VM - Scalar Multiplication Lesson
Scalar Multiplication
Scalar multiplication just means to multiply a matrix by a constant.
For instance, let A=[1−2504−1] then if we want to find 2A, we would multiply each entry by 2.
So, 2A=[2−41008−2]
Try a few problems to see if you've got it:
A=[−479032]B=[368−1−20]C=[10542−27−31]D=[7−205336−14]
Try these problems to see if you get the correct solution.
- Problem: -3B
- Solution:
−3B=[−9−18−24360]
- Problem: (1/2)C
- Solution:
12C=[1205221−172−3212]
- Problem: 3C+2D
- Solution:
3C+2D=[17−4152212033−1111]
- Problem: 4B - A
- Solution:
4B−A=[161723−4−11−2]
Application:
The revenue generated by sales in the Vancouver and Quebec branches of the A-Plus auto parts store was as follows:
Vancouver | Quebec | |
Wiper Blades | 140 | 150 |
Clean Fluid (bottles) | 30 | 36 |
Floor Mats | 96 | 48 |
If the Canadian dollar was worth $0.65 U.S. at the time, compute the revenue in U.S. dollars.
We need to multiply each revenue figure by 0.65. Let A be the matrix of revenue figures in Canadian dollars:
The revenue figures in U.S. dollars are then given by the scalar multiple
In other words, in U.S. dollars, $91 worth of wiper blades was sold in Vancouver, $68.25 worth of wiper blades was sold in Quebec, and so on.
Another example:
The A-Plus auto parts store had the following sales in its Vancouver store:
Vancouver | |
---|---|
Wiper Blades | 20 |
Clean Fluid (bottles) | 10 |
Floor Mats | 8 |
We need to multiply each sales figure by the corresponding price and then add the resulting revenue figures. We represent the sales by a column vector, as suggested by the table.
We put the selling prices in a row vector.
We can now compute the total revenue as the product
So, the sale of these items generated a total revenue of $266.00.
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