A florist is making 5 identical bouquets for a wedding. The florist has $610 to spend in total. Each bouquet must have 24 flowers. Roses cost $6 each, tulips cost $4 each, and lilies cost $3 each. The number of roses in each bouquet is twice the combined number of tulips and lilies. We need to find the number of roses, tulips, and lilies in each bouquet using Cramer's rule, finding the inverse of a matrix, and converting a matrix to an upper triangular matrix.

1. Problem Description

A florist is making 5 identical bouquets for a wedding. The florist has $610 to spend in total. Each bouquet must have 24 flowers. Roses cost$6 each, tulips cost $4 each, and lilies cost$3 each. The number of roses in each bouquet is twice the combined number of tulips and lilies. We need to find the number of roses, tulips, and lilies in each bouquet using Cramer's rule, finding the inverse of a matrix, and converting a matrix to an upper triangular matrix.

2. Solution Steps

Let $r$ be the number of roses, $t$ be the number of tulips, and $l$ be the number of lilies in each bouquet.

Since there are 5 identical bouquets, the cost for each bouquet is $610 / 5 =$122$.

We have the following equations:

1. The total number of flowers in each bouquet is 24:

$r + t + l = 24$

2. The cost of each bouquet is $122:

$6r + 4t + 3l = 122$

3. The number of roses is twice the number of tulips and lilies combined:

$r = 2(t + l)$, which can be rewritten as $r - 2t - 2l = 0$

Now we have a system of three linear equations with three variables:

$r + t + l = 24$

$6r + 4t + 3l = 122$

$r - 2t - 2l = 0$

We can represent this system of equations in matrix form as $Ax = b$, where:

$A = \begin{bmatrix} 1 & 1 & 1 \\ 6 & 4 & 3 \\ 1 & -2 & -2 \end{bmatrix}$, $x = \begin{bmatrix} r \\ t \\ l \end{bmatrix}$, and $b = \begin{bmatrix} 24 \\ 122 \\ 0 \end{bmatrix}$

1. Cramer's Rule:

First, we find the determinant of matrix A:

$det(A) = 1(4(-2) - 3(-2)) - 1(6(-2) - 3(1)) + 1(6(-2) - 4(1)) = 1(-8 + 6) - 1(-12 - 3) + 1(-12 - 4) = -2 + 15 - 16 = -3$

Next, we find the determinants of the matrices formed by replacing each column of A with the vector b:

$det(A_r) = \begin{vmatrix} 24 & 1 & 1 \\ 122 & 4 & 3 \\ 0 & -2 & -2 \end{vmatrix} = 24(4(-2) - 3(-2)) - 1(122(-2) - 3(0)) + 1(122(-2) - 4(0)) = 24(-2) - (-244) + (-244) = -48 + 244 - 244 = -48$

$r = det(A_r) / det(A) = -48 / -3 = 16$

$det(A_t) = \begin{vmatrix} 1 & 24 & 1 \\ 6 & 122 & 3 \\ 1 & 0 & -2 \end{vmatrix} = 1(122(-2) - 3(0)) - 24(6(-2) - 3(1)) + 1(6(0) - 122(1)) = -244 - 24(-15) + (-122) = -244 + 360 - 122 = -8 + 360 - 122 = -8 - 122 + 360 = -38 + 360 = -8-122+360= -366+360 = -6$

$t = det(A_t) / det(A) = -6 / -3 = 2$

$det(A_l) = \begin{vmatrix} 1 & 1 & 24 \\ 6 & 4 & 122 \\ 1 & -2 & 0 \end{vmatrix} = 1(4(0) - 122(-2)) - 1(6(0) - 122(1)) + 24(6(-2) - 4(1)) = 1(244) - (-122) + 24(-16) = 244 + 122 - 384 = 366 - 384 = -18$

$l = det(A_l) / det(A) = -18 / -3 = 6$

Therefore, $r = 16, t = 2, l = 6$

2. Finding the Inverse:

To find the inverse of matrix A, we can use the formula $A^{-1} = adj(A) / det(A)$, where $adj(A)$ is the adjugate (transpose of the cofactor matrix) of A.

We already found $det(A) = -3$.

The cofactor matrix is:

$C = \begin{bmatrix} -2 & 15 & -16 \\ 0 & -3 & 3 \\ -1 & 3 & -2 \end{bmatrix}$

The adjugate matrix is the transpose of the cofactor matrix:

$adj(A) = \begin{bmatrix} -2 & 0 & -1 \\ 15 & -3 & 3 \\ -16 & 3 & -2 \end{bmatrix}$

The inverse matrix is:

$A^{-1} = \frac{1}{-3} \begin{bmatrix} -2 & 0 & -1 \\ 15 & -3 & 3 \\ -16 & 3 & -2 \end{bmatrix} = \begin{bmatrix} 2/3 & 0 & 1/3 \\ -5 & 1 & -1 \\ 16/3 & -1 & 2/3 \end{bmatrix}$

Now, we can find the solution vector $x$ by multiplying the inverse matrix by the vector $b$:

$x = A^{-1} b = \begin{bmatrix} 2/3 & 0 & 1/3 \\ -5 & 1 & -1 \\ 16/3 & -1 & 2/3 \end{bmatrix} \begin{bmatrix} 24 \\ 122 \\ 0 \end{bmatrix} = \begin{bmatrix} (2/3)(24) + 0(122) + (1/3)(0) \\ -5(24) + 1(122) - 1(0) \\ (16/3)(24) - 1(122) + (2/3)(0) \end{bmatrix} = \begin{bmatrix} 16 \\ -120 + 122 \\ 128 - 122 \end{bmatrix} = \begin{bmatrix} 16 \\ 2 \\ 6 \end{bmatrix}$

Therefore, $r = 16, t = 2, l = 6$

3. Converting to Upper Triangular Matrix:

The original system of equations is:

$r + t + l = 24$

$6r + 4t + 3l = 122$

$r - 2t - 2l = 0$

We can represent this system of equations in matrix form as:

$\begin{bmatrix} 1 & 1 & 1 & | & 24 \\ 6 & 4 & 3 & | & 122 \\ 1 & -2 & -2 & | & 0 \end{bmatrix}$

Subtract 6 times the first row from the second row:

$\begin{bmatrix} 1 & 1 & 1 & | & 24 \\ 0 & -2 & -3 & | & -22 \\ 1 & -2 & -2 & | & 0 \end{bmatrix}$

Subtract the first row from the third row:

$\begin{bmatrix} 1 & 1 & 1 & | & 24 \\ 0 & -2 & -3 & | & -22 \\ 0 & -3 & -3 & | & -24 \end{bmatrix}$

Multiply the second row by -1/2:

$\begin{bmatrix} 1 & 1 & 1 & | & 24 \\ 0 & 1 & 3/2 & | & 11 \\ 0 & -3 & -3 & | & -24 \end{bmatrix}$

Add 3 times the second row to the third row:

$\begin{bmatrix} 1 & 1 & 1 & | & 24 \\ 0 & 1 & 3/2 & | & 11 \\ 0 & 0 & 3/2 & | & 9 \end{bmatrix}$

Now we have an upper triangular matrix. We can solve for $l$ from the third row:

$(3/2)l = 9 \implies l = 9 * (2/3) = 6$

Substitute $l = 6$ into the second row:

$t + (3/2)(6) = 11 \implies t + 9 = 11 \implies t = 2$

Substitute $t = 2$ and $l = 6$ into the first row:

$r + 2 + 6 = 24 \implies r + 8 = 24 \implies r = 16$

Therefore, $r = 16, t = 2, l = 6$

Comparison:

All three methods yield the same results: $r = 16$, $t = 2$, and $l = 6$.

3. Final Answer

Roses: 16

Tulips: 2

Lilies: 6