Chapter 8.3, Problem 71E

a.

To determine

The system of the linear equations.

Answer to Problem 71E

  $R+L+I=120,2.5R+4L+2I=300,$ and $R=2(L+I)$ .

Explanation of Solution

Given information:

A florist is creating $10$ centerpieces for the tables at a wedding reception. 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 $12$ flower, with twice as many roses as the number of irises and lilies combined.

  $(a)$ Write a system of linear equations that represents the situation.

Calculation:

From the given information,florist is creating $10$ centerpieces for the tables at a wedding reception. 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 $12$ flower, with twice as many roses as the number of irises and lilies combined.

Let cost of roses $=R$

Let cost of lilies $=L$

Let cost of irises $=I$

Total number of flowers $=120$

  $R+L+I=\mathrm{120..............}(1)$

Total price of the flower must be $\$300$ :

  $2.5R+4L+2I=\mathrm{300...............}(2)$

At last there must be twice as many roses as irises and lilies combined:

  $R=2(L+I)\mathrm{.............}(3)$

Thus, the system of equations are $R+L+I=120,2.5R+4L+2I=300,$ and $R=2(L+I)$ .

b.

To determine

The matrix equation that correspond to the system.

Answer to Problem 71E

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

Explanation of Solution

Given information:

A florist is creating $10$ centerpieces for the tables at a wedding reception. 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 $12$ flower, with twice as many roses as the number of irises and lilies combined.

  $(b)$ Write a matrix equation that correspond to your system.

Calculation:

From the given information,florist is creating $10$ centerpieces for the tables at a wedding reception. 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 $12$ flower, with twice as many roses as the number of irises and lilies combined.

Let cost of roses $=R$

Let cost of lilies $=L$

Let cost of irises $=I$

Total number of flowers $=120$

  $R+L+I=\mathrm{120..............}(1)$

Total price of the flower must be $\$300$ :

  $2.5R+4L+2I=\mathrm{300...............}(2)$

At last there must be twice as many roses as irises and lilies combined:

  $R=2(L+I)\mathrm{.............}(3)$

From the system of equations from the part (a), write the matrix equation which correspond to this system:

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

Thus, the system of matrix equations are $\left[\begin{array}{ccc}1& 1& 1\\ 2.5& 4& 2\\ 1& -2& -2\end{array}\right]\left[\begin{array}{c}R\\ L\\ I\end{array}\right]=\left[\begin{array}{c}120\\ 300\\ 0\end{array}\right]$ .

c.

To determine

The system of linear equations by using an inverse of matrix.

Answer to Problem 71E

  $R=80,L=10,$ and $z=30$ .

Explanation of Solution

Given information:

A florist is creating $10$ centerpieces for the tables at a wedding reception. 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 $12$ flower, with twice as many roses as the number of irises and lilies combined.

  $(c)$ 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.

Calculation:

From the given information,florist is creating $10$ centerpieces for the tables at a wedding reception. 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 $12$ flower, with twice as many roses as the number of irises and lilies combined.

Let cost of roses $=R$

Let cost of lilies $=L$

Let cost of irises $=I$

Total number of flowers $=120$

  $R+L+I=\mathrm{120..............}(1)$

Total price of the flower must be $\$300$ :

  $2.5R+4L+2I=\mathrm{300...............}(2)$

At last there must be twice as many roses as irises and lilies combined:

  $R=2(L+I)\mathrm{.............}(3)$

From the system of equations from the part (a), write the matrix equation which correspond to this system:

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

Now, find the inverse of the matrix of the linear equations by using Gauss-Jordan elimination method:

To find the system of equations by writing the system in the matrix form $AX=B$ .

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

Find the determinant of $A$ :

  $\begin{array}{l}\text{det }A=\left|\begin{array}{ccc}1& 1& 1\\ 2.5& 4& 2\\ 1& -2& -2\end{array}\right|\\ =1\cdot 4\cdot (-2)+1\cdot 2\cdot 1+1\cdot 2\cdot 5\cdot (-2)-1\cdot 4\cdot 1-1\cdot 2\cdot (-2)-1\cdot 2.5\cdot (-2)\\ =-8+2-5-4+4+5\\ =-6\end{array}$

The determinant of $A$ is not zero,therefore the inverse matrix $A^{-1}$ exist. To calculate the inverse matrix find additional minors and cofactors of matrix $A$ :

  $A^{-1}=\frac{C^T}{\det A}=\left(\begin{array}{ccc}\frac{2}{3}& 0& \frac{1}{3}\\ -\frac{7}{6}& \frac{1}{2}& -\frac{1}{12}\\ \frac{3}{2}& -\frac{1}{2}& -\frac{1}{4}\end{array}\right)$

Find the solution:

  $\begin{array}{l}X=A^{-1}\cdot B\\ =\left(\begin{array}{ccc}\frac{2}{3}& 0& \frac{1}{3}\\ -\frac{7}{6}& \frac{1}{2}& -\frac{1}{12}\\ \frac{3}{2}& -\frac{1}{2}& -\frac{1}{4}\end{array}\right)\left(\begin{array}{c}120\\ 300\\ 0\end{array}\right)\\ =\left(\begin{array}{ccc}\frac{2}{3}\cdot 120& 0\cdot 300& \frac{1}{3}\cdot 0\\ -\frac{7}{6}\cdot 120& \frac{1}{2}\cdot 300& -\frac{1}{12}\cdot 0\\ \frac{3}{2}\cdot 120& -\frac{1}{2}\cdot 300& -\frac{1}{4}\cdot 0\end{array}\right)\end{array}$

  $\begin{array}{l}=\left(\begin{array}{l}80+0+0\\ -140+150+0\\ 180-150+0\end{array}\right)\\ =\left(\begin{array}{c}80\\ \begin{array}{l}10\\ 30\end{array}\end{array}\right)\end{array}$

Therefore, the solutions are $R=80,L=10,$ and $z=30$ .