在摊位 A，一个鸡肉派售价 $1.50，一个蘑菇派售价$ 1.30，一个金枪鱼派售价 $1.80。在摊位 B，一个鸡肉派贵 $0.20，一个蘑菇派便宜$ 0.30，一个金枪鱼派便宜 $0.10。这些信息可以用矩阵 P 表示（见图表描述）。

((a))
西蒙购买了 50 个鸡肉派和 20 个金枪鱼派。

艾薇购买了 40 个鸡肉派、20 个蘑菇派和 30 个金枪鱼派。

用一个 3 × 2 矩阵 Q 表示他们的购买数量。

答案 Q =
((b))
计算矩阵 R = PQ。

答案 R =
((c))
利用你在 (b) 中的答案，说明西蒙是从摊位 A 还是摊位 B 购买更好。

答案

摊位 ... 因为 ...

...
((d))
摊位 B 所有派实行 7 折（优惠 30%）促销，而摊位 A 的派价上涨了 10%。

利用你在 (b) 中的答案或用其他方法，计算西蒙和艾薇购买派所需支付的最低总金额。

答案 $

矩阵表示矩阵加法与乘法矩阵数据分析百分比增减数系与四则运算计算器使用不等号 English

它不会直接告诉你答案，而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

2((a))

1. Arrange the purchases in a $3\times2$ matrix using the order Chicken, Mushroom, Tuna for the rows and Simon, Ivy for the columns.

Since $P$ has columns in the order $C, M, T$ , the rows of $Q$ should be in the order $C, M, T$ so that $PQ$ is defined.

Simon buys $50$ Chicken, $0$ Mushroom and $20$ Tuna pies.

Ivy buys $40$ Chicken, $20$ Mushroom and $30$ Tuna pies.

$Q=\begin{pmatrix}50&40\\0&20\\20&30\end{pmatrix}$

答案

$Q=\begin{pmatrix}50&40\\0&20\\20&30\end{pmatrix}$

2((b))

1. Set up the matrix multiplication using the given matrix $P$ and the purchase matrix $Q$ from part (a).

$P=\begin{pmatrix}1.5&1.3&1.8\\0.2&-0.3&-0.1\end{pmatrix},\quad Q=\begin{pmatrix}50&40\\0&20\\20&30\end{pmatrix}$

Since $P$ is $2\times3$ and $Q$ is $3\times2$ , $R=PQ$ will be a $2\times2$ matrix.

2. Calculate each entry of $R$ by row-by-column multiplication.

R=\begin{pmatrix} 1.5(50)+1.3(0)+1.8(20) & 1.5(40)+1.3(20)+1.8(30)\\ 0.2(50)+(-0.3)(0)+(-0.1)(20) & 0.2(40)+(-0.3)(20)+(-0.1)(30) \end{pmatrix}

R=\begin{pmatrix} 75+0+36 & 60+26+54\\ 10+0-2 & 8-6-3 \end{pmatrix}

3. Simplify the entries to obtain the final matrix.

R=\begin{pmatrix} 111 & 140\\ 8 & -1 \end{pmatrix}

答案

$R=\begin{pmatrix}111&134\\8&-4\end{pmatrix}$

2((c))

1. Interpret Simon's entries in the first column of $R$.

From part (b),
$R=\begin{pmatrix}111&134\\8&-4\end{pmatrix}.$
For Simon, the first column is
$\begin{pmatrix}111\\8\end{pmatrix}.$
This means Simon pays $111 at Stall A, and Stall B costs $8 more than Stall A for his purchase.

2. Decide which stall is better for Simon.

Since Stall B costs $8 more than Stall A, Simon should buy from Stall A.

答案

Stall A because the corresponding value for Simon in the second row of $R$ is $8$ , so Stall B costs $8 more.

2((d))

1. Find the original total cost at Stall B for Simon and Ivy using matrix R.

From part (b),
$R=\begin{pmatrix}111&134\\8&-4\end{pmatrix}.$
The first row gives the total cost at Stall A, and the second row gives the change from Stall A to Stall B.

Simon at Stall B:
$111+8=119.$
Ivy at Stall B:
$134+(-4)=130.$

2. Calculate the new costs after Stall A increases prices by 10% and Stall B gives 30% discount.

Stall A increases by $10\%$ , so multiply by $1.10$ .

Simon at Stall A:
$111\times1.10=122.10.$
Ivy at Stall A:
$134\times1.10=147.40.$

Stall B gives $30\%$ off, so multiply by $0.70$ .

Simon at Stall B:
$119\times0.70=83.30.$
Ivy at Stall B:
$130\times0.70=91.00.$

3. Choose the lower cost for each person and add them to get the lowest total amount.

For Simon, the lower cost is from Stall B:
$83.30<122.10.$

For Ivy, the lower cost is also from Stall B:
$91.00<147.40.$

Lowest total amount:
$83.30+91.00=174.30.$

答案

$174.30

AI 分析，可能出错

用 AI 批改，圈出写错的那一步，并自动将你的薄弱点存入错题本，形成你的错因库。

完整解法

2((a))

2((b))

2((c))

2((d))

同一知识点的其他题