Matrices: Multiplication Word Problems

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Question 1 of 3

1. Question

A stationery shop sells the items below. Find the total cost of these items:

Round your answer to two decimal places

`$` (22.10, 22.1)

Hint

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Correct

Fantastic!

Incorrect

Multiplying Two Matrices

`[[a,b,c]]xx[[p],[q],[r]]=ap+bq+cr`

Two matrices can be multiplied only if the number of columns (`n`) in the first matrix is equal to the number of rows
(`m`) in the second matrix.

To get the total cost, form the Quantity and Cost matrices, then multiply them.

List each item’s quantity in `1` row

Pens: `6`

Erasers: `2`

Gluesticks: `4`

Pencils: `5`

Quantity Matrix:

`[[6,2,4,5]]`

List each item’s cost in `1` column

Pen: `$``1.60`

Eraser: `$``0.80`

Gluestick: `$``1.35`

Pencil: `$``1.10`

Cost Matrix:

\begin{bmatrix}
\color{#9a00c7}{1.60} \\
\color{#9a00c7}{0.80} \\
\color{#9a00c7}{1.35} \\
\color{#9a00c7}{1.10}
\end{bmatrix}

Next, proceed with multiplying the two matrices

		`[[6,2,4,5]]times``[[1.60],[0.80],[1.35],[1.10]]`

	`=`	\begin{bmatrix} (6\cdot\color{#9a00c7}{1.60})+(2\cdot\color{#9a00c7}{0.80})+(4\cdot\color{#9a00c7}{1.35})+(5\cdot\color{#9a00c7}{1.10}) \end{bmatrix}

	`=`	`[9.60+1.60+5.40+5.50]`

	`=`	`[22.10]`

Therefore, the total cost of the items is `$22.10`

`$22.10`

Question 2 of 3

2. Question

Bakeries A and B buy sacks of flours, boxes of yeast and cans of baking powder from the same vendor on a weekly basis. The quantities and costs are listed below. Find the total weekly cost for each bakery in matrix format.

\begin{matrix} \mathsf{A} \\ \mathsf{B} \end{matrix}\begin{bmatrix} 244 \\ 275 \end{bmatrix}
\begin{matrix} \mathsf{A} \\ \mathsf{B} \end{matrix}\begin{bmatrix} 275 \\ 244 \end{bmatrix}
\begin{matrix} \mathsf{A} \\ \mathsf{B} \end{matrix}\begin{bmatrix} 240 \\ 265 \end{bmatrix}
\begin{matrix} \mathsf{A} \\ \mathsf{B} \end{matrix}\begin{bmatrix} 265 \\ 240 \end{bmatrix}

Hint

Help Video

Correct

Keep Going!

Incorrect

Multiplying Two Matrices

`[[a,b,c]]xx[[p],[q],[r]]=ap+bq+cr`

Two matrices can be multiplied only if the number of columns (`n`) in the first matrix is equal to the number of rows
(`m`) in the second matrix.

To get the total cost, form the Quantity and Cost matrices, then multiply them.

List each bakery’s item quantity in `2` rows

Bakery A: `15` Flour, `2` Yeast, `6` Baking Powder

Bakery B: `17` Flour, `3` Yeast, `4` Baking Powder

Quantity Matrix:

\begin{matrix}
\mathsf{A} \\
\mathsf{B}
\end{matrix}\begin{bmatrix}
15 & 2 & 6 \\
17 & 3 & 4
\end{bmatrix}

List each item’s cost in `1` column

Flour: `$``12`

Yeast: `$``17`

Baking Powder: `$``5`

Cost Matrix:

\begin{bmatrix}
\color{#9a00c7}{12} \\
\color{#9a00c7}{17} \\
\color{#9a00c7}{5}
\end{bmatrix}

Next, proceed with multiplying the two matrices

		\begin{matrix} \mathsf{A} \\ \mathsf{B} \end{matrix}\begin{bmatrix} 15 & 2 & 6 \\ 17 & 3 & 4 \end{bmatrix}$$\times$$\begin{bmatrix} \color{#9a00c7}{12} \\ \color{#9a00c7}{17} \\ \color{#9a00c7}{5} \end{bmatrix}

	`=`	\begin{matrix} \mathsf{A} \\ \mathsf{B} \end{matrix}\begin{bmatrix} (15\cdot\color{#9a00c7}{12})+(2\cdot\color{#9a00c7}{17})+(6\cdot\color{#9a00c7}{5}) \\ (17\cdot\color{#9a00c7}{12})+(3\cdot\color{#9a00c7}{17})+(4\cdot\color{#9a00c7}{5}) \end{bmatrix}

	`=`	\begin{matrix} \mathsf{A} \\ \mathsf{B} \end{matrix}\begin{bmatrix} 180+34+30 \\ 204+51+20 \end{bmatrix}

	`=`	\begin{matrix} \mathsf{A} \\ \mathsf{B} \end{matrix}\begin{bmatrix} 244 \\ 275 \end{bmatrix}

Therefore, the total cost of the items for Bakery A is `$244` and for Bakery B, it is `$275`

\begin{matrix}
\mathsf{A} \\
\mathsf{B}
\end{matrix}\begin{bmatrix}
244 \\
275
\end{bmatrix}

Question 3 of 3

3. Question

Three tennis players A, B and C submitted an equipment list. Their lists and the cost of each item is below. Find the total equipment cost for each tennis player in matrix format.

\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}\begin{bmatrix} 1827.50 \\ 1998 \\ 1933.50 \end{bmatrix}
\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}\begin{bmatrix} 1998 \\ 1827.50 \\ 1933.50 \end{bmatrix}
\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}\begin{bmatrix} 1990 \\ 1825.50 \\ 1933.50 \end{bmatrix}
\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}\begin{bmatrix} 1998.50 \\ 1820 \\ 1931.50 \end{bmatrix}

Hint

Help Video

Correct

Nice Job!

Incorrect

Multiplying Two Matrices

`[[a,b,c]]xx[[p],[q],[r]]=ap+bq+cr`

Two matrices can be multiplied only if the number of columns (`n`) in the first matrix is equal to the number of rows
(`m`) in the second matrix.

To get the total cost, form the Quantity and Cost matrices, then multiply them.

List each player’s equipment quantity in `3` rows

Player A: `7` Racquets, `45` Balls, `5` Shoes

Player B: `9` Racquets, `50` Balls, `4` Shoes

Player C: `6` Racquets, `35` Balls, `7` Shoes

Quantity Matrix:

\begin{matrix}
\mathsf{A} \\
\mathsf{B} \\
\mathsf{C}
\end{matrix}\begin{bmatrix}
7 & 45 & 5 \\
9 & 50 & 4 \\
6 & 35 & 7
\end{bmatrix}

List each item’s cost in `1` column

Racquets: `$``149`

Balls: `$``2.10`

Shoes: `$``138`

Cost Matrix:

\begin{bmatrix}
\color{#9a00c7}{149} \\
\color{#9a00c7}{2.10} \\
\color{#9a00c7}{138}
\end{bmatrix}

Next, proceed with multiplying the two matrices

		\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}\begin{bmatrix} 7 & 45 & 5 \\ 9 & 50 & 4 \\ 6 & 35 & 7 \end{bmatrix}$$\times$$\begin{bmatrix} \color{#9a00c7}{149} \\ \color{#9a00c7}{2.10} \\ \color{#9a00c7}{138} \end{bmatrix}

	`=`	\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}\begin{bmatrix} (7\cdot\color{#9a00c7}{149})+(45\cdot\color{#9a00c7}{2.10})+(5\cdot\color{#9a00c7}{138}) \\ (9\cdot\color{#9a00c7}{149})+(50\cdot\color{#9a00c7}{2.10})+(4\cdot\color{#9a00c7}{138}) \\ (6\cdot\color{#9a00c7}{149})+(35\cdot\color{#9a00c7}{2.10})+(7\cdot\color{#9a00c7}{138}) \end{bmatrix}

	`=`	\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}\begin{bmatrix} 1043+94.50+690 \\ 1341+105+552 \\ 894+73.50+966 \end{bmatrix}

	`=`	\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}\begin{bmatrix} 1827.50 \\ 1998 \\ 1933.50 \end{bmatrix}

Therefore, the total equipment cost for Players A, B and C is `$1827.50`, `$1998` and `$1933.50` respectively.

\begin{matrix}
\mathsf{A} \\
\mathsf{B} \\
\mathsf{C}
\end{matrix}\begin{bmatrix}
1827.50 \\
1998 \\
1933.50
\end{bmatrix}

Quizzes

Add & Subtract Matrices 1
Add & Subtract Matrices 2
Add & Subtract Matrices 3
Multiply Matrices 1
Multiply Matrices 2
Matrices: Multiplication Word Problems
Determinant of a Matrix
Inverse of a Matrix
Matrices: Systems of Equations 1
Matrices: Systems of Equations 2
Gauss Jordan Elimination
Cramer’s Rule