A florist offers three sizes of flower arrangements containing roses, daisies, and crysan- themums. Each small arrangement contains one rose, three daisies, and three crysan- themums. Each medium...
A florist offers three sizes of flower arrangements containing roses, daisies, and crysan- themums. Each small arrangement contains one rose, three daisies, and three crysan- themums. Each medium arrangement contains two roses, four daisies, and six crysan- themums. Each large arrangement contains four roses, eight daisies, and six crysan- themums. One day the florist noted that a total of 24 roses, 50 daisies, and 48 crysan- themums had been used in filling orders for these three types of arrangements. (a) Write down a system of linear equations that describes this information. (b) Use Gaussian elimination to reduce the corresponding augmented matrix to row echelon form. (c) Use back-substitution to find the numbers of each type of arrangement that were made.
To write the system of equations, assign variables. Let x be the number of small arrangements, y the number of medium, and z the number of large arrangements.
The first equation involves roses. Each small requires 1 rose, mediums take 2, and large require 4. There are a total of 24 roses so:
x+2y+4z=24 In a similar vein we get the other equations, so the system is:
The augmented matrix is:
A matrix is row reduced if in each row the first nonzero element is 1, and the entries in the column below the one are all zeros. For a 3x4 matrix, the reduced form looks like:
`[[1,x,x,x],[0,1,x,x],[0,0,1,x]]` where x is some real number.
Use row operations to reduce the matrix:
`[[1,2,4,24],[0,-2,-4,-22],[0,0,-6,-24]]` Add R2-3R1 for row 2 and R3-3R1 for row 3.
`[[1,2,4,24],[0,1,2,11],[0,0,1,4]]` Multiply row 2 by -1/2 and row 3 by -1/6
This is in row reduced form.
z=4 from the last row.
y+8=11 ==> y=3 (Substitute z=4 into y+2z=11)
x+6+16=24 ==> x=2
So there were 2 small, 3 medium, and 4 large.
** Note that instead of back substitution, we could have put the matrix in reduced row echelon form -- entries above the leading 1 are converted to zero:
`[[1,2,4,24],[0,1,2,11],[0,0,1,4]] -> [[1,0,0,2],[0,1,2,11],[0,0,1,4]] -> [[1,0,0,2],[0,1,0,3],[0,0,1,4]]`
Now you can read the solution -- x=2,y=3,z=4 as above.