a) The number of each type of radiotors required by each type of house * b) the cost of each radiator * c) the number of each type of house *
a) the no. of each type of radiators required by each type of house can be shown as a 3X4 matrix
3 = types of house
4 = types of radiators
so the matrix shud be
A B C D
small (3 2 2 0)
med. (2 0 3 3)
large (0 1 6 3)
the second matrix can shud have 4 rows because the 1st one has 4 columns
2nd matrix i.e cost of each radiator can be represnted this way with 4 rows
(30)
(40)
(50)
(80)
first multiply the 1st and 2nd matrix
(3 2 2 0) * (30) = (270) Cost for 1 small
(2 0 3 3) (40) (450) Medium
(0 1 6 3) (50) (580) Large
(80)
once u multiply these matrices u get a matrix which represents the total cost for 1 of each type of house..but there is not 1 of each..there are many..so v need one more matrix which shows the number of houses for small, med. or large
as the matrix we just got has 1 column the new matrix to be multiplied with shud have 1 row which will represent the number of house.
that can be
(20 30 15)
so the final answer can be the matrix multiplication of
(270) * (20 30 15) = (27600)
(450)
(580)
i guess the markscheme missed the "0" ciz the answer says only 2760 whereas it shud be 27600
check it algebra way too, u shud get 27600
