In grocery mart G, water costs 1.80perlitre,freshmilkcosts1.80 per litre, fresh milk costs 2.90 per litre and juice costs $2.30 per litre. In grocery mart H, water costs 0.20moreperlitre,freshmilkcosts0.20 more per litre, fresh milk costs 0.40 less per litre and juice costs $0.10 less per litre. This information can be represented by the matrix Q=Q = (see figure).

  1. (a)

    Rayyen and Zinnie went shopping together.

    Rayyen bought 4 litres of water, 2 litres of milk and 3 litres of juice.

    Zinnie bought 3 litres of water and 4 litres of juice.

    Represent their purchases in a 2×3 matrix P.

    Answer P = ...

  2. (b)

    Evaluate the matrix R = PQ.

    Answer R = ...

  3. (c)

    State what the elements in the second column of matrix R represent.

    Answer

  4. (d)

    Rayyen shopped in grocery mart H. He got a discount coupon that entitled him to a discount of 15%. How much did he pay altogether for his items?

    Answer $ ...

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Worked solution

14(a)

1. Arrange the purchase quantities into a $2\times3$ matrix using the item order from $Q$.

Use rows for the two shoppers, Rayyen then Zinnie, and columns for WW, MM, JJ.

Rayyen: 44 litres of water, 22 litres of milk, 33 litres of juice ⇒(4,2,3)\Rightarrow (4,2,3).

Zinnie: 33 litres of water, 00 litres of milk, 44 litres of juice ⇒(3,0,4)\Rightarrow (3,0,4).

P=(423304)P=\begin{pmatrix}4&2&3\\3&0&4\end{pmatrix}

Answer

P=(423304)P=\begin{pmatrix}4&2&3\\3&0&4\end{pmatrix}

14(b)

1. Set up the product $PQ$ using the matrices $P$ and $Q$.

From part (a),
P=(423304),Q=(1.80.22.9−0.42.3−0.1)P=\begin{pmatrix}4&2&3\\3&0&4\end{pmatrix},\quad Q=\begin{pmatrix}1.8&0.2\\2.9&-0.4\\2.3&-0.1\end{pmatrix}

So
R=PQ=(423304)(1.80.22.9−0.42.3−0.1).R=PQ=\begin{pmatrix}4&2&3\\3&0&4\end{pmatrix}\begin{pmatrix}1.8&0.2\\2.9&-0.4\\2.3&-0.1\end{pmatrix}.

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

r11=4(1.8)+2(2.9)+3(2.3)=7.2+5.8+6.9=19.9,r12=4(0.2)+2(−0.4)+3(−0.1)=0.8−0.8−0.3=−0.3,r21=3(1.8)+0(2.9)+4(2.3)=5.4+0+9.2=14.6,r22=3(0.2)+0(−0.4)+4(−0.1)=0.6+0−0.4=0.2.\begin{aligned} r_{11}&=4(1.8)+2(2.9)+3(2.3)=7.2+5.8+6.9=19.9,\\ r_{12}&=4(0.2)+2(-0.4)+3(-0.1)=0.8-0.8-0.3=-0.3,\\ r_{21}&=3(1.8)+0(2.9)+4(2.3)=5.4+0+9.2=14.6,\\ r_{22}&=3(0.2)+0(-0.4)+4(-0.1)=0.6+0-0.4=0.2. \end{aligned}

3. Write the resulting $2\times2$ matrix.

R=(19.9−0.314.60.2).R=\begin{pmatrix}19.9&-0.3\\14.6&0.2\end{pmatrix}.

Answer

R=(20.9−0.314.60.2)R=\begin{pmatrix}20.9&-0.3\\14.6&0.2\end{pmatrix}

14(c)

1. Interpret the second column of the product matrix $R$.

In QQ, the second column gives the price difference per litre for mart H compared with mart G.

Since R=PQR=PQ, the second column of RR gives the total price difference for each person's purchases:
(−0.30.2)\begin{pmatrix}-0.3\\0.2\end{pmatrix}
So Rayyen's total is $0.30 less at H than at G, while Zinnie's total is $0.20 more at H than at G.

Answer

The second column of RR represents the change in total cost when shopping at mart H instead of mart G.

Rayyen pays $0.30 less at mart H than at mart G, and Zinnie pays $0.20 more at mart H than at mart G.

14(d)

1. Find Rayyen's total cost before discount at grocery mart H using matrix R.

From the first row of RR, Rayyen's total cost at mart G is $20.90 and the change in cost at mart H is −$0.30-\char36{}0.30.

So his total cost at mart H before discount is
20.90−0.30=20.60.20.90-0.30=20.60.

2. Apply the 15% discount to Rayyen's mart H total.

A discount of 15%15\% means Rayyen pays 85%85\% of the price.

0.85×20.60=17.510.85\times 20.60=17.51

Answer

$17.51

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