# High School: Number and Quantity

### Vector and Matrix Quantities HSN-VM.C.12

12. Work with 2 × 2 matrices as transformations of the plane and interpret the absolute value of the determinant in terms of area.

Students should appreciate the ease with which they can find the determinant of a 2 × 2 matrix. Seriously. It's super easy. The determinant of matrix:

Let's say we have a parallelogram with its vertices at coordinates (0, 0), (a, b), (a + c, b + d), and (c, d). Then the area of that parallelogram is the absolute value of the determinant of the matrix:

Deep breaths. It's easier than it looks.

If we have a parallelogram with vertices at O (0, 0), P (2, 4), Q (3, 9), and R (1, 5), we can make the matrix:

Its determinant is 2 × 5 – 1 × 4 = 10 – 4 = 6. And the absolute value of 6 is also 6. (Duh.) So the area of parallelogram OPQR is 6 units2.

Students should also know that you aren't making this up. If they don't believe you, make them check by calculating the area of that same parallelogram the old-fashioned way. Make them use an abacus, while you're at it. That'll teach them to doubt you.

#### Drills

1. What is the determinant of the following matrix?

20

To find the determinant of a 2 × 2 matrix, follow the rule adbc, or 12 × 5 – 10 × 4 = 60 – 40 = 20. Easier than juggling one ball, isn't it?

2. What is the determinant of the following matrix?

-18

The determinant is the difference between the product of the top left and bottom right numbers and the top right and bottom left numbers. Or in other words, 3 × 4 – 5 × 6 = 12 – 30 = -18. Seriously, how much easier could it get?

3. What is the absolute value of the determinant of the following matrix?

24

The absolute value of the determinant is |abbc|, not |ab| – |bc| or |ab| + |-bc|. Got that? So the determinant is |3 × 7 – 9 × 5| = |21 – 45| = |-24|. The absolute value of -24 is just 24. That means our answer is (A).

4. Why does the area formula need to include absolute value?

Because area is always positive

Without the absolute value in the formula, some problems would yield a negative area. Negative areas can't exist. We can subtract areas from each other, sure. But negative areas as their own separate things? Nope.

5. Find the area of the parallelogram with coordinates (0, 0), (5, 3), (6, 5), and (1, 2).

7 units^2

Before we go into area-calculating, we should make sure this parallelogram follows our rules (the ones where the third point has to have coordinates that are the sums of the other two non-origin coordinates). Since it does (5 + 1 = 6 and 3 + 2 = 5), we can make the matrix . The absolute value of the determinant is |5 × 2 – 1 × 3| = |10 – 3| = 7. And there we go.

6. Three points of a parallelogram are (0, 0), (2, 9), and (3, 7). If we can calculate its area using the following matrix, what is the coordinate of its last point?

(5, 16)

For our matrix-parallelogram-determinant rule, we need the four points of the parallelogram to be (0, 0), (a, b), (c, d), and (a + c, b + d). We're given that a = 2, b = 9, c = 3, and d = 7, so our resulting coordinate is (2 + 3, 9 + 7), or (5, 16).

7. What are the coordinates of a parallelogram represented by the following matrix?

(0, 0), (5, 5), (1, 2), (6, 7)

Our matrix is in the form  and our points have to be (0, 0), (a, b), (c, d), and (a + c, b + d). That means our points are (5, 5) and (1, 2). One of the points is always (0, 0), and the other is the addition of the two coordinates. That means (B) is the right answer.

8. Find the area of a parallelogram with points (0, 0), (3, 4), (9, 8), and (6, 4).

12 units^2

Given the points we have, we can set up the matrix . The absolute value of the determinant will give us the area of the parallelogram. That would be |12 – 24| = |-12| = 12.

9. Can we calculate the area of a parallelogram with points (1, 2), (5, 7), (6, 2), and (10, 7) using the matrix and determinant method?

Yes, if we just subtract the x coordinates by 1 and all the y coordinates by 2

Subtracting 1 from each x coordinate and 2 from each y coordinate will shift the parallelogram on the coordinate plane. It won't distort the shape or change the area. When we do that, we have the points (0, 0), (4, 5), (5, 0), and (9, 5). Since these points match the criteria for using the matrix and determinant method, we can use it.

10. What is the area of a parallelogram with the points (1, 2), (5, 7), (6, 2), and (10, 7)?