Question

Given A= and B= , find AB. Picture Below:

Answer 1

@e.mccormick @jim_thompson5910 @satellite73

Answer 2

Have you done any matrix multiplication before?

Answer 3

yes but I may need a refresher

Answer 4

OK.... having some computer issues. Switched machines (again).

Answer 5

There is refresher! =)

Answer 6

\[\left[\begin{matrix}8 & -1 &-1 \\ -5 & 2 &9\end{matrix}\right]\] this is what i got

Answer 7

Yah, I get that too! So, do you like my image? First person I tried it on. Had to do this by hand a few times, so I made a picture for it all.

Answer 8

Yes I did! thank you!!

Answer 9

Evaluate the determinant of the matrix.

Answer 10

I got -108 @e.mccormick

Answer 11

-54-54 yah. -108.

Answer 12

Can you help me with some more please? @e.mccormick

Answer 13

Sure.

Answer 14

What is the area of the triangle with vertices (–1, –4), (–3, –2), and (–4, –2)

Answer 15

1.5?

Answer 16

You did the three columns with the last as 1, took the determinant, and that is base*height, so took half that?

Answer 17

wait what?

Answer 18

let me put it in a calculator then. LOL

Answer 19

hmmm.. I got a different answer. How did you do this one?

Answer 20

i meant 21.5

Answer 21

Still not what I got. What method are you using to find this? I am working it a different way to confirm as well, but it will take a bit (Heron's Fromula).

Answer 22

i'm not sure how to explain it

Answer 23

Well, what was the setup?

Answer 24

my choices are 1, 1.5, 21.5 and 23

Answer 25

I see my answer in there, but I want to get the how down right. What did you do for a setup?

Answer 26

i was like adding them together idk i confused myself

Answer 27

OK, have you ever heard of testing htree points to see if they are on the same line? The method I used is related to that. So ifyou know one, you pretty much know the other.

Answer 28

For the three points, \((x_1,y_1),~(x_2,y_2),~(x_3,y_3)\) the test for linearity is:

\[\left|\begin{matrix}
x_1&y_1&1\\
x_2&y_2&1\\
x_3&y_3&1
\end{matrix}\right|\]Where the bars, |A|, mean take the determinant. If \(|A|=0\), then the three points are on a line. If \(|A|\ne 0\) then \(|A|=bh\) where bh is the base times the height. The area of a triangle is \(\frac{1}{2}bh\).