Let roses be denoted by #x#, carnations by#y# and tulips by #z#.
Given that:
#3x_1 + 2 y_1 + 1 z_1 = $14#-------equation(1)
#6 x_2+2y_2+ 6z_2= $38#---------equation(2)
#1x_3+ 12y_3+ 1z_3= $18#-------equation (3)
Method 1:
Matrix A uses the coefficients from the x, y, and z terms
Matrix B uses the constants
Matrix C contains the solutions to the system of equations
Matrix A
#| x_1,y_1,z_1|#
#| x_2,y_2,z_2|#
#| x_3,y_3,z_3|#
#| 3,2,1|#
#| 6,2,6|# = A
#| 1,12,1|#
Matrix B :
#|C_1|#
#|C_2|# = B
#|C_3|#
#|14|#
#|38|# = B
#|18|#
use calculator to perform
#A^(−1) ⋅B# to obtain matrix # C#
Matrix C :
#|x|#
#|y|# = C
#|z|#
Matrix C :
#|3|#
#|1|# = C
#|3|#
OR Method 2:
Begin by creating the augmented matrix, or a matrix with the #x,y,and z# coefficients on the left side of a vertical bar, and the constant values on the right side:
#[ [3,2,1,|,14], [6,2,6,|,38] , [1,12,1,|,18] ]#
We now perform row operations on the 3 rows in order to reduce the left portion of the augmented matrix to the identity form:
#[ [1,0,0], [0,1,0], [0,0,1] ]#
We can do any of the following:
-Swap any two rows
-Multiply a single row by a non-zero constant value
-Add/Subtract a constant multiple of a row to another row:
Using the notation #Rn# to refer to row# n#, here's the steps we can use:
#(R_1) / 3 → R_1# (divide row 1 by 3)
#[ [1,2/3,1/3,|,14/3], [6,2,6,|,38] , [1,12,1,|,18] ]#
#(R_2)- 6 R_1 → R_2#(multiply row 1 by 6 and subtract it from row2);
#R_3 - 1 R_1 → R_3#(multiply row1 by 1 and subtract it from row3):
#[ [1,2/3,1/3,|,14/3], [0,-2,4,|,10] , [0,34/3,2/3,|,40/3] ]#
#R_2 / -2 → R_2# (divide the row2 by -2):
#[ [1,2/3,1/3,|,14/3], [0,1,-2,|,-5] , [0,34/3,2/3,|,40/3] ]#
#R_1 -2/3 R_2 → R_1# (multiply row2 by2/3 and subtract it from row1);
# R_3-34/3R2→ R3# (multiply 2 row by 34/ 3 and subtract it from 3 row):
#[ [1,0,5/3,|,8], [0,1,-2,|,-5] , [0,0,70/3,|,70] ]#
#R_3 / 70/ 3 → R_3 #(divide the row 3 by 70/ 3 ):
#[ [1,0,5/3,|,8], [0,1,-2,|,-5] , [0,0,1,|,3] ]#
#R_1- 5/ 3R_3 → R_1 #(multiply row 3 by 5/3 and subtract it from row 1);
#R_2 + 2R_3 → R_2# (multiply row 3 by 2 and add it to row 2):
#[ [1,0,0,|,3], [0,1,0,|,1] , [0,0,1,|,3] ]#
#{(x=3),(y=1),(z=3):}#
Cost of Roses = #$3#
Cost of Carnation = # $1#
Cost of Tulip =# $1#
Cross check:
#3*3 + 2*1 + 1*3 = 9 + 2 + 3 = 14#
#6*3 + 2*1 + 6*3 = 18 + 2 + 18 = 38#
#1*3 + 12*1 + 1*3 = 3 + 12 + 3 = 18#
