Question

matrix inversion problem...

Answer 1

i really have no idea what to do here.

Answer 2

Been a while since I've done matrix multiplication, but I think if
\[Ax = b\]
and you know the inverse matrix of A, A^-1, then you can find x like so: \[A ^{-1} Ax = A ^{-1}b\] And A times it's inverse is just the identity matrix, which leaves\[x = A ^{-1 }b\]
And you're given both A^-1 and b, so you can find x.

Answer 3

yeah you're right about that (i have it in my notes)... that's all i was shown before actually doing this problem.

Answer 4

Do you know how to multiply A^-1 and b? It's just regular matrix multiplication. If you're stuck i can show you how

Answer 5

yeah... im stuck... havent done this before...

Answer 6

im trying to use my graphing calc to punch it in but its not working right

Answer 7

http://www.youtube.com/watch?v=IRdVU8tyRDU

This should show you step by step better than I'll be able to with paper.

Answer 8

o thats all i needa do?

Answer 9

not bad

Answer 10

where did they get the -13 here?

Answer 11

@agent0smith

Answer 12

if you want to find A^-1, then you need to start with attaching the matrix

1 0 0
0 1 0
0 0 1

onto matrix A

then row reducing until the matrix A turns into the 3x3 identity matrix

Answer 13

its already a 3 by 3

Answer 14

@KatClaire

Answer 15

times each side by the inverse of A

Answer 16

so then x=the inverse of A time b

Answer 17

can you show me what you're saying using the numbers?

Answer 18

haha yeah sorry hold on

Answer 19

So you times each side by the inverse of A
\[A^{-1}Ax=A^{-1}b\]
say you times 3 by it's inverse 1/3, that equals1
so timing each side by the inverse of A you are freeing up your x

Answer 20

so...\[x=bA^{1}\]
then just multiply out the two matrices given
\[x=\left[\begin{matrix}6 & -9 & -4 \\ 3 & 4 & -3 \\ -3 & -7 & 4\end{matrix}\right] \left[\begin{matrix}5 \\ 5 \\ 9\end{matrix}\right]\]

Answer 21

and I know you can multiply that out :P

Answer 22

i wasnt asking about that problem lol
i was asking about the 2nd one i attached

Answer 23

where did they get the -13?

Answer 24

gahhhh I never did an inverse of a 3x3 so idk

Answer 25

@agent0smith
Do you know??

Answer 26

have a look at this page

http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi?cmd=inv;m1_1=1;m1_2=-2;m1_3=-1;m2_1=0;m2_2=1;m2_3=7;m3_1=0;m3_2=0;m3_3=1;curn=3;curn=3;submit21=Submit&hideops=0

and it should explain everything