneous equations.             1 3432 334 3 xy z xyz xyz           Solution: 111 1 343 2 334 3 x y z                         71 1 31 0 30 1 1 2 3 x y z                          8 1 6 x y z           [3] Question 5 [2012 EOY, Q1] marks Write down the inverse of the matrix  27 31 0      and use this to solve the simultaneous  equations  72 8 0 , 31 01 1 0 . yx xy       Solution:  127 1 07 1 31 0 32 20 (21)                                              10 7 32    27 8 31 0 1 1 10 7 8 3 32 1 1 2 x y x y                                        [4]

Question 6 [2013 EOY, Q3] marks (i) Find 1A and 1B , the inverses of the matrices      43 22   A     and      54 11      B .   Solution: 11 11 . 5 14 and 12 1 5         AB     [2] (ii) Use 1A and 1B from (i), to find the matrix M                           43 54 40 22 11 62   M  .   Solution: 11 . 5 4 014 12 6 2 1 5              M 53 1 4 84 1 5                    25 41 2                                           [3]

Question 7 [2014 EOY, Q4] marks Two outlets of a local bakery sold the following types of cakes on a particular day. Pandan cake Chocolate cake Outlet A 100 120 Outlet B 70 85 The costs of baking a pandan cake and a chocolate cake are $5 and $6 respectively. In both outlets, each pa ndan cake was sold at $ x, and each chocolate cake was sold at $y. It is given that 100 120 70 85  P , 5 6  Q and x y  R . (i) Write down an expression connecting P, Q and R, such that it represents the profit earned by each outlet on that particular day. Solution: P(R Q)   or  100 120 5 70 85 6 x y           or  100 120 5 70 85 6 x y                       [1] (ii) If the profits earned by Outlet A and Outlet B on that particular day were $1710 and $1205 respectively, find the selling price for each type of cake using the inverse matrix method. Solution: 1 100 120 5 1710 70 85 6 1205 5 100 120 1710 6 70 85 1205 5 85 120 17101 6 70 100 1205100 57 . 5 68 12.5, 14 x y x y x y x y xy                                                 The price of pandan cake is $12.50 each, and the price of chocolate cake is $14 each.     [4]

Question 8 [2015 EOY, Q7] marks 1 A factory makes tables and chairs. The following table shows the breakdown of the amount of time taken, as well as the amount of wood and paint needed to produce each item. Time taken (hours) Wood (number of blocks) Paint (number of tins) Table 6 4 3 Chair 3 3 2 Labour costs $10 per hour, while wood costs $3 per block and each tin of paint costs $5. It is given that 643 332   A= , 10 3 5    B= and C=A B . (i) Evaluate C and explain what the numbers in C represent. Solution: 87 49  C  the numbers represent the cost of producing one table and one chair ($87 and $49  respectively). [2] (ii) If the factory makes 20 tables and 50 chairs, represent the information as a row matrix D. Evaluate DC and explain what the answer represents. Solution: (20 50) 4190DD C = , it represents the total cost ($4190) of producing 20 tables and  50 chairs. [3]