Matrices: Multiplication Word Problems

Question 1 of 3

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)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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$

Question 2 of 3

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.

1. $\begin{matrix} \mathsf{A} \\ \mathsf{B} \end{matrix}$ $\begin{bmatrix} 265 \\ 240 \end{bmatrix}$
2. $\begin{matrix} \mathsf{A} \\ \mathsf{B} \end{matrix}$ $\begin{bmatrix} 244 \\ 275 \end{bmatrix}$
3. $\begin{matrix} \mathsf{A} \\ \mathsf{B} \end{matrix}$ $\begin{bmatrix} 275 \\ 244 \end{bmatrix}$
4. $\begin{matrix} \mathsf{A} \\ \mathsf{B} \end{matrix}$ $\begin{bmatrix} 240 \\ 265 \end{bmatrix}$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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

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.

1. $\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}$ $\begin{bmatrix} 1990 \\ 1825.50 \\ 1933.50 \end{bmatrix}$
2. $\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}$ $\begin{bmatrix} 1998 \\ 1827.50 \\ 1933.50 \end{bmatrix}$
3. $\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}$ $\begin{bmatrix} 1998.50 \\ 1820 \\ 1931.50 \end{bmatrix}$
4. $\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}$ $\begin{bmatrix} 1827.50 \\ 1998 \\ 1933.50 \end{bmatrix}$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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.

$\begin{matrix} \mathsf{A} \\ \mathsf{B} \\ \mathsf{C} \end{matrix}$

$\begin{bmatrix} 1827.50 \\ 1998 \\ 1933.50 \end{bmatrix}$

Matrices: Multiplication Word Problems

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Fantastic!

Multiplying Two Matrices

2. Question

Hint

Keep Going!

Multiplying Two Matrices

3. Question

Hint

Nice Job!

Multiplying Two Matrices

Quizzes