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Ryan, Michelle, and Erwin spent [tex]$\$[/tex]13.50[tex]$, $[/tex]\[tex]$16.50$[/tex], and [tex]$\$[/tex]14[tex]$, respectively, at an amusement park. Ryan bought three tickets for the Ferris wheel and two tickets for the water slide. Michelle bought one ticket for the Ferris wheel and four tickets for the merry-go-round. Erwin bought three tickets for the Ferris wheel, one ticket for the water slide, and one ticket for the merry-go-round.

Let $[/tex]x[tex]$, $[/tex]y[tex]$, and $[/tex]z[tex]$ represent the ticket cost for the Ferris wheel, water slide, and merry-go-round, respectively. Identify the column entries that belong in the matrix equation that models this situation, $[/tex]AX = B[tex]$.

Matrix $[/tex]A[tex]$:

\[
\begin{pmatrix}
3 & 2 & 0 \\
1 & 0 & 4 \\
3 & 1 & 1
\end{pmatrix}
\]

Matrix $[/tex]B[tex]$:

\[
\begin{pmatrix}
13.50 \\
16.50 \\
14.00
\end{pmatrix}
\]

Solve for $[/tex]X[tex]$:

\[
A^{-1}B = X
\]

Where $[/tex]X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$.

Sagot :

GinnyAnsweridn
To determine the costs of the Ferris wheel, water slide, and merry-go-round tickets given the expenses and the purchases by Ryan, Michelle, and Erwin, we need to use a system of linear equations. Let's set up the equations and the corresponding matrices to solve for [tex]\(x\)[/tex] (cost of the Ferris wheel ticket), [tex]\(y\)[/tex] (cost of the water slide ticket), and [tex]\(z\)[/tex] (cost of the merry-go-round ticket).

Given:
- Ryan spent [tex]\( \$ 13.50 \)[/tex].
- Michelle spent [tex]\( \$ 16.50 \)[/tex].
- Erwin spent [tex]\( \$ 14 \)[/tex].

Ryan bought:
- 3 Ferris wheel tickets
- 2 water slide tickets
- 0 merry-go-round tickets

Michelle bought:
- 1 Ferris wheel ticket
- 0 water slide tickets
- 4 merry-go-round tickets

Erwin bought:
- 3 Ferris wheel tickets
- 1 water slide ticket
- 1 merry-go-round ticket

We can form the following system of linear equations:

1. [tex]\( 3x + 2y + 0z = 13.50 \)[/tex]
2. [tex]\( 1x + 0y + 4z = 16.50 \)[/tex]
3. [tex]\( 3x + 1y + 1z = 14 \)[/tex]

These equations can be written in matrix form as:
[tex]\[ A \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 13.50 \\ 16.50 \\ 14 \end{bmatrix} \][/tex]

Where:
[tex]\[ A = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 0 & 4 \\ 3 & 1 & 1 \end{bmatrix} \][/tex]
[tex]\[ B = \begin{bmatrix} 13.50 \\ 16.50 \\ 14 \end{bmatrix} \][/tex]

To solve for [tex]\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \)[/tex], we need to find the inverse of matrix [tex]\( A \)[/tex] and multiply it by matrix [tex]\( B \)[/tex]:

[tex]\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = A^{-1} B \][/tex]

Given the final results from solving this equation:
[tex]\[ X = \begin{bmatrix} 2.5 \\ 3.0 \\ 3.5 \end{bmatrix} \][/tex]

Therefore, the costs of the tickets are:
- Ferris wheel ticket cost, [tex]\( x \)[/tex]: [tex]\( \$ 2.50 \)[/tex]
- Water slide ticket cost, [tex]\( y \)[/tex]: [tex]\( \$ 3.00 \)[/tex]
- Merry-go-round ticket cost, [tex]\( z \)[/tex]: [tex]\( \$ 3.50 \)[/tex]

Thus, the values for [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex] are:
[tex]\[ x = 2.5, \quad y = 3.0, \quad z = 3.5 \][/tex]
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