Solution:
Let the price of mango, strawberry and chocolate flavors be x, y, and z respectively.
Thus,
[tex]\begin{gathered} mango\Rightarrow x \\ strwberry\Rightarrow y \\ chocolate\Rightarrow z \end{gathered}[/tex]
From the table below:
Given that the total sales in shift 1 is $419, $649 in shift 2, and $96 in shift 3, this implies that
[tex]\begin{gathered} 16x+13y+29z=419\text{ ----- equation 1} \\ 26x+19y+45z\text{ = 649 ----- equation 2} \\ 6x+11y+0z\text{ = 96 ------ equation 3} \end{gathered}[/tex]
From the above system of equations, we can resolve into matrices of the form:
[tex]A\cdot X=B[/tex]
This gives:
[tex]\begin{bmatrix}{16} & {13} & {29} \\ {26} & {19} & {45} \\ {6} & {11} & {0}\end{bmatrix}\times\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {z} & {} & {}\end{bmatrix}=\begin{bmatrix}{419} & {} & {} \\ {649} & {} & {} \\ {96} & {} & {}\end{bmatrix}[/tex]
Where:
[tex]\begin{gathered} A=\begin{bmatrix}{16} & {13} & {29} \\ {26} & {19} & {45} \\ {6} & {11} & {0}\end{bmatrix} \\ X=\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {z} & {} & {}\end{bmatrix} \\ B=\begin{bmatrix}{419} & {} & {} \\ {649} & {} & {} \\ {96} & {} & {}\end{bmatrix} \end{gathered}[/tex]
To solve for the unknowns, we use the Cramer's rule expressed s
[tex]\begin{gathered} x=\frac{D_x}{D} \\ y=\frac{D_y}{D} \\ z=\frac{D_z}{D} \\ where\text{ D is the determinant of the matrix A} \end{gathered}[/tex]
To evaluate the price of one chocolate, which is z, we have
[tex]\begin{gathered} z=\frac{D_z}{D}=\frac{det\begin{bmatrix}{16} & {13} & {419} \\ {26} & {19} & {649} \\ {6} & {11} & {96}\end{bmatrix}}{det\text{ A}} \\ =\frac{16\times\det\begin{pmatrix}19 & 649 \\ 11 & 96\end{pmatrix}-13\times\det\begin{pmatrix}26 & 649 \\ 6 & 96\end{pmatrix}+419\times\det\begin{pmatrix}26 & 19 \\ 6 & 11\end{pmatrix}}{578} \\ =\frac{16(-5315)-13(-1398)+419(172)}{578} \\ =\frac{5202}{578} \\ \Rightarrow z=9 \end{gathered}[/tex]
Hence, the price of one chocolate is
[tex]\$9[/tex]
The correct option is D
