Chapter 8.3, Problem 65E

i.

To determine

To calculate: A florist is creating 10 centerpieces. Roses cost $2.50 each, lilies cost $4 each, and irises cost $2 each. The customer has a budget of $300 allocated for the centerpieces and wants each centerpiece to contain 12 flowers, with twice as many roses as the number of irises and lilies combined. Write a system of linear equations that represents the situation. Then write a matrix equation that corresponds to your system.

Answer to Problem 65E

The systems of equations are

  $\begin{array}{l}R+L+I=120\\ 2.5R+4L+2I=300\\ 2\left(L+I\right)=R\end{array}$

Explanation of Solution

Let R be the number of roses, L be the number of lilies and I be the number of irises. First we know that the total number of flowers is 120.

  $R+L+I=120\cdots \left(i\right)$

The budget of flowers is $300 and roses cost $2.50 each, lilies cost $4 each, and irises cost $2 each.

  $2.5R+4L+2I=300\cdots \left(ii\right)$

Now, the last condition is that there should be twice as many roses as the number of irises and lilies combined.

  $2\left(L+I\right)=R\cdots \left(iii\right)$

Then from above 3 equations a system of matrix is formed

  $\begin{array}{l}R+L+I=120\\ 2.5R+4L+2I=300\\ 2\left(L+I\right)=R\end{array}$

ii.

To determine

To calculate: A florist is creating 10 centerpieces. Roses cost $2.50 each, lilies cost $4 each, and irises cost $2 each. The customer has a budget of $300 allocated for the centerpieces and wants each centerpiece to contain 12 flowers, with twice as many roses as the number of irises and lilies combined. Solve your system of linear equations using an inverse matrix. Find the number of flowers of each type that the florist can use to create the 10 centerpieces.

Answer to Problem 65E

  $X=\left[\begin{array}{c}80\\ 10\\ 30\end{array}\right]=\left[\begin{array}{c}R\\ L\\ I\end{array}\right]$

Hence Roses=80, Lily=10, and Iris=30.

Explanation of Solution

From part (a) we have

  $\begin{array}{l}R+L+I=120\\ 2.5R+4L+2I=300\\ 2\left(L+I\right)=R\end{array}$

The above system of equation can be written as $AX=B$

  $\left[\begin{array}{ccc}1& 1& 1\\ 2.5& 4& 2\\ 1& -2& -2\end{array}\right]\left[\begin{array}{c}R\\ L\\ I\end{array}\right]=\left[\begin{array}{c}120\\ 300\\ 0\end{array}\right]$

Now find the inverse of matrix A

Solve above system of equation using Gauss-Jordan elimination method $\left(\left[A|I\right]\to \left[I|A^{-1}\right]\right)$

Further

  $\begin{array}{l}{R}_3\to -\frac{1}{4}{R}_3\\ \left[\begin{array}{ccc}1& 0& \frac{4}{3}\\ 0& 1& -\frac{1}{3}\\ 0& 0& 1\end{array}|\begin{array}{ccc}\frac{8}{3}& -\frac{2}{3}& 0\\ -\frac{5}{2}& \frac{2}{3}& 0\\ -\frac{3}{2}& -\frac{1}{2}& -\frac{1}{4}\end{array}\right]\\ R_1\to R_1-\frac{4}{3}{R}_3\\ \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& -\frac{1}{3}\\ 0& 0& 1\end{array}|\begin{array}{ccc}\frac{2}{3}& 0& \frac{1}{3}\\ -\frac{5}{2}& \frac{2}{3}& 0\\ -\frac{3}{2}& -\frac{1}{2}& -\frac{1}{4}\end{array}\right]\\ R_2\to R_2+\frac{1}{3}{R}_3\\ \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}\frac{2}{3}& 0& \frac{1}{3}\\ -\frac{7}{6}& \frac{1}{2}& -\frac{1}{12}\\ -\frac{3}{2}& -\frac{1}{2}& -\frac{1}{4}\end{array}\right]\end{array}$

  $A^{-1}=\left[\begin{array}{ccc}\frac{2}{3}& 0& \frac{1}{3}\\ -\frac{7}{6}& \frac{1}{2}& -\frac{1}{12}\\ \frac{3}{2}& -\frac{1}{2}& -\frac{1}{4}\end{array}\right]$

As $AX=B\Rightarrow X=A^{-1}B$

  $\begin{array}{l}X=A^{-1}B=\left[\begin{array}{ccc}\frac{2}{3}& 0& \frac{1}{3}\\ -\frac{7}{6}& \frac{1}{2}& -\frac{1}{12}\\ \frac{3}{2}& -\frac{1}{2}& -\frac{1}{4}\end{array}\right]\left[\begin{array}{c}120\\ 300\\ 0\end{array}\right]\\ X=\left[\begin{array}{c}80\\ 10\\ 30\end{array}\right]=\left[\begin{array}{c}R\\ L\\ I\end{array}\right]\end{array}$