The matrix A below shows the prices of football match tickets for seats in Category 1 (Cat 1), Category 2 (Cat 2) and Category 3 (Cat 3). The prices are given in dollars.

  1. (a)

    There are 300 Cat 1 seats, 500 Cat 2 seats and 1000 Cat 3 seats. Represent this information in a 3×1 matrix, B.

    Answer B = ...

  2. (b)

    Evaluate the matrix X = (0.5 0.5) AB.

    Answer X = ...

  3. (c)

    State what the element of X represents.

    Answer ...

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Worked solution

16(a)

1. Represent the seat numbers as a 3×1 column matrix in category order

Cat 1, Cat 2, Cat 3 correspond to the three columns of matrix AA, so
B=(Cat 1 seatsCat 2 seatsCat 3 seats)=(3005001000).B=\begin{pmatrix}\text{Cat 1 seats}\\\text{Cat 2 seats}\\\text{Cat 3 seats}\end{pmatrix}=\begin{pmatrix}300\\500\\1000\end{pmatrix}.

Answer

B=(3005001000)B=\begin{pmatrix}300\\500\\1000\end{pmatrix}

16(b)

1. Form and evaluate the product $AB$ using the ticket prices and seat numbers.

Using B=(3005001000),B=\begin{pmatrix}300\\500\\1000\end{pmatrix},
AB=(8042201206230)(3005001000)AB=\begin{pmatrix}80&42&20\\120&62&30\end{pmatrix}\begin{pmatrix}300\\500\\1000\end{pmatrix}
=(80(300)+42(500)+20(1000)120(300)+62(500)+30(1000))=\begin{pmatrix}80(300)+42(500)+20(1000)\\120(300)+62(500)+30(1000)\end{pmatrix}
=(6500097000).=\begin{pmatrix}65000\\97000\end{pmatrix}.

2. Multiply the row matrix $(0.5\quad 0.5)$ by $AB$ to find $X$.

X=(0.50.5)(6500097000)X=\begin{pmatrix}0.5&0.5\end{pmatrix}\begin{pmatrix}65000\\97000\end{pmatrix}
=0.5(65000)+0.5(97000)=0.5(65000)+0.5(97000)
=81000.=81000.

Answer

X=81000X=81000

16(c)

1. Interpret the element of $X$ in context.

From part (b), X=(0.50.5)AB=81000X=(0.5\quad 0.5)AB=81000. This is half the semi-final total takings plus half the final total takings, so it represents the average total ticket sales revenue per match.

Answer

XX represents the average total amount of money collected from ticket sales for one match, assuming all the seats are sold.

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