## vera_ewing one year ago Math question

1. vera_ewing

2. jim_thompson5910

They want you to multiply 3 by that matrix

3. jim_thompson5910

Multiply the 3 by each term inside the matrix $\large \color{red}{3*}\left[\begin{array}{cccc} -1 & 5 & 5 & -2\\ 2 & 5 & 3 & -4 \end{array}\right] = \left[\begin{array}{cccc} \color{red}{3*}(-1) & \color{red}{3*}5 & \color{red}{3*}5 & \color{red}{3*}(-2)\\ \color{red}{3*}2 & \color{red}{3*}5 & \color{red}{3*}3 & \color{red}{3*}(-4) \end{array}\right] =??$ Note: the final result will be a 2x4 matrix

4. vera_ewing

-3, 15, 15, -6 6, 15, 15, -12

5. jim_thompson5910

very good

6. jim_thompson5910

the first column -1, 2 turns into -3, 6 so it's the same as multiplying the point (-1,2) by 3 to get (-3,6). Dilating it out 3 times further away from the origin

7. vera_ewing

Ok, so how can I tell which graph it is?

8. jim_thompson5910

each column of the matrix represents a point or ordered pair

9. jim_thompson5910

eg: first column -1,2 means we have the point (-1,2)

10. jim_thompson5910

11. vera_ewing

A.

12. vera_ewing

B.

13. vera_ewing

C.

14. vera_ewing

D.

15. vera_ewing

@jim_thompson5910 It's A?

16. jim_thompson5910

one moment

17. jim_thompson5910

Probably the best thing you can do is just graph each point. What I did was graph using geogebra. That program has a lot of interactivity and flexibility, which is why I like it So you'd graph each column as its own point eg: W = (-1,2), X = (5,5), Y = (5,3), etc etc see the attached as for what you'd get in the graph

18. vera_ewing

Thanks so much!

19. jim_thompson5910

no problem