Question

Matrices Help please

Answer 1

I was thinking A

Answer 2

last one

Answer 3

multiply matrix by 6

Answer 4

But don't they have to flip?

Answer 5

after that u neeed to do so

Answer 6

\[A = \left[\begin{matrix}a & b \\ c& d\end{matrix}\right]\]\[A^{-1} = \frac{ 1 }{ det(A) }\left[\begin{matrix}d & -b \\ -c & a\end{matrix}\right]\]\[\det(A) = ad - bc\]

Answer 7

i thought u will do the remain

Answer 8

@Euler271 right approch....but in 2 by 2 matrix we can find inverse without this

Answer 9

um. no you can't..

Answer 10

what you're probably doing is THAT but in your head

Answer 11

alright i got for the first step
3 0
1 2

Answer 12

@Euler271 leave it .....solve by ur method

Answer 13

would you flip it diagonally?

Answer 14

teach a man to fish.. @gorv ...

Answer 15

yep

Answer 16

ok in 2 by 2

Answer 17

we interchange the first diagonal element

Answer 18

and multiply second diagonal element by -1 thats it buddy @Euler271

Answer 19

try this on asked Q

Answer 20

after muiltiplicatoin with 6

Answer 21

heheheheheehehehehehhehehehehehehehehhheheeheh

Answer 22

so would it be the 1st one?

Answer 23

\[\left[\begin{matrix}2 & 0 \\ -1 & 3\end{matrix}\right]\]

Answer 24

only reason you multiply by 6 is because 1/detA = 6.

such a kupo nut...

Answer 25

why doesn't the zero flip

Answer 26

now interchange 2 and 3

Answer 27

@DarkBlueChocobo

if you reall wanna learn. look at what i wrote.

Answer 28

you don't always have multiple choice..... @gorv ...

Answer 29

and multiply other diagonal by -1

Answer 30

\[\left[\begin{matrix}3 & 0 \\ 1 & 3\end{matrix}\right]\]

Answer 31

again but why doesn't the zero move?

Answer 32

because it is not part of the matrix's main diagonal which goes from top left to bottom right

Answer 33

oh ok thank you

Answer 34

kweh! :)